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Riesz representation theorem

From Wikipedia, the free encyclopedia
Theorem about the dual of a Hilbert space
This article is about a theorem concerning the dual of a Hilbert space. For the theorems relating linear functionals to measures, seeRiesz–Markov–Kakutani representation theorem. For other theorems, seeRiesz theorem.

TheRiesz representation theorem, sometimes called theRiesz–Fréchet representation theorem afterFrigyes Riesz andMaurice René Fréchet, establishes an important connection between aHilbert space and itscontinuous dual space. If the underlyingfield is thereal numbers, the two areisometricallyisomorphic; if the underlying field is thecomplex numbers, the two are isometricallyanti-isomorphic. The (anti-)isomorphism is a particularnatural isomorphism.

Preliminaries and notation

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LetH{\displaystyle H} be aHilbert space over a fieldF,{\displaystyle \mathbb {F} ,} whereF{\displaystyle \mathbb {F} } is either the real numbersR{\displaystyle \mathbb {R} } or the complex numbersC.{\displaystyle \mathbb {C} .} IfF=C{\displaystyle \mathbb {F} =\mathbb {C} } (resp. ifF=R{\displaystyle \mathbb {F} =\mathbb {R} }) thenH{\displaystyle H} is called acomplex Hilbert space (resp. areal Hilbert space). Every real Hilbert space can be extended to be adense subset of a unique (up tobijectiveisometry) complex Hilbert space, called itscomplexification, which is why Hilbert spaces are often automatically assumed to be complex. Real and complex Hilbert spaces have in common many, but by no means all, properties and results/theorems.

This article is intended for bothmathematicians andphysicists and will describe the theorem for both. In both mathematics and physics, if a Hilbert space is assumed to be real (that is, ifF=R{\displaystyle \mathbb {F} =\mathbb {R} }) then this will usually be made clear. Often in mathematics, and especially in physics, unless indicated otherwise, "Hilbert space" is usually automatically assumed to mean "complex Hilbert space." Depending on the author, in mathematics, "Hilbert space" usually means either (1) a complex Hilbert space, or (2) a realor complex Hilbert space.

Linear and antilinear maps

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By definition, anantilinear map (also called aconjugate-linear map)f:HY{\displaystyle f:H\to Y} is a map betweenvector spaces that isadditive:f(x+y)=f(x)+f(y) for all x,yH,{\displaystyle f(x+y)=f(x)+f(y)\quad {\text{ for all }}x,y\in H,}andantilinear (also calledconjugate-linear orconjugate-homogeneous):f(cx)=c¯f(x) for all xH and all scalar cF,{\displaystyle f(cx)={\overline {c}}f(x)\quad {\text{ for all }}x\in H{\text{ and all scalar }}c\in \mathbb {F} ,}wherec¯{\displaystyle {\overline {c}}} is the conjugate of the complex numberc=a+bi{\displaystyle c=a+bi}, given byc¯=abi{\displaystyle {\overline {c}}=a-bi}.

In contrast, a mapf:HY{\displaystyle f:H\to Y} islinear if it is additive andhomogeneous:f(cx)=cf(x) for all xH and all scalars cF.{\displaystyle f(cx)=cf(x)\quad {\text{ for all }}x\in H\quad {\text{ and all scalars }}c\in \mathbb {F} .}

Every constant0{\displaystyle 0} map is always both linear and antilinear. IfF=R{\displaystyle \mathbb {F} =\mathbb {R} } then the definitions of linear maps and antilinear maps are completely identical. A linear map from a Hilbert space into aBanach space (or more generally, from any Banach space into anytopological vector space) iscontinuousif and only if it isbounded; the same is true of antilinear maps. Theinverse of any antilinear (resp. linear) bijection is again an antilinear (resp. linear) bijection. The composition of twoantilinear maps is alinear map.

Continuous dual and anti-dual spaces

Afunctional onH{\displaystyle H} is a functionHF{\displaystyle H\to \mathbb {F} } whosecodomain is the underlying scalar fieldF.{\displaystyle \mathbb {F} .} Denote byH{\displaystyle H^{*}} (resp. byH¯){\displaystyle {\overline {H}}^{*})} the set of all continuous linear (resp. continuous antilinear) functionals onH,{\displaystyle H,} which is called the(continuous) dual space (resp. the(continuous) anti-dual space) ofH.{\displaystyle H.}[1] IfF=R{\displaystyle \mathbb {F} =\mathbb {R} } then linear functionals onH{\displaystyle H} are the same as antilinear functionals and consequently, the same is true for such continuous maps: that is,H=H¯.{\displaystyle H^{*}={\overline {H}}^{*}.}

One-to-one correspondence between linear and antilinear functionals

Given any functionalf : HF,{\displaystyle f~:~H\to \mathbb {F} ,} theconjugate off{\displaystyle f} is the functionalf¯:HFhf(h)¯.{\displaystyle {\begin{alignedat}{4}{\overline {f}}:\,&H&&\to \,&&\mathbb {F} \\&h&&\mapsto \,&&{\overline {f(h)}}.\\\end{alignedat}}}

This assignment is most useful whenF=C{\displaystyle \mathbb {F} =\mathbb {C} } because ifF=R{\displaystyle \mathbb {F} =\mathbb {R} } thenf=f¯{\displaystyle f={\overline {f}}} and the assignmentff¯{\displaystyle f\mapsto {\overline {f}}} reduces down to theidentity map.

The assignmentff¯{\displaystyle f\mapsto {\overline {f}}} defines an antilinearbijective correspondence from the set of

all functionals (resp. all linear functionals, all continuous linear functionalsH{\displaystyle H^{*}}) onH,{\displaystyle H,}

onto the set of

all functionals (resp. allantilinear functionals, all continuousantilinear functionalsH¯{\displaystyle {\overline {H}}^{*}}) onH.{\displaystyle H.}

Mathematics vs. physics notations and definitions of inner product

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TheHilbert spaceH{\displaystyle H} has an associatedinner productH×HF{\displaystyle H\times H\to \mathbb {F} } valued inH{\displaystyle H}'s underlying scalar fieldF{\displaystyle \mathbb {F} } that is linear in one coordinate and antilinear in the other (as specified below).IfH{\displaystyle H} is a complex Hilbert space (F=C{\displaystyle \mathbb {F} =\mathbb {C} }), then there is a crucial difference between the notations prevailing in mathematics versus physics, regarding which of the two variables is linear.However, for real Hilbert spaces (F=R{\displaystyle \mathbb {F} =\mathbb {R} }), the inner product is asymmetric map that is linear in each coordinate (bilinear), so there can be no such confusion.

Inmathematics, the inner product on a Hilbert spaceH{\displaystyle H} is often denoted by,{\displaystyle \left\langle \cdot \,,\cdot \right\rangle } or,H{\displaystyle \left\langle \cdot \,,\cdot \right\rangle _{H}} while inphysics, thebra–ket notation{\displaystyle \left\langle \cdot \mid \cdot \right\rangle } orH{\displaystyle \left\langle \cdot \mid \cdot \right\rangle _{H}} is typically used. In this article, these two notations will be related by the equality:

x,y:=yx for all x,yH.{\displaystyle \left\langle x,y\right\rangle :=\left\langle y\mid x\right\rangle \quad {\text{ for all }}x,y\in H.}These have the following properties:

  1. The map,{\displaystyle \left\langle \cdot \,,\cdot \right\rangle } islinear in its first coordinate; equivalently, the map{\displaystyle \left\langle \cdot \mid \cdot \right\rangle } islinear in its second coordinate. That is, for fixedyH,{\displaystyle y\in H,} the mapy=,y:HF{\displaystyle \left\langle \,y\mid \cdot \,\right\rangle =\left\langle \,\cdot \,,y\,\right\rangle :H\to \mathbb {F} } withhyh=h,y{\textstyle h\mapsto \left\langle \,y\mid h\,\right\rangle =\left\langle \,h,y\,\right\rangle }is a linear functional onH.{\displaystyle H.} This linear functional is continuous, soy=,yH.{\displaystyle \left\langle \,y\mid \cdot \,\right\rangle =\left\langle \,\cdot ,y\,\right\rangle \in H^{*}.}
  2. The map,{\displaystyle \left\langle \cdot \,,\cdot \right\rangle } isantilinear in itssecond coordinate; equivalently, the map{\displaystyle \left\langle \cdot \mid \cdot \right\rangle } isantilinear in itsfirst coordinate. That is, for fixedyH,{\displaystyle y\in H,} the mapy=y,:HF{\displaystyle \left\langle \,\cdot \mid y\,\right\rangle =\left\langle \,y,\cdot \,\right\rangle :H\to \mathbb {F} } withhhy=y,h{\textstyle h\mapsto \left\langle \,h\mid y\,\right\rangle =\left\langle \,y,h\,\right\rangle }is an antilinear functional onH.{\displaystyle H.} This antilinear functional is continuous, soy=y,H¯.{\displaystyle \left\langle \,\cdot \mid y\,\right\rangle =\left\langle \,y,\cdot \,\right\rangle \in {\overline {H}}^{*}.}

In computations, one must consistently use either the mathematics notation,{\displaystyle \left\langle \cdot \,,\cdot \right\rangle }, which is (linear, antilinear); or the physics notation{\displaystyle \left\langle \cdot \mid \cdot \right\rangle }, which is (antilinear | linear).

Canonical norm and inner product on the dual space and anti-dual space

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Ifx=y{\displaystyle x=y} thenxx=x,x{\displaystyle \langle \,x\mid x\,\rangle =\langle \,x,x\,\rangle } is a non-negative real number and the mapx:=x,x=xx{\displaystyle \|x\|:={\sqrt {\langle x,x\rangle }}={\sqrt {\langle x\mid x\rangle }}}

defines acanonical norm onH{\displaystyle H} that makesH{\displaystyle H} into anormed space.[1] As with all normed spaces, the (continuous) dual spaceH{\displaystyle H^{*}} carries a canonical norm, called thedual norm, that is defined by[1]fH := supx1,xH|f(x)| for every fH.{\displaystyle \|f\|_{H^{*}}~:=~\sup _{\|x\|\leq 1,x\in H}|f(x)|\quad {\text{ for every }}f\in H^{*}.}

The canonical norm on the (continuous)anti-dual spaceH¯,{\displaystyle {\overline {H}}^{*},} denoted byfH¯,{\displaystyle \|f\|_{{\overline {H}}^{*}},} is defined by using this same equation:[1]fH¯ := supx1,xH|f(x)| for every fH¯.{\displaystyle \|f\|_{{\overline {H}}^{*}}~:=~\sup _{\|x\|\leq 1,x\in H}|f(x)|\quad {\text{ for every }}f\in {\overline {H}}^{*}.}

This canonical norm onH{\displaystyle H^{*}} satisfies theparallelogram law, which means that thepolarization identity can be used to define acanonical inner product onH,{\displaystyle H^{*},} which this article will denote by the notationsf,gH:=gfH,{\displaystyle \left\langle f,g\right\rangle _{H^{*}}:=\left\langle g\mid f\right\rangle _{H^{*}},} where this inner product turnsH{\displaystyle H^{*}} into a Hilbert space. There are now two ways of defining a norm onH:{\displaystyle H^{*}:} the norm induced by this inner product (that is, the norm defined byff,fH{\displaystyle f\mapsto {\sqrt {\left\langle f,f\right\rangle _{H^{*}}}}}) and the usualdual norm (defined as the supremum over the closedunit ball). These norms are the same; explicitly, this means that the following holds for everyfH:{\displaystyle f\in H^{*}:}supx1,xH|f(x)|=fH = f,fH = ffH.{\displaystyle \sup _{\|x\|\leq 1,x\in H}|f(x)|=\|f\|_{H^{*}}~=~{\sqrt {\langle f,f\rangle _{H^{*}}}}~=~{\sqrt {\langle f\mid f\rangle _{H^{*}}}}.}

As will be described later, the Riesz representation theorem can be used to give an equivalent definition of the canonical norm and the canonical inner product onH.{\displaystyle H^{*}.}

The same equations that were used above can also be used to define a norm and inner product onH{\displaystyle H}'santi-dual spaceH¯.{\displaystyle {\overline {H}}^{*}.}[1]

Canonical isometry between the dual and antidual

Thecomplex conjugatef¯{\displaystyle {\overline {f}}} of a functionalf,{\displaystyle f,} which was defined above, satisfiesfH = f¯H¯ and g¯H = gH¯{\displaystyle \|f\|_{H^{*}}~=~\left\|{\overline {f}}\right\|_{{\overline {H}}^{*}}\quad {\text{ and }}\quad \left\|{\overline {g}}\right\|_{H^{*}}~=~\|g\|_{{\overline {H}}^{*}}}for everyfH{\displaystyle f\in H^{*}} and everygH¯.{\displaystyle g\in {\overline {H}}^{*}.} This says exactly that the canonical antilinearbijection defined byCong:HH¯ff¯{\displaystyle {\begin{alignedat}{4}\operatorname {Cong} :\;&&H^{*}&&\;\to \;&{\overline {H}}^{*}\\[0.3ex]&&f&&\;\mapsto \;&{\overline {f}}\\\end{alignedat}}}as well as its inverseCong1 : H¯H{\displaystyle \operatorname {Cong} ^{-1}~:~{\overline {H}}^{*}\to H^{*}} are antilinearisometries and consequently alsohomeomorphisms. The inner products on the dual spaceH{\displaystyle H^{*}} and the anti-dual spaceH¯,{\displaystyle {\overline {H}}^{*},} denoted respectively by,H{\displaystyle \langle \,\cdot \,,\,\cdot \,\rangle _{H^{*}}} and,H¯,{\displaystyle \langle \,\cdot \,,\,\cdot \,\rangle _{{\overline {H}}^{*}},} are related byf¯|g¯H¯=f|gH¯=g|fH for all f,gH{\displaystyle \langle \,{\overline {f}}\,|\,{\overline {g}}\,\rangle _{{\overline {H}}^{*}}={\overline {\langle \,f\,|\,g\,\rangle _{H^{*}}}}=\langle \,g\,|\,f\,\rangle _{H^{*}}\qquad {\text{ for all }}f,g\in H^{*}}andf¯|g¯H=f|gH¯¯=g|fH¯ for all f,gH¯.{\displaystyle \langle \,{\overline {f}}\,|\,{\overline {g}}\,\rangle _{H^{*}}={\overline {\langle \,f\,|\,g\,\rangle _{{\overline {H}}^{*}}}}=\langle \,g\,|\,f\,\rangle _{{\overline {H}}^{*}}\qquad {\text{ for all }}f,g\in {\overline {H}}^{*}.}

IfF=R{\displaystyle \mathbb {F} =\mathbb {R} } thenH=H¯{\displaystyle H^{*}={\overline {H}}^{*}} and this canonical mapCong:HH¯{\displaystyle \operatorname {Cong} :H^{*}\to {\overline {H}}^{*}} reduces down to the identity map.

Riesz representation theorem

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Two vectorsx{\displaystyle x} andy{\displaystyle y} areorthogonal ifx,y=0,{\displaystyle \langle x,y\rangle =0,} which happens if and only ifyy+sx{\displaystyle \|y\|\leq \|y+sx\|} for all scalarss.{\displaystyle s.}[2] Theorthogonal complement of a subsetXH{\displaystyle X\subseteq H} isX:={yH:y,x=0 for all xX},{\displaystyle X^{\bot }:=\{\,y\in H:\langle y,x\rangle =0{\text{ for all }}x\in X\,\},}which is always aclosed vector subspace ofH.{\displaystyle H.} TheHilbert projection theorem guarantees that for anynonempty closedconvex subsetC{\displaystyle C} of aHilbert space there exists a unique vectormC{\displaystyle m\in C} such thatm=infcCc;{\displaystyle \|m\|=\inf _{c\in C}\|c\|;} that is,mC{\displaystyle m\in C} is the (unique)global minimum point of the functionC[0,){\displaystyle C\to [0,\infty )} defined bycc.{\displaystyle c\mapsto \|c\|.}

Statement

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Riesz representation theoremLetH{\displaystyle H} be aHilbert space whoseinner productx,y{\displaystyle \left\langle x,y\right\rangle } is linear in itsfirst argument andantilinear in its second argument and letyx:=x,y{\displaystyle \langle y\mid x\rangle :=\langle x,y\rangle } be the corresponding physics notation. For every continuous linear functionalφH,{\displaystyle \varphi \in H^{*},} there exists a unique vectorfφH,{\displaystyle f_{\varphi }\in H,} called theRiesz representation ofφ,{\displaystyle \varphi ,} such that[3]φ(x)=x,fφ=fφx for all xH.{\displaystyle \varphi (x)=\left\langle x,f_{\varphi }\right\rangle =\left\langle f_{\varphi }\mid x\right\rangle \quad {\text{ for all }}x\in H.}

Importantly forcomplex Hilbert spaces,fφ{\displaystyle f_{\varphi }} is always located in theantilinear coordinate of the inner product.[note 1]

Furthermore, the length of the representation vector is equal to the norm of the functional:fφH=φH,{\displaystyle \left\|f_{\varphi }\right\|_{H}=\|\varphi \|_{H^{*}},} andfφ{\displaystyle f_{\varphi }} is the unique vectorfφ(kerφ){\displaystyle f_{\varphi }\in \left(\ker \varphi \right)^{\bot }} withφ(fφ)=φ2.{\displaystyle \varphi \left(f_{\varphi }\right)=\|\varphi \|^{2}.} It is also the unique element of minimum norm inC:=φ1(φ2){\displaystyle C:=\varphi ^{-1}\left(\|\varphi \|^{2}\right)}; that is to say,fφ{\displaystyle f_{\varphi }} is the unique element ofC{\displaystyle C} satisfyingfφ=infcCc.{\displaystyle \left\|f_{\varphi }\right\|=\inf _{c\in C}\|c\|.}Moreover, any non-zeroq(kerφ){\displaystyle q\in (\ker \varphi )^{\bot }} can be written asq=(q2/φ(q)¯) fφ.{\displaystyle q=\left(\|q\|^{2}/\,{\overline {\varphi (q)}}\right)\ f_{\varphi }.}

CorollaryThecanonical map fromH{\displaystyle H} into its dualH{\displaystyle H^{*}}[1] is theinjectiveantilinear operatorisometry[note 2][1]Φ:HHy,y=y|{\displaystyle {\begin{alignedat}{4}\Phi :\;&&H&&\;\to \;&H^{*}\\[0.3ex]&&y&&\;\mapsto \;&\langle \,\cdot \,,y\rangle =\langle y|\,\cdot \,\rangle \\\end{alignedat}}}The Riesz representation theorem states that this map issurjective (and thusbijective) whenH{\displaystyle H} is complete and that its inverse is thebijectiveisometric antilinear isomorphismΦ1:HHφfφ.{\displaystyle {\begin{alignedat}{4}\Phi ^{-1}:\;&&H^{*}&&\;\to \;&H\\[0.3ex]&&\varphi &&\;\mapsto \;&f_{\varphi }\\\end{alignedat}}.}Consequently,every continuous linear functional on the Hilbert spaceH{\displaystyle H} can be written uniquely in the formy|{\displaystyle \langle y\,|\,\cdot \,\rangle }[1] wherey|H=yH{\displaystyle \|\langle y\,|\cdot \rangle \|_{H^{*}}=\|y\|_{H}} for everyyH.{\displaystyle y\in H.} The assignmentyy,=|y{\displaystyle y\mapsto \langle y,\cdot \rangle =\langle \cdot \,|\,y\rangle } can also be viewed as a bijectivelinear isometryHH¯{\displaystyle H\to {\overline {H}}^{*}} into theanti-dual space ofH,{\displaystyle H,}[1] which is thecomplex conjugate vector space of thecontinuous dual spaceH.{\displaystyle H^{*}.}

The inner products onH{\displaystyle H} andH{\displaystyle H^{*}} are related byΦh,ΦkH=h,k¯H=k,hH for all h,kH{\displaystyle \left\langle \Phi h,\Phi k\right\rangle _{H^{*}}={\overline {\langle h,k\rangle }}_{H}=\langle k,h\rangle _{H}\quad {\text{ for all }}h,k\in H}and similarly,Φ1φ,Φ1ψH=φ,ψ¯H=ψ,φH for all φ,ψH.{\displaystyle \left\langle \Phi ^{-1}\varphi ,\Phi ^{-1}\psi \right\rangle _{H}={\overline {\langle \varphi ,\psi \rangle }}_{H^{*}}=\left\langle \psi ,\varphi \right\rangle _{H^{*}}\quad {\text{ for all }}\varphi ,\psi \in H^{*}.}

The setC:=φ1(φ2){\displaystyle C:=\varphi ^{-1}\left(\|\varphi \|^{2}\right)} satisfiesC=fφ+kerφ{\displaystyle C=f_{\varphi }+\ker \varphi } andCfφ=kerφ{\displaystyle C-f_{\varphi }=\ker \varphi } so whenfφ0{\displaystyle f_{\varphi }\neq 0} thenC{\displaystyle C} can be interpreted as being theaffine hyperplane[note 3] that is parallel to the vector subspacekerφ{\displaystyle \ker \varphi } and containsfφ.{\displaystyle f_{\varphi }.}

ForyH,{\displaystyle y\in H,} the physics notation for the functionalΦ(y)H{\displaystyle \Phi (y)\in H^{*}} is the bray|,{\displaystyle \langle y|,} where explicitly this means thaty|:=Φ(y),{\displaystyle \langle y|:=\Phi (y),} which complements the ket notation|y{\displaystyle |y\rangle } defined by|y:=y.{\displaystyle |y\rangle :=y.} In the mathematical treatment ofquantum mechanics, the theorem can be seen as a justification for the popularbra–ket notation. The theorem says that, every braψ|{\displaystyle \langle \psi \,|} has a corresponding ket|ψ,{\displaystyle |\,\psi \rangle ,} and the latter is unique.

Historically, the theorem is often attributed simultaneously toRiesz andFréchet in 1907 (see references).

Proof[4]

LetF{\displaystyle \mathbb {F} } denote the underlying scalar field ofH.{\displaystyle H.}

Proof of norm formula:

FixyH.{\displaystyle y\in H.} DefineΛ:HF{\displaystyle \Lambda :H\to \mathbb {F} } byΛ(z):=y|z,{\displaystyle \Lambda (z):=\langle \,y\,|\,z\,\rangle ,} which is a linear functional onH{\displaystyle H} sincez{\displaystyle z} is in the linear argument. By theCauchy–Schwarz inequality,|Λ(z)|=|y|z|yz{\displaystyle |\Lambda (z)|=|\langle \,y\,|\,z\,\rangle |\leq \|y\|\|z\|}which shows thatΛ{\displaystyle \Lambda } is bounded (equivalently,continuous) and thatΛy.{\displaystyle \|\Lambda \|\leq \|y\|.} It remains to show thatyΛ.{\displaystyle \|y\|\leq \|\Lambda \|.} By usingy{\displaystyle y} in place ofz,{\displaystyle z,} it follows thaty2=y|y=Λ(y)=|Λ(y)|Λy{\displaystyle \|y\|^{2}=\langle \,y\,|\,y\,\rangle =\Lambda (y)=|\Lambda (y)|\leq \|\Lambda \|\|y\|}(the equalityΛ(y)=|Λ(y)|{\displaystyle \Lambda (y)=|\Lambda (y)|} holds becauseΛ(y)=y20{\displaystyle \Lambda (y)=\|y\|^{2}\geq 0} is real and non-negative). Thus thatΛ=y.{\displaystyle \|\Lambda \|=\|y\|.}{\displaystyle \blacksquare }

The proof above did not use the fact thatH{\displaystyle H} iscomplete, which shows that the formula for the normy|H=yH{\displaystyle \|\langle \,y\,|\,\cdot \,\rangle \|_{H^{*}}=\|y\|_{H}} holds more generally for allinner product spaces.


Proof that a Riesz representation ofφ{\displaystyle \varphi } is unique:

Supposef,gH{\displaystyle f,g\in H} are such thatφ(z)=f|z{\displaystyle \varphi (z)=\langle \,f\,|\,z\,\rangle } andφ(z)=g|z{\displaystyle \varphi (z)=\langle \,g\,|\,z\,\rangle } for allzH.{\displaystyle z\in H.} Thenfg|z=f|zg|z=φ(z)φ(z)=0 for all zH{\displaystyle \langle \,f-g\,|\,z\,\rangle =\langle \,f\,|\,z\,\rangle -\langle \,g\,|\,z\,\rangle =\varphi (z)-\varphi (z)=0\quad {\text{ for all }}z\in H}which shows thatΛ:=fg|{\displaystyle \Lambda :=\langle \,f-g\,|\,\cdot \,\rangle } is the constant0{\displaystyle 0} linear functional. Consequently0=fg|=fg,{\displaystyle 0=\|\langle \,f-g\,|\,\cdot \,\rangle \|=\|f-g\|,} which implies thatfg=0.{\displaystyle f-g=0.}{\displaystyle \blacksquare }


Proof that a vectorfφ{\displaystyle f_{\varphi }} representingφ{\displaystyle \varphi } exists:

LetK:=kerφ:={mH:φ(m)=0}.{\displaystyle K:=\ker \varphi :=\{m\in H:\varphi (m)=0\}.} IfK=H{\displaystyle K=H} (or equivalently, ifφ=0{\displaystyle \varphi =0}) then takingfφ:=0{\displaystyle f_{\varphi }:=0} completes the proof so assume thatKH{\displaystyle K\neq H} andφ0.{\displaystyle \varphi \neq 0.} The continuity ofφ{\displaystyle \varphi } implies thatK{\displaystyle K} is a closed subspace ofH{\displaystyle H} (becauseK=φ1({0}){\displaystyle K=\varphi ^{-1}(\{0\})} and{0}{\displaystyle \{0\}} is a closed subset ofF{\displaystyle \mathbb {F} }). LetK:={vH : v|k=0  for all kK}{\displaystyle K^{\bot }:=\{v\in H~:~\langle \,v\,|\,k\,\rangle =0~{\text{ for all }}k\in K\}} denote theorthogonal complement ofK{\displaystyle K} inH.{\displaystyle H.}BecauseK{\displaystyle K} is closed andH{\displaystyle H} is a Hilbert space,[note 4]H{\displaystyle H} can be written as the direct sumH=KK{\displaystyle H=K\oplus K^{\bot }}[note 5] (a proof of this is given in the article on theHilbert projection theorem). BecauseKH,{\displaystyle K\neq H,} there exists some non-zeropK.{\displaystyle p\in K^{\bot }.} For anyhH,{\displaystyle h\in H,}φ[(φh)p(φp)h] = φ[(φh)p]φ[(φp)h] = (φh)φp(φp)φh=0,{\displaystyle \varphi [(\varphi h)p-(\varphi p)h]~=~\varphi [(\varphi h)p]-\varphi [(\varphi p)h]~=~(\varphi h)\varphi p-(\varphi p)\varphi h=0,}which shows that(φh)p(φp)h  kerφ=K,{\displaystyle (\varphi h)p-(\varphi p)h~\in ~\ker \varphi =K,} where nowpK{\displaystyle p\in K^{\bot }} implies0=p|(φh)p(φp)h = p|(φh)pp|(φp)h = (φh)p|p(φp)p|h.{\displaystyle 0=\langle \,p\,|\,(\varphi h)p-(\varphi p)h\,\rangle ~=~\langle \,p\,|\,(\varphi h)p\,\rangle -\langle \,p\,|\,(\varphi p)h\,\rangle ~=~(\varphi h)\langle \,p\,|\,p\,\rangle -(\varphi p)\langle \,p\,|\,h\,\rangle .} Solving forφh{\displaystyle \varphi h} shows thatφh=(φp)p|hp2=φp¯p2p|h for every hH,{\displaystyle \varphi h={\frac {(\varphi p)\langle \,p\,|\,h\,\rangle }{\|p\|^{2}}}=\left\langle \,{\frac {\overline {\varphi p}}{\|p\|^{2}}}p\,{\Bigg |}\,h\,\right\rangle \quad {\text{ for every }}h\in H,}which proves that the vectorfφ:=φp¯p2p{\displaystyle f_{\varphi }:={\frac {\overline {\varphi p}}{\|p\|^{2}}}p} satisfiesφh=fφ|h for every hH.{\displaystyle \varphi h=\langle \,f_{\varphi }\,|\,h\,\rangle {\text{ for every }}h\in H.}

Applying the norm formula that was proved above withy:=fφ{\displaystyle y:=f_{\varphi }} shows thatφH=fφ|H=fφH.{\displaystyle \|\varphi \|_{H^{*}}=\left\|\left\langle \,f_{\varphi }\,|\,\cdot \,\right\rangle \right\|_{H^{*}}=\left\|f_{\varphi }\right\|_{H}.} Also, the vectoru:=pp{\displaystyle u:={\frac {p}{\|p\|}}} has normu=1{\displaystyle \|u\|=1} and satisfiesfφ:=φ(u)¯u.{\displaystyle f_{\varphi }:={\overline {\varphi (u)}}u.}{\displaystyle \blacksquare }


It can now be deduced thatK{\displaystyle K^{\bot }} is1{\displaystyle 1}-dimensional whenφ0.{\displaystyle \varphi \neq 0.} LetqK{\displaystyle q\in K^{\bot }} be any non-zero vector. Replacingp{\displaystyle p} withq{\displaystyle q} in the proof above shows that the vectorg:=φq¯q2q{\displaystyle g:={\frac {\overline {\varphi q}}{\|q\|^{2}}}q} satisfiesφ(h)=g|h{\displaystyle \varphi (h)=\langle \,g\,|\,h\,\rangle } for everyhH.{\displaystyle h\in H.} The uniqueness of the (non-zero) vectorfφ{\displaystyle f_{\varphi }} representingφ{\displaystyle \varphi } implies thatfφ=g,{\displaystyle f_{\varphi }=g,} which in turn implies thatφq¯0{\displaystyle {\overline {\varphi q}}\neq 0} andq=q2φq¯fφ.{\displaystyle q={\frac {\|q\|^{2}}{\overline {\varphi q}}}f_{\varphi }.} Thus every vector inK{\displaystyle K^{\bot }} is a scalar multiple offφ.{\displaystyle f_{\varphi }.}{\displaystyle \blacksquare }

The formulas for the inner products follow from thepolarization identity.

Observations

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IfφH{\displaystyle \varphi \in H^{*}} thenφ(fφ)=fφ,fφ=fφ2=φ2.{\displaystyle \varphi \left(f_{\varphi }\right)=\left\langle f_{\varphi },f_{\varphi }\right\rangle =\left\|f_{\varphi }\right\|^{2}=\|\varphi \|^{2}.} So in particular,φ(fφ)0{\displaystyle \varphi \left(f_{\varphi }\right)\geq 0} is always real and furthermore,φ(fφ)=0{\displaystyle \varphi \left(f_{\varphi }\right)=0} if and only iffφ=0{\displaystyle f_{\varphi }=0} if and only ifφ=0.{\displaystyle \varphi =0.}

Linear functionals as affine hyperplanes

A non-trivial continuous linear functionalφ{\displaystyle \varphi } is often interpreted geometrically by identifying it with the affine hyperplaneA:=φ1(1){\displaystyle A:=\varphi ^{-1}(1)} (the kernelkerφ=φ1(0){\displaystyle \ker \varphi =\varphi ^{-1}(0)} is also often visualized alongsideA:=φ1(1){\displaystyle A:=\varphi ^{-1}(1)} although knowingA{\displaystyle A} is enough to reconstructkerφ{\displaystyle \ker \varphi } because ifA={\displaystyle A=\varnothing } thenkerφ=H{\displaystyle \ker \varphi =H} and otherwisekerφ=AA{\displaystyle \ker \varphi =A-A}). In particular, the norm ofφ{\displaystyle \varphi } should somehow be interpretable as the "norm of the hyperplaneA{\displaystyle A}". Whenφ0{\displaystyle \varphi \neq 0} then the Riesz representation theorem provides such an interpretation ofφ{\displaystyle \|\varphi \|} in terms of the affine hyperplane[note 3]A:=φ1(1){\displaystyle A:=\varphi ^{-1}(1)} as follows: using the notation from the theorem's statement, fromφ20{\displaystyle \|\varphi \|^{2}\neq 0} it follows thatC:=φ1(φ2)=φ2φ1(1)=φ2A{\displaystyle C:=\varphi ^{-1}\left(\|\varphi \|^{2}\right)=\|\varphi \|^{2}\varphi ^{-1}(1)=\|\varphi \|^{2}A} and soφ=fφ=infcCc{\displaystyle \|\varphi \|=\left\|f_{\varphi }\right\|=\inf _{c\in C}\|c\|} impliesφ=infaAφ2a{\displaystyle \|\varphi \|=\inf _{a\in A}\|\varphi \|^{2}\|a\|} and thusφ=1infaAa.{\displaystyle \|\varphi \|={\frac {1}{\inf _{a\in A}\|a\|}}.} This can also be seen by applying theHilbert projection theorem toA{\displaystyle A} and concluding that the global minimum point of the mapA[0,){\displaystyle A\to [0,\infty )} defined byaa{\displaystyle a\mapsto \|a\|} isfφφ2A.{\displaystyle {\frac {f_{\varphi }}{\|\varphi \|^{2}}}\in A.} The formulas1infaAa=supaA1a{\displaystyle {\frac {1}{\inf _{a\in A}\|a\|}}=\sup _{a\in A}{\frac {1}{\|a\|}}}provide the promised interpretation of the linear functional's normφ{\displaystyle \|\varphi \|} entirely in terms of its associated affine hyperplaneA=φ1(1){\displaystyle A=\varphi ^{-1}(1)} (because with this formula, knowing only thesetA{\displaystyle A} is enough to describe the norm of its associated linearfunctional). Defining1:=0,{\displaystyle {\frac {1}{\infty }}:=0,} theinfimum formulaφ=1infaφ1(1)a{\displaystyle \|\varphi \|={\frac {1}{\inf _{a\in \varphi ^{-1}(1)}\|a\|}}}will also hold whenφ=0.{\displaystyle \varphi =0.} When the supremum is taken inR{\displaystyle \mathbb {R} } (as is typically assumed), then the supremum of the empty set issup={\displaystyle \sup \varnothing =-\infty } but if the supremum is taken in the non-negative reals[0,){\displaystyle [0,\infty )} (which is theimage/range of the norm{\displaystyle \|\,\cdot \,\|} whendimH>0{\displaystyle \dim H>0}) then this supremum is insteadsup=0,{\displaystyle \sup \varnothing =0,} in which case the supremum formulaφ=supaφ1(1)1a{\displaystyle \|\varphi \|=\sup _{a\in \varphi ^{-1}(1)}{\frac {1}{\|a\|}}} will also hold whenφ=0{\displaystyle \varphi =0} (although the atypical equalitysup=0{\displaystyle \sup \varnothing =0} is usually unexpected and so risks causing confusion).

Constructions of the representing vector

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Using the notation from the theorem above, several ways of constructingfφ{\displaystyle f_{\varphi }} fromφH{\displaystyle \varphi \in H^{*}} are now described. Ifφ=0{\displaystyle \varphi =0} thenfφ:=0{\displaystyle f_{\varphi }:=0}; in other words,f0=0.{\displaystyle f_{0}=0.}

This special case ofφ=0{\displaystyle \varphi =0} is henceforth assumed to be known, which is why some of the constructions given below start by assumingφ0.{\displaystyle \varphi \neq 0.}

Orthogonal complement of kernel

Ifφ0{\displaystyle \varphi \neq 0} then for any0u(kerφ),{\displaystyle 0\neq u\in (\ker \varphi )^{\bot },}fφ:=φ(u)¯uu2.{\displaystyle f_{\varphi }:={\frac {{\overline {\varphi (u)}}u}{\|u\|^{2}}}.}

Ifu(kerφ){\displaystyle u\in (\ker \varphi )^{\bot }} is aunit vector (meaningu=1{\displaystyle \|u\|=1}) thenfφ:=φ(u)¯u{\displaystyle f_{\varphi }:={\overline {\varphi (u)}}u}(this is true even ifφ=0{\displaystyle \varphi =0} because in this casefφ=φ(u)¯u=0¯u=0{\displaystyle f_{\varphi }={\overline {\varphi (u)}}u={\overline {0}}u=0}). Ifu{\displaystyle u} is a unit vector satisfying the above condition then the same is true ofu,{\displaystyle -u,} which is also a unit vector in(kerφ).{\displaystyle (\ker \varphi )^{\bot }.} However,φ(u)¯(u)=φ(u)¯u=fφ{\displaystyle {\overline {\varphi (-u)}}(-u)={\overline {\varphi (u)}}u=f_{\varphi }} so both these vectors result in the samefφ.{\displaystyle f_{\varphi }.}

Orthogonal projection onto kernel

IfxH{\displaystyle x\in H} is such thatφ(x)0{\displaystyle \varphi (x)\neq 0} and ifxK{\displaystyle x_{K}} is theorthogonal projection ofx{\displaystyle x} ontokerφ{\displaystyle \ker \varphi } then[proof 1]fφ=φ2φ(x)(xxK).{\displaystyle f_{\varphi }={\frac {\|\varphi \|^{2}}{\varphi (x)}}\left(x-x_{K}\right).}

Orthonormal basis

Given anorthonormal basis{ei}iI{\displaystyle \left\{e_{i}\right\}_{i\in I}} ofH{\displaystyle H} and a continuous linear functionalφH,{\displaystyle \varphi \in H^{*},} the vectorfφH{\displaystyle f_{\varphi }\in H} can be constructed uniquely byfφ=iIφ(ei)¯ei{\displaystyle f_{\varphi }=\sum _{i\in I}{\overline {\varphi \left(e_{i}\right)}}e_{i}} where all but at most countably manyφ(ei){\displaystyle \varphi \left(e_{i}\right)} will be equal to0{\displaystyle 0} and where the value offφ{\displaystyle f_{\varphi }} does not actually depend on choice of orthonormal basis (that is, using any other orthonormal basis forH{\displaystyle H} will result in the same vector). IfyH{\displaystyle y\in H} is written asy=iIaiei{\displaystyle y=\sum _{i\in I}a_{i}e_{i}} thenφ(y)=iIφ(ei)ai=fφ|y{\displaystyle \varphi (y)=\sum _{i\in I}\varphi \left(e_{i}\right)a_{i}=\langle f_{\varphi }|y\rangle }andfφ2=φ(fφ)=iIφ(ei)φ(ei)¯=iI|φ(ei)|2=φ2.{\displaystyle \left\|f_{\varphi }\right\|^{2}=\varphi \left(f_{\varphi }\right)=\sum _{i\in I}\varphi \left(e_{i}\right){\overline {\varphi \left(e_{i}\right)}}=\sum _{i\in I}\left|\varphi \left(e_{i}\right)\right|^{2}=\|\varphi \|^{2}.}

If the orthonormal basis{ei}iI={ei}i=1{\displaystyle \left\{e_{i}\right\}_{i\in I}=\left\{e_{i}\right\}_{i=1}^{\infty }} is a sequence then this becomesfφ=φ(e1)¯e1+φ(e2)¯e2+{\displaystyle f_{\varphi }={\overline {\varphi \left(e_{1}\right)}}e_{1}+{\overline {\varphi \left(e_{2}\right)}}e_{2}+\cdots } and ifyH{\displaystyle y\in H} is written asy=iIaiei=a1e1+a2e2+{\displaystyle y=\sum _{i\in I}a_{i}e_{i}=a_{1}e_{1}+a_{2}e_{2}+\cdots } thenφ(y)=φ(e1)a1+φ(e2)a2+=fφ|y.{\displaystyle \varphi (y)=\varphi \left(e_{1}\right)a_{1}+\varphi \left(e_{2}\right)a_{2}+\cdots =\langle f_{\varphi }|y\rangle .}

Example in finite dimensions using matrix transformations

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Consider the special case ofH=Cn{\displaystyle H=\mathbb {C} ^{n}} (wheren>0{\displaystyle n>0} is aninteger) with the standard inner productzw:=z¯Tw for all w,zH{\displaystyle \langle z\mid w\rangle :={\overline {\,{\vec {z}}\,\,}}^{\operatorname {T} }{\vec {w}}\qquad {\text{ for all }}\;w,z\in H} wherew and z{\displaystyle w{\text{ and }}z} are represented ascolumn matricesw:=[w1wn]{\displaystyle {\vec {w}}:={\begin{bmatrix}w_{1}\\\vdots \\w_{n}\end{bmatrix}}} andz:=[z1zn]{\displaystyle {\vec {z}}:={\begin{bmatrix}z_{1}\\\vdots \\z_{n}\end{bmatrix}}} with respect to the standard orthonormal basise1,,en{\displaystyle e_{1},\ldots ,e_{n}} onH{\displaystyle H} (here,ei{\displaystyle e_{i}} is1{\displaystyle 1} at itsi{\displaystyle i}th coordinate and0{\displaystyle 0} everywhere else; as usual,H{\displaystyle H^{*}} will now be associated with thedual basis) and wherez¯T:=[z1¯,,zn¯]{\displaystyle {\overline {\,{\vec {z}}\,}}^{\operatorname {T} }:=\left[{\overline {z_{1}}},\ldots ,{\overline {z_{n}}}\right]} denotes theconjugate transpose ofz.{\displaystyle {\vec {z}}.} LetφH{\displaystyle \varphi \in H^{*}} be any linear functional and letφ1,,φnC{\displaystyle \varphi _{1},\ldots ,\varphi _{n}\in \mathbb {C} } be the unique scalars such thatφ(w1,,wn)=φ1w1++φnwn for all w:=(w1,,wn)H,{\displaystyle \varphi \left(w_{1},\ldots ,w_{n}\right)=\varphi _{1}w_{1}+\cdots +\varphi _{n}w_{n}\qquad {\text{ for all }}\;w:=\left(w_{1},\ldots ,w_{n}\right)\in H,} where it can be shown thatφi=φ(ei){\displaystyle \varphi _{i}=\varphi \left(e_{i}\right)} for alli=1,,n.{\displaystyle i=1,\ldots ,n.} Then the Riesz representation ofφ{\displaystyle \varphi } is the vectorfφ := φ1¯e1++φn¯en = (φ1¯,,φn¯)H.{\displaystyle f_{\varphi }~:=~{\overline {\varphi _{1}}}e_{1}+\cdots +{\overline {\varphi _{n}}}e_{n}~=~\left({\overline {\varphi _{1}}},\ldots ,{\overline {\varphi _{n}}}\right)\in H.}To see why, identify every vectorw=(w1,,wn){\displaystyle w=\left(w_{1},\ldots ,w_{n}\right)} inH{\displaystyle H} with the column matrixw:=[w1wn]{\displaystyle {\vec {w}}:={\begin{bmatrix}w_{1}\\\vdots \\w_{n}\end{bmatrix}}} so thatfφ{\displaystyle f_{\varphi }} is identified withfφ:=[φ1¯φn¯]=[φ(e1)¯φ(en)¯].{\displaystyle {\vec {f_{\varphi }}}:={\begin{bmatrix}{\overline {\varphi _{1}}}\\\vdots \\{\overline {\varphi _{n}}}\end{bmatrix}}={\begin{bmatrix}{\overline {\varphi \left(e_{1}\right)}}\\\vdots \\{\overline {\varphi \left(e_{n}\right)}}\end{bmatrix}}.} As usual, also identify the linear functionalφ{\displaystyle \varphi } with itstransformation matrix, which is therow matrixφ:=[φ1,,φn]{\displaystyle {\vec {\varphi }}:=\left[\varphi _{1},\ldots ,\varphi _{n}\right]} so thatfφ:=φ¯T{\displaystyle {\vec {f_{\varphi }}}:={\overline {\,{\vec {\varphi }}\,\,}}^{\operatorname {T} }} and the functionφ{\displaystyle \varphi } is the assignmentwφw,{\displaystyle {\vec {w}}\mapsto {\vec {\varphi }}\,{\vec {w}},} where the right hand side ismatrix multiplication. Then for allw=(w1,,wn)H,{\displaystyle w=\left(w_{1},\ldots ,w_{n}\right)\in H,}φ(w)=φ1w1++φnwn=[φ1,,φn][w1wn]=[φ1¯φn¯]¯Tw=fφ¯Tw=fφw,{\displaystyle \varphi (w)=\varphi _{1}w_{1}+\cdots +\varphi _{n}w_{n}=\left[\varphi _{1},\ldots ,\varphi _{n}\right]{\begin{bmatrix}w_{1}\\\vdots \\w_{n}\end{bmatrix}}={\overline {\begin{bmatrix}{\overline {\varphi _{1}}}\\\vdots \\{\overline {\varphi _{n}}}\end{bmatrix}}}^{\operatorname {T} }{\vec {w}}={\overline {\,{\vec {f_{\varphi }}}\,\,}}^{\operatorname {T} }{\vec {w}}=\left\langle \,\,f_{\varphi }\,\mid \,w\,\right\rangle ,}which shows thatfφ{\displaystyle f_{\varphi }} satisfies the defining condition of the Riesz representation ofφ.{\displaystyle \varphi .} The bijective antilinear isometryΦ:HH{\displaystyle \Phi :H\to H^{*}} defined in the corollary to the Riesz representation theorem is the assignment that sendsz=(z1,,zn)H{\displaystyle z=\left(z_{1},\ldots ,z_{n}\right)\in H} to the linear functionalΦ(z)H{\displaystyle \Phi (z)\in H^{*}} onH{\displaystyle H} defined byw=(w1,,wn)  zw=z1¯w1++zn¯wn,{\displaystyle w=\left(w_{1},\ldots ,w_{n}\right)~\mapsto ~\langle \,z\,\mid \,w\,\rangle ={\overline {z_{1}}}w_{1}+\cdots +{\overline {z_{n}}}w_{n},} where under the identification of vectors inH{\displaystyle H} with column matrices and vector inH{\displaystyle H^{*}} with row matrices,Φ{\displaystyle \Phi } is just the assignmentz=[z1zn]  z¯T=[z1¯,,zn¯].{\displaystyle {\vec {z}}={\begin{bmatrix}z_{1}\\\vdots \\z_{n}\end{bmatrix}}~\mapsto ~{\overline {\,{\vec {z}}\,}}^{\operatorname {T} }=\left[{\overline {z_{1}}},\ldots ,{\overline {z_{n}}}\right].} As described in the corollary,Φ{\displaystyle \Phi }'s inverseΦ1:HH{\displaystyle \Phi ^{-1}:H^{*}\to H} is the antilinear isometryφfφ,{\displaystyle \varphi \mapsto f_{\varphi },} which was just shown above to be:φ  fφ := (φ(e1)¯,,φ(en)¯);{\displaystyle \varphi ~\mapsto ~f_{\varphi }~:=~\left({\overline {\varphi \left(e_{1}\right)}},\ldots ,{\overline {\varphi \left(e_{n}\right)}}\right);} where in terms of matrices,Φ1{\displaystyle \Phi ^{-1}} is the assignmentφ=[φ1,,φn]  φ¯T=[φ1¯φn¯].{\displaystyle {\vec {\varphi }}=\left[\varphi _{1},\ldots ,\varphi _{n}\right]~\mapsto ~{\overline {\,{\vec {\varphi }}\,\,}}^{\operatorname {T} }={\begin{bmatrix}{\overline {\varphi _{1}}}\\\vdots \\{\overline {\varphi _{n}}}\end{bmatrix}}.} Thus in terms of matrices, each ofΦ:HH{\displaystyle \Phi :H\to H^{*}} andΦ1:HH{\displaystyle \Phi ^{-1}:H^{*}\to H} is just the operation ofconjugate transpositionvv¯T{\displaystyle {\vec {v}}\mapsto {\overline {\,{\vec {v}}\,}}^{\operatorname {T} }} (although between different spaces of matrices: ifH{\displaystyle H} is identified with the space of all column (respectively, row) matrices thenH{\displaystyle H^{*}} is identified with the space of all row (respectively, column matrices).

This example used the standard inner product, which is the mapzw:=z¯Tw,{\displaystyle \langle z\mid w\rangle :={\overline {\,{\vec {z}}\,\,}}^{\operatorname {T} }{\vec {w}},} but if a different inner product is used, such aszwM:=z¯TMw{\displaystyle \langle z\mid w\rangle _{M}:={\overline {\,{\vec {z}}\,\,}}^{\operatorname {T} }\,M\,{\vec {w}}\,} whereM{\displaystyle M} is anyHermitianpositive-definite matrix, or if a different orthonormal basis is used then the transformation matrices, and thus also the above formulas, will be different.

Relationship with the associated real Hilbert space

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See also:Complexification

Assume thatH{\displaystyle H} is a complex Hilbert space with inner product.{\displaystyle \langle \,\cdot \mid \cdot \,\rangle .} When the Hilbert spaceH{\displaystyle H} is reinterpreted as a real Hilbert space then it will be denoted byHR,{\displaystyle H_{\mathbb {R} },} where the (real) inner-product onHR{\displaystyle H_{\mathbb {R} }} is the real part ofH{\displaystyle H}'s inner product; that is:x,yR:=rex,y.{\displaystyle \langle x,y\rangle _{\mathbb {R} }:=\operatorname {re} \langle x,y\rangle .}

The norm onHR{\displaystyle H_{\mathbb {R} }} induced by,R{\displaystyle \langle \,\cdot \,,\,\cdot \,\rangle _{\mathbb {R} }} is equal to the original norm onH{\displaystyle H} and the continuous dual space ofHR{\displaystyle H_{\mathbb {R} }} is the set of allreal-valued boundedR{\displaystyle \mathbb {R} }-linear functionals onHR{\displaystyle H_{\mathbb {R} }} (see the article about thepolarization identity for additional details about this relationship). LetψR:=reψ{\displaystyle \psi _{\mathbb {R} }:=\operatorname {re} \psi } andψi:=imψ{\displaystyle \psi _{i}:=\operatorname {im} \psi } denote the real and imaginary parts of a linear functionalψ,{\displaystyle \psi ,} so thatψ=reψ+iimψ=ψR+iψi.{\displaystyle \psi =\operatorname {re} \psi +i\operatorname {im} \psi =\psi _{\mathbb {R} }+i\psi _{i}.} The formulaexpressing a linear functional in terms of its real part isψ(h)=ψR(h)iψR(ih) for hH,{\displaystyle \psi (h)=\psi _{\mathbb {R} }(h)-i\psi _{\mathbb {R} }(ih)\quad {\text{ for }}h\in H,}whereψi(h)=iψR(ih){\displaystyle \psi _{i}(h)=-i\psi _{\mathbb {R} }(ih)} for allhH.{\displaystyle h\in H.} It follows thatkerψR=ψ1(iR),{\displaystyle \ker \psi _{\mathbb {R} }=\psi ^{-1}(i\mathbb {R} ),} and thatψ=0{\displaystyle \psi =0} if and only ifψR=0.{\displaystyle \psi _{\mathbb {R} }=0.} It can also be shown thatψ=ψR=ψi{\displaystyle \|\psi \|=\left\|\psi _{\mathbb {R} }\right\|=\left\|\psi _{i}\right\|} whereψR:=suph1|ψR(h)|{\displaystyle \left\|\psi _{\mathbb {R} }\right\|:=\sup _{\|h\|\leq 1}\left|\psi _{\mathbb {R} }(h)\right|} andψi:=suph1|ψi(h)|{\displaystyle \left\|\psi _{i}\right\|:=\sup _{\|h\|\leq 1}\left|\psi _{i}(h)\right|} are the usualoperator norms. In particular, a linear functionalψ{\displaystyle \psi } is bounded if and only if its real partψR{\displaystyle \psi _{\mathbb {R} }} is bounded.

Representing a functional and its real part

The Riesz representation of a continuous linear functionφ{\displaystyle \varphi } on a complex Hilbert space is equal to the Riesz representation of its real partreφ{\displaystyle \operatorname {re} \varphi } on its associated real Hilbert space.

Explicitly, letφH{\displaystyle \varphi \in H^{*}} and as above, letfφH{\displaystyle f_{\varphi }\in H} be the Riesz representation ofφ{\displaystyle \varphi } obtained in(H,,,),{\displaystyle (H,\langle ,\cdot ,\cdot \rangle ),} so it is the unique vector that satisfiesφ(x)=fφx{\displaystyle \varphi (x)=\left\langle f_{\varphi }\mid x\right\rangle } for allxH.{\displaystyle x\in H.} The real part ofφ{\displaystyle \varphi } is a continuous real linear functional onHR{\displaystyle H_{\mathbb {R} }} and so the Riesz representation theorem may be applied toφR:=reφ{\displaystyle \varphi _{\mathbb {R} }:=\operatorname {re} \varphi } and the associated real Hilbert space(HR,,,R){\displaystyle \left(H_{\mathbb {R} },\langle ,\cdot ,\cdot \rangle _{\mathbb {R} }\right)} to produce its Riesz representation, which will be denoted byfφR.{\displaystyle f_{\varphi _{\mathbb {R} }}.} That is,fφR{\displaystyle f_{\varphi _{\mathbb {R} }}} is the unique vector inHR{\displaystyle H_{\mathbb {R} }} that satisfiesφR(x)=fφRxR{\displaystyle \varphi _{\mathbb {R} }(x)=\left\langle f_{\varphi _{\mathbb {R} }}\mid x\right\rangle _{\mathbb {R} }} for allxH.{\displaystyle x\in H.} The conclusion isfφR=fφ.{\displaystyle f_{\varphi _{\mathbb {R} }}=f_{\varphi }.} This follows from the main theorem becausekerφR=φ1(iR){\displaystyle \ker \varphi _{\mathbb {R} }=\varphi ^{-1}(i\mathbb {R} )} and ifxH{\displaystyle x\in H} thenfφxR=refφx=reφ(x)=φR(x){\displaystyle \left\langle f_{\varphi }\mid x\right\rangle _{\mathbb {R} }=\operatorname {re} \left\langle f_{\varphi }\mid x\right\rangle =\operatorname {re} \varphi (x)=\varphi _{\mathbb {R} }(x)} and consequently, ifmkerφR{\displaystyle m\in \ker \varphi _{\mathbb {R} }} thenfφmR=0,{\displaystyle \left\langle f_{\varphi }\mid m\right\rangle _{\mathbb {R} }=0,} which shows thatfφ(kerφR)R.{\displaystyle f_{\varphi }\in (\ker \varphi _{\mathbb {R} })^{\perp _{\mathbb {R} }}.} Moreover,φ(fφ)=φ2{\displaystyle \varphi (f_{\varphi })=\|\varphi \|^{2}} being a real number implies thatφR(fφ)=reφ(fφ)=φ2.{\displaystyle \varphi _{\mathbb {R} }(f_{\varphi })=\operatorname {re} \varphi (f_{\varphi })=\|\varphi \|^{2}.}In other words, in the theorem and constructions above, ifH{\displaystyle H} is replaced with its real Hilbert space counterpartHR{\displaystyle H_{\mathbb {R} }} and ifφ{\displaystyle \varphi } is replaced withreφ{\displaystyle \operatorname {re} \varphi } thenfφ=freφ.{\displaystyle f_{\varphi }=f_{\operatorname {re} \varphi }.} This means that vectorfφ{\displaystyle f_{\varphi }} obtained by using(HR,,,R){\displaystyle \left(H_{\mathbb {R} },\langle ,\cdot ,\cdot \rangle _{\mathbb {R} }\right)} and the real linear functionalreφ{\displaystyle \operatorname {re} \varphi } is the equal to the vector obtained by using the origin complex Hilbert space(H,,,){\displaystyle \left(H,\left\langle ,\cdot ,\cdot \right\rangle \right)} and original complex linear functionalφ{\displaystyle \varphi } (with identical norm values as well).

Furthermore, ifφ0{\displaystyle \varphi \neq 0} thenfφ{\displaystyle f_{\varphi }} is perpendicular tokerφR{\displaystyle \ker \varphi _{\mathbb {R} }} with respect to,R{\displaystyle \langle \cdot ,\cdot \rangle _{\mathbb {R} }} where the kernel ofφ{\displaystyle \varphi } is be aproper subspace of the kernel of its real partφR.{\displaystyle \varphi _{\mathbb {R} }.} Assume now thatφ0.{\displaystyle \varphi \neq 0.} ThenfφkerφR{\displaystyle f_{\varphi }\not \in \ker \varphi _{\mathbb {R} }} becauseφR(fφ)=φ(fφ)=φ20{\displaystyle \varphi _{\mathbb {R} }\left(f_{\varphi }\right)=\varphi \left(f_{\varphi }\right)=\|\varphi \|^{2}\neq 0} andkerφ{\displaystyle \ker \varphi } is a proper subset ofkerφR.{\displaystyle \ker \varphi _{\mathbb {R} }.} The vector subspacekerφ{\displaystyle \ker \varphi } has real codimension1{\displaystyle 1} inkerφR,{\displaystyle \ker \varphi _{\mathbb {R} },} whilekerφR{\displaystyle \ker \varphi _{\mathbb {R} }} hasreal codimension1{\displaystyle 1} inHR,{\displaystyle H_{\mathbb {R} },} andfφ,kerφRR=0.{\displaystyle \left\langle f_{\varphi },\ker \varphi _{\mathbb {R} }\right\rangle _{\mathbb {R} }=0.} That is,fφ{\displaystyle f_{\varphi }} is perpendicular tokerφR{\displaystyle \ker \varphi _{\mathbb {R} }} with respect to,R.{\displaystyle \langle \cdot ,\cdot \rangle _{\mathbb {R} }.}

Canonical injections into the dual and anti-dual

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Induced linear map into anti-dual

The map defined by placingy{\displaystyle y} into thelinear coordinate of the inner product and letting the variablehH{\displaystyle h\in H} vary over theantilinear coordinate results in anantilinear functional:y=y,:HF defined by hhy=y,h.{\displaystyle \langle \,\cdot \mid y\,\rangle =\langle \,y,\cdot \,\rangle :H\to \mathbb {F} \quad {\text{ defined by }}\quad h\mapsto \langle \,h\mid y\,\rangle =\langle \,y,h\,\rangle .}

This map is an element ofH¯,{\displaystyle {\overline {H}}^{*},} which is the continuousanti-dual space ofH.{\displaystyle H.} Thecanonical map fromH{\displaystyle H} into its anti-dualH¯{\displaystyle {\overline {H}}^{*}}[1] is thelinear operatorInHH¯:HH¯yy=y,{\displaystyle {\begin{alignedat}{4}\operatorname {In} _{H}^{{\overline {H}}^{*}}:\;&&H&&\;\to \;&{\overline {H}}^{*}\\[0.3ex]&&y&&\;\mapsto \;&\langle \,\cdot \mid y\,\rangle =\langle \,y,\cdot \,\rangle \\[0.3ex]\end{alignedat}}}which is also aninjectiveisometry.[1] TheFundamental theorem of Hilbert spaces, which is related to Riesz representation theorem, states that this map is surjective (and thusbijective). Consequently, every antilinear functional onH{\displaystyle H} can be written (uniquely) in this form.[1]

IfCong:HH¯{\displaystyle \operatorname {Cong} :H^{*}\to {\overline {H}}^{*}} is the canonicalantilinearbijectiveisometryff¯{\displaystyle f\mapsto {\overline {f}}} that was defined above, then the following equality holds:Cong  InHH = InHH¯.{\displaystyle \operatorname {Cong} ~\circ ~\operatorname {In} _{H}^{H^{*}}~=~\operatorname {In} _{H}^{{\overline {H}}^{*}}.}

Extending the bra–ket notation to bras and kets

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Main article:Bra–ket notation

Let(H,,H){\displaystyle \left(H,\langle \cdot ,\cdot \rangle _{H}\right)} be a Hilbert space and as before, lety|xH:=x,yH.{\displaystyle \langle y\,|\,x\rangle _{H}:=\langle x,y\rangle _{H}.} LetΦ:HHggH=,gH{\displaystyle {\begin{alignedat}{4}\Phi :\;&&H&&\;\to \;&H^{*}\\[0.3ex]&&g&&\;\mapsto \;&\left\langle \,g\mid \cdot \,\right\rangle _{H}=\left\langle \,\cdot ,g\,\right\rangle _{H}\\\end{alignedat}}}which is a bijective antilinear isometry that satisfies(Φh)g=hgH=g,hH for all g,hH.{\displaystyle (\Phi h)g=\langle h\mid g\rangle _{H}=\langle g,h\rangle _{H}\quad {\text{ for all }}g,h\in H.}

Bras

Given a vectorhH,{\displaystyle h\in H,} leth|{\displaystyle \langle h\,|} denote the continuous linear functionalΦh{\displaystyle \Phi h}; that is,h| := Φh{\displaystyle \langle h\,|~:=~\Phi h} so that this functionalh|{\displaystyle \langle h\,|} is defined byghgH.{\displaystyle g\mapsto \left\langle \,h\mid g\,\right\rangle _{H}.} This map was denoted byh{\displaystyle \left\langle h\mid \cdot \,\right\rangle } earlier in this article.

The assignmenthh|{\displaystyle h\mapsto \langle h|} is just the isometric antilinear isomorphismΦ : HH,{\displaystyle \Phi ~:~H\to H^{*},} which is why cg+h| = c¯g + h| {\displaystyle ~\langle cg+h\,|~=~{\overline {c}}\langle g\mid ~+~\langle h\,|~} holds for allg,hH{\displaystyle g,h\in H} and all scalarsc.{\displaystyle c.} The result of plugging some givengH{\displaystyle g\in H} into the functionalh|{\displaystyle \langle h\,|} is the scalarh|gH=g,hH,{\displaystyle \langle h\,|\,g\rangle _{H}=\langle g,h\rangle _{H},} which may be denoted byhg.{\displaystyle \langle h\mid g\rangle .}[note 6]

Bra of a linear functional

Given a continuous linear functionalψH,{\displaystyle \psi \in H^{*},} letψ{\displaystyle \langle \psi \mid } denote the vectorΦ1ψH{\displaystyle \Phi ^{-1}\psi \in H}; that is,ψ := Φ1ψ.{\displaystyle \langle \psi \mid ~:=~\Phi ^{-1}\psi .}

The assignmentψψ{\displaystyle \psi \mapsto \langle \psi \mid } is just the isometric antilinear isomorphismΦ1 : HH,{\displaystyle \Phi ^{-1}~:~H^{*}\to H,} which is why cψ+ϕ = c¯ψ + ϕ {\displaystyle ~\langle c\psi +\phi \mid ~=~{\overline {c}}\langle \psi \mid ~+~\langle \phi \mid ~} holds for allϕ,ψH{\displaystyle \phi ,\psi \in H^{*}} and all scalarsc.{\displaystyle c.}

The defining condition of the vectorψ|H{\displaystyle \langle \psi |\in H} is the technically correct but unsightly equalityψgH = ψg for all gH,{\displaystyle \left\langle \,\langle \psi \mid \,\mid g\right\rangle _{H}~=~\psi g\quad {\text{ for all }}g\in H,}which is why the notationψg{\displaystyle \left\langle \psi \mid g\right\rangle } is used in place ofψgH=g,ψH.{\displaystyle \left\langle \,\langle \psi \mid \,\mid g\right\rangle _{H}=\left\langle g,\,\langle \psi \mid \right\rangle _{H}.} With this notation, the defining condition becomesψg = ψg for all gH.{\displaystyle \left\langle \psi \mid g\right\rangle ~=~\psi g\quad {\text{ for all }}g\in H.}

Kets

For any given vectorgH,{\displaystyle g\in H,} the notation|g{\displaystyle |\,g\rangle } is used to denoteg{\displaystyle g}; that is,g:=g.{\displaystyle \mid g\rangle :=g.}

The assignmentg|g{\displaystyle g\mapsto |\,g\rangle } is just the identity mapIdH:HH,{\displaystyle \operatorname {Id} _{H}:H\to H,} which is why cg+h = cg + h {\displaystyle ~\mid cg+h\rangle ~=~c\mid g\rangle ~+~\mid h\rangle ~} holds for allg,hH{\displaystyle g,h\in H} and all scalarsc.{\displaystyle c.}

The notationhg{\displaystyle \langle h\mid g\rangle } andψg{\displaystyle \langle \psi \mid g\rangle } is used in place ofhgH = g,hH{\displaystyle \left\langle h\mid \,\mid g\rangle \,\right\rangle _{H}~=~\left\langle \mid g\rangle ,h\right\rangle _{H}} andψgH = g,ψH,{\displaystyle \left\langle \psi \mid \,\mid g\rangle \,\right\rangle _{H}~=~\left\langle g,\,\langle \psi \mid \right\rangle _{H},} respectively. As expected, ψg=ψg {\displaystyle ~\langle \psi \mid g\rangle =\psi g~} and hg {\displaystyle ~\langle h\mid g\rangle ~} really is just the scalar hgH = g,hH.{\displaystyle ~\langle h\mid g\rangle _{H}~=~\langle g,h\rangle _{H}.}

Adjoints and transposes

[edit]

LetA:HZ{\displaystyle A:H\to Z} be acontinuous linear operator betweenHilbert spaces(H,,H){\displaystyle \left(H,\langle \cdot ,\cdot \rangle _{H}\right)} and(Z,,Z).{\displaystyle \left(Z,\langle \cdot ,\cdot \rangle _{Z}\right).} As before, letyxH:=x,yH{\displaystyle \langle y\mid x\rangle _{H}:=\langle x,y\rangle _{H}} andyxZ:=x,yZ.{\displaystyle \langle y\mid x\rangle _{Z}:=\langle x,y\rangle _{Z}.}

Denote byΦH:HHggH and ΦZ:ZZyyZ{\displaystyle {\begin{alignedat}{4}\Phi _{H}:\;&&H&&\;\to \;&H^{*}\\[0.3ex]&&g&&\;\mapsto \;&\langle \,g\mid \cdot \,\rangle _{H}\\\end{alignedat}}\quad {\text{ and }}\quad {\begin{alignedat}{4}\Phi _{Z}:\;&&Z&&\;\to \;&Z^{*}\\[0.3ex]&&y&&\;\mapsto \;&\langle \,y\mid \cdot \,\rangle _{Z}\\\end{alignedat}}}the usual bijective antilinear isometries that satisfy:(ΦHg)h=ghH for all g,hH and (ΦZy)z=yzZ for all y,zZ.{\displaystyle \left(\Phi _{H}g\right)h=\langle g\mid h\rangle _{H}\quad {\text{ for all }}g,h\in H\qquad {\text{ and }}\qquad \left(\Phi _{Z}y\right)z=\langle y\mid z\rangle _{Z}\quad {\text{ for all }}y,z\in Z.}

Definition of the adjoint

[edit]
Main articles:Hermitian adjoint andConjugate transpose

For everyzZ,{\displaystyle z\in Z,} the scalar-valued mapzA()Z{\displaystyle \langle z\mid A(\cdot )\rangle _{Z}}[note 7] onH{\displaystyle H} defined byhzAhZ=Ah,zZ{\displaystyle h\mapsto \langle z\mid Ah\rangle _{Z}=\langle Ah,z\rangle _{Z}}

is a continuous linear functional onH{\displaystyle H} and so by the Riesz representation theorem, there exists a unique vector inH,{\displaystyle H,} denoted byAz,{\displaystyle A^{*}z,} such thatzA()Z=AzH,{\displaystyle \langle z\mid A(\cdot )\rangle _{Z}=\left\langle A^{*}z\mid \cdot \,\right\rangle _{H},} or equivalently, such thatzAhZ=AzhH for all hH.{\displaystyle \langle z\mid Ah\rangle _{Z}=\left\langle A^{*}z\mid h\right\rangle _{H}\quad {\text{ for all }}h\in H.}

The assignmentzAz{\displaystyle z\mapsto A^{*}z} thus induces a functionA:ZH{\displaystyle A^{*}:Z\to H} called theadjoint ofA:HZ{\displaystyle A:H\to Z} whose defining condition iszAhZ=AzhH for all hH and all zZ.{\displaystyle \langle z\mid Ah\rangle _{Z}=\left\langle A^{*}z\mid h\right\rangle _{H}\quad {\text{ for all }}h\in H{\text{ and all }}z\in Z.} The adjointA:ZH{\displaystyle A^{*}:Z\to H} is necessarily acontinuous (equivalently, abounded)linear operator.

IfH{\displaystyle H} is finite dimensional with the standard inner product and ifM{\displaystyle M} is thetransformation matrix ofA{\displaystyle A} with respect to the standard orthonormal basis thenM{\displaystyle M}'sconjugate transposeMT¯{\displaystyle {\overline {M^{\operatorname {T} }}}} is the transformation matrix of the adjointA.{\displaystyle A^{*}.}

Adjoints are transposes

[edit]
Main article:Transpose of a linear map
See also:Transpose

It is also possible to define thetranspose oralgebraic adjoint ofA:HZ,{\displaystyle A:H\to Z,} which is the maptA:ZH{\displaystyle {}^{t}A:Z^{*}\to H^{*}} defined by sending a continuous linear functionalsψZ{\displaystyle \psi \in Z^{*}} totA(ψ):=ψA,{\displaystyle {}^{t}A(\psi ):=\psi \circ A,} where thecompositionψA{\displaystyle \psi \circ A} is always a continuous linear functional onH{\displaystyle H} and it satisfiesA=tA{\displaystyle \|A\|=\left\|{}^{t}A\right\|} (this is true more generally, whenH{\displaystyle H} andZ{\displaystyle Z} are merelynormed spaces).[5] So for example, ifzZ{\displaystyle z\in Z} thentA{\displaystyle {}^{t}A} sends the continuous linear functionalzZZ{\displaystyle \langle z\mid \cdot \rangle _{Z}\in Z^{*}} (defined onZ{\displaystyle Z} bygzgZ{\displaystyle g\mapsto \langle z\mid g\rangle _{Z}}) to the continuous linear functionalzA()ZH{\displaystyle \langle z\mid A(\cdot )\rangle _{Z}\in H^{*}} (defined onH{\displaystyle H} byhzA(h)Z{\displaystyle h\mapsto \langle z\mid A(h)\rangle _{Z}});[note 7] using bra-ket notation, this can be written astAz = zA{\displaystyle {}^{t}A\langle z\mid ~=~\langle z\mid A} where the juxtaposition ofz{\displaystyle \langle z\mid } withA{\displaystyle A} on the right hand side denotes function composition:HAZzC.{\displaystyle H\xrightarrow {A} Z\xrightarrow {\langle z\mid } \mathbb {C} .}

The adjointA:ZH{\displaystyle A^{*}:Z\to H} is actually just to the transposetA:ZH{\displaystyle {}^{t}A:Z^{*}\to H^{*}}[2] when the Riesz representation theorem is used to identifyZ{\displaystyle Z} withZ{\displaystyle Z^{*}} andH{\displaystyle H} withH.{\displaystyle H^{*}.}

Explicitly, the relationship between the adjoint and transpose is:

tA  ΦZ = ΦH  A{\displaystyle {}^{t}A~\circ ~\Phi _{Z}~=~\Phi _{H}~\circ ~A^{*}}
Adjoint-transpose

which can be rewritten as:A = ΦH1  tA  ΦZ and tA = ΦH  A  ΦZ1.{\displaystyle A^{*}~=~\Phi _{H}^{-1}~\circ ~{}^{t}A~\circ ~\Phi _{Z}\quad {\text{ and }}\quad {}^{t}A~=~\Phi _{H}~\circ ~A^{*}~\circ ~\Phi _{Z}^{-1}.}

Proof

To show thattA  ΦZ = ΦH  A,{\displaystyle {}^{t}A~\circ ~\Phi _{Z}~=~\Phi _{H}~\circ ~A^{*},} fixzZ.{\displaystyle z\in Z.} The definition oftA{\displaystyle {}^{t}A} implies(tAΦZ)z=tA(ΦZz)=(ΦZz)A{\displaystyle \left({}^{t}A\circ \Phi _{Z}\right)z={}^{t}A\left(\Phi _{Z}z\right)=\left(\Phi _{Z}z\right)\circ A} so it remains to show that(ΦZz)A=ΦH(Az).{\displaystyle \left(\Phi _{Z}z\right)\circ A=\Phi _{H}\left(A^{*}z\right).} IfhH{\displaystyle h\in H} then((ΦZz)A)h=(ΦZz)(Ah)=zAhZ=AzhH=(ΦH(Az))h,{\displaystyle \left(\left(\Phi _{Z}z\right)\circ A\right)h=\left(\Phi _{Z}z\right)(Ah)=\langle z\mid Ah\rangle _{Z}=\langle A^{*}z\mid h\rangle _{H}=\left(\Phi _{H}(A^{*}z)\right)h,} as desired.{\displaystyle \blacksquare }

Alternatively, the value of the left and right hand sides of (Adjoint-transpose) at any givenzZ{\displaystyle z\in Z} can be rewritten in terms of the inner products as:(tA  ΦZ)z=zA()Z and (ΦH  A)z=AzH{\displaystyle \left({}^{t}A~\circ ~\Phi _{Z}\right)z=\langle z\mid A(\cdot )\rangle _{Z}\quad {\text{ and }}\quad \left(\Phi _{H}~\circ ~A^{*}\right)z=\langle A^{*}z\mid \cdot \,\rangle _{H}}so thattA  ΦZ = ΦH  A{\displaystyle {}^{t}A~\circ ~\Phi _{Z}~=~\Phi _{H}~\circ ~A^{*}} holds if and only ifzA()Z=AzH{\displaystyle \langle z\mid A(\cdot )\rangle _{Z}=\langle A^{*}z\mid \cdot \,\rangle _{H}} holds; but the equality on the right holds by definition ofAz.{\displaystyle A^{*}z.} The defining condition ofAz{\displaystyle A^{*}z} can also be writtenzA = Az{\displaystyle \langle z\mid A~=~\langle A^{*}z\mid } if bra-ket notation is used.

Descriptions of self-adjoint, normal, and unitary operators

[edit]

AssumeZ=H{\displaystyle Z=H} and letΦ:=ΦH=ΦZ.{\displaystyle \Phi :=\Phi _{H}=\Phi _{Z}.} LetA:HH{\displaystyle A:H\to H} be a continuous (that is, bounded) linear operator.

Whether or notA:HH{\displaystyle A:H\to H} isself-adjoint,normal, orunitary depends entirely on whether or notA{\displaystyle A} satisfies certain defining conditions related to its adjoint, which was shown by (Adjoint-transpose) to essentially be just the transposetA:HH.{\displaystyle {}^{t}A:H^{*}\to H^{*}.} Because the transpose ofA{\displaystyle A} is a map between continuous linear functionals, these defining conditions can consequently be re-expressed entirely in terms of linear functionals, as the remainder of subsection will now describe in detail. The linear functionals that are involved are the simplest possible continuous linear functionals onH{\displaystyle H} that can be defined entirely in terms ofA,{\displaystyle A,} the inner product{\displaystyle \langle \,\cdot \mid \cdot \,\rangle } onH,{\displaystyle H,} and some given vectorhH.{\displaystyle h\in H.} Specifically, these areAh{\displaystyle \left\langle Ah\mid \cdot \,\right\rangle } andhA(){\displaystyle \langle h\mid A(\cdot )\rangle }[note 7] whereAh=Φ(Ah)=(ΦA)h and hA()=(tAΦ)h.{\displaystyle \left\langle Ah\mid \cdot \,\right\rangle =\Phi (Ah)=(\Phi \circ A)h\quad {\text{ and }}\quad \langle h\mid A(\cdot )\rangle =\left({}^{t}A\circ \Phi \right)h.}

Self-adjoint operators

See also:Self-adjoint operator,Hermitian matrix, andSymmetric matrix

A continuous linear operatorA:HH{\displaystyle A:H\to H} is calledself-adjoint if it is equal to its own adjoint; that is, ifA=A.{\displaystyle A=A^{*}.} Using (Adjoint-transpose), this happens if and only if:ΦA=tAΦ{\displaystyle \Phi \circ A={}^{t}A\circ \Phi }where this equality can be rewritten in the following two equivalent forms:A=Φ1tAΦ or tA=ΦAΦ1.{\displaystyle A=\Phi ^{-1}\circ {}^{t}A\circ \Phi \quad {\text{ or }}\quad {}^{t}A=\Phi \circ A\circ \Phi ^{-1}.}

Unraveling notation and definitions produces the following characterization of self-adjoint operators in terms of the aforementioned continuous linear functionals:A{\displaystyle A} is self-adjoint if and only if for allzH,{\displaystyle z\in H,} the linear functionalzA(){\displaystyle \langle z\mid A(\cdot )\rangle }[note 7] is equal to the linear functionalAz{\displaystyle \langle Az\mid \cdot \,\rangle }; that is, if and only if

zA()=Az for all zH{\displaystyle \langle z\mid A(\cdot )\rangle =\langle Az\mid \cdot \,\rangle \quad {\text{ for all }}z\in H}
Self-adjointness functionals

where if bra-ket notation is used, this iszA = Az for all zH.{\displaystyle \langle z\mid A~=~\langle Az\mid \quad {\text{ for all }}z\in H.}

Normal operators

See also:Normal operator andNormal matrix

A continuous linear operatorA:HH{\displaystyle A:H\to H} is callednormal ifAA=AA,{\displaystyle AA^{*}=A^{*}A,} which happens if and only if for allz,hH,{\displaystyle z,h\in H,}AAzh=AAzh.{\displaystyle \left\langle AA^{*}z\mid h\right\rangle =\left\langle A^{*}Az\mid h\right\rangle .}

Using (Adjoint-transpose) and unraveling notation and definitions produces[proof 2] the following characterization of normal operators in terms of inner products of continuous linear functionals:A{\displaystyle A} is a normal operator if and only if

AhAzH = h|A()zA()H for all z,hH{\displaystyle \left\langle \,\langle Ah\mid \cdot \,\rangle \mid \langle Az\mid \cdot \,\rangle \,\right\rangle _{H^{*}}~=~\left\langle \,\langle h|A(\cdot )\rangle \mid \langle z\mid A(\cdot )\rangle \,\right\rangle _{H^{*}}\quad {\text{ for all }}z,h\in H}
Normality functionals

where the left hand side is also equal toAhAz¯H=AzAhH.{\displaystyle {\overline {\langle Ah\mid Az\rangle }}_{H}=\langle Az\mid Ah\rangle _{H}.} The left hand side of this characterization involvesonly linear functionals of the formAh{\displaystyle \langle Ah\mid \cdot \,\rangle } while the right hand side involvesonly linear functions of the formhA(){\displaystyle \langle h\mid A(\cdot )\rangle } (defined as above[note 7]). So in plain English, characterization (Normality functionals) says that an operator isnormal when the inner product of any two linear functions of the first form is equal to the inner product of their second form (using the same vectorsz,hH{\displaystyle z,h\in H} for both forms).In other words, if it happens to be the case (and whenA{\displaystyle A} is injective or self-adjoint, it is) that the assignment of linear functionalsAh  h|A(){\displaystyle \langle Ah\mid \cdot \,\rangle ~\mapsto ~\langle h|A(\cdot )\rangle } is well-defined (or alternatively, ifh|A()  Ah{\displaystyle \langle h|A(\cdot )\rangle ~\mapsto ~\langle Ah\mid \cdot \,\rangle } is well-defined) whereh{\displaystyle h} ranges overH,{\displaystyle H,} thenA{\displaystyle A} is a normal operator if and only if this assignment preserves the inner product onH.{\displaystyle H^{*}.}

The fact that every self-adjoint bounded linear operator is normal follows readily by direct substitution ofA=A{\displaystyle A^{*}=A} into either side ofAA=AA.{\displaystyle A^{*}A=AA^{*}.} This same fact also follows immediately from the direct substitution of the equalities (Self-adjointness functionals) into either side of (Normality functionals).

Alternatively, for a complex Hilbert space, the continuous linear operatorA{\displaystyle A} is a normal operator if and only ifAz=Az{\displaystyle \|Az\|=\left\|A^{*}z\right\|} for everyzH,{\displaystyle z\in H,}[2] which happens if and only ifAzH=z|A()H for every zH.{\displaystyle \|Az\|_{H}=\|\langle z\,|\,A(\cdot )\rangle \|_{H^{*}}\quad {\text{ for every }}z\in H.}

Unitary operators

See also:Unitary transformation andUnitary matrix

An invertible bounded linear operatorA:HH{\displaystyle A:H\to H} is said to beunitary if its inverse is its adjoint:A1=A.{\displaystyle A^{-1}=A^{*}.} By using (Adjoint-transpose), this is seen to be equivalent toΦA1=tAΦ.{\displaystyle \Phi \circ A^{-1}={}^{t}A\circ \Phi .} Unraveling notation and definitions, it follows thatA{\displaystyle A} is unitary if and only ifA1z=zA() for all zH.{\displaystyle \langle A^{-1}z\mid \cdot \,\rangle =\langle z\mid A(\cdot )\rangle \quad {\text{ for all }}z\in H.}

The fact that a bounded invertible linear operatorA:HH{\displaystyle A:H\to H} is unitary if and only ifAA=IdH{\displaystyle A^{*}A=\operatorname {Id} _{H}} (or equivalently,tAΦA=Φ{\displaystyle {}^{t}A\circ \Phi \circ A=\Phi }) produces another (well-known) characterization: an invertible bounded linear mapA{\displaystyle A} is unitary if and only ifAzA()=z for all zH.{\displaystyle \langle Az\mid A(\cdot )\,\rangle =\langle z\mid \cdot \,\rangle \quad {\text{ for all }}z\in H.}

BecauseA:HH{\displaystyle A:H\to H} is invertible (and so in particular a bijection), this is also true of the transposetA:HH.{\displaystyle {}^{t}A:H^{*}\to H^{*}.} This fact also allows the vectorzH{\displaystyle z\in H} in the above characterizations to be replaced withAz{\displaystyle Az} orA1z,{\displaystyle A^{-1}z,} thereby producing many more equalities. Similarly,{\displaystyle \,\cdot \,} can be replaced withA(){\displaystyle A(\cdot )} orA1().{\displaystyle A^{-1}(\cdot ).}

See also

[edit]

Citations

[edit]
  1. ^abcdefghijklTrèves 2006, pp. 112–123.
  2. ^abcRudin 1991, pp. 306–312.
  3. ^Roman 2008, p. 351 Theorem 13.32
  4. ^Rudin 1991, pp. 307−309.
  5. ^Rudin 1991, pp. 92–115.

Notes

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  1. ^IfF=R{\displaystyle \mathbb {F} =\mathbb {R} } then the inner product will be symmetric so it does not matter which coordinate of the inner product the elementy{\displaystyle y} is placed into because the same map will result. But ifF=C{\displaystyle \mathbb {F} =\mathbb {C} } then except for the constant0{\displaystyle 0} map,antilinear functionals onH{\displaystyle H} are completely distinct fromlinear functionals onH,{\displaystyle H,} which makes the coordinate thaty{\displaystyle y} is placed into isvery important. For a non-zeroyH{\displaystyle y\in H} to induce alinear functional (rather than anantilinear functional),y{\displaystyle y}must be placed into theantilinear coordinate of the inner product. If it is incorrectly placed into the linear coordinate instead of the antilinear coordinate then the resulting map will be the antilinear maphy,h=hy,{\displaystyle h\mapsto \langle y,h\rangle =\langle h\mid y\rangle ,} which isnot a linear functional onH{\displaystyle H} and so it willnot be an element of the continuous dual spaceH.{\displaystyle H^{*}.}
  2. ^This means that for all vectorsyH:{\displaystyle y\in H:} (1)Φ:HH{\displaystyle \Phi :H\to H^{*}} isinjective. (2) Thenorms ofy{\displaystyle y} andΦ(y){\displaystyle \Phi (y)} are the same:Φ(y)=y.{\displaystyle \|\Phi (y)\|=\|y\|.} (3)Φ{\displaystyle \Phi } is anadditive map, meaning thatΦ(x+y)=Φ(x)+Φ(y){\displaystyle \Phi (x+y)=\Phi (x)+\Phi (y)} for allx,yH.{\displaystyle x,y\in H.} (4)Φ{\displaystyle \Phi } isconjugate homogeneous:Φ(sy)=s¯Φ(y){\displaystyle \Phi (sy)={\overline {s}}\Phi (y)} for all scalarss.{\displaystyle s.} (5)Φ{\displaystyle \Phi } isreal homogeneous:Φ(ry)=rΦ(y){\displaystyle \Phi (ry)=r\Phi (y)} for all real numbersrR.{\displaystyle r\in \mathbb {R} .}
  3. ^abThis footnote explains how to define - using onlyH{\displaystyle H}'s operations - addition and scalar multiplication of affine hyperplanes so that these operations correspond to addition and scalar multiplication of linear functionals. LetH{\displaystyle H} be any vector space and letH#{\displaystyle H^{\#}} denote itsalgebraic dual space. LetA:={φ1(1):φH#}{\displaystyle {\mathcal {A}}:=\left\{\varphi ^{-1}(1):\varphi \in H^{\#}\right\}} and let^{\displaystyle \,{\hat {\cdot }}\,} and+^{\displaystyle \,{\hat {+}}\,} denote the (unique) vector space operations onA{\displaystyle {\mathcal {A}}} that make the bijectionI:H#A{\displaystyle I:H^{\#}\to {\mathcal {A}}} defined byφφ1(1){\displaystyle \varphi \mapsto \varphi ^{-1}(1)} into avector space isomorphism. Note thatφ1(1)={\displaystyle \varphi ^{-1}(1)=\varnothing } if and only ifφ=0,{\displaystyle \varphi =0,} so{\displaystyle \varnothing } is the additive identity of(A,+^,^){\displaystyle \left({\mathcal {A}},{\hat {+}},{\hat {\cdot }}\right)} (because this is true ofI1()=0{\displaystyle I^{-1}(\varnothing )=0} inH#{\displaystyle H^{\#}} andI{\displaystyle I} is a vector space isomorphism). For everyAA,{\displaystyle A\in {\mathcal {A}},} letkerA=H{\displaystyle \ker A=H} ifA={\displaystyle A=\varnothing } and letkerA=AA{\displaystyle \ker A=A-A} otherwise; ifA=I(φ)=φ1(1){\displaystyle A=I(\varphi )=\varphi ^{-1}(1)} thenkerA=kerφ{\displaystyle \ker A=\ker \varphi } so this definition is consistent with the usual definition of the kernel of a linear functional. Say thatA,BA{\displaystyle A,B\in {\mathcal {A}}} areparallel ifkerA=kerB,{\displaystyle \ker A=\ker B,} where ifA{\displaystyle A} andB{\displaystyle B} are not empty then this happens if and only if the linear functionalsI1(A){\displaystyle I^{-1}(A)} andI1(B){\displaystyle I^{-1}(B)} are non-zero scalar multiples of each other. The vector space operations on the vector space of affine hyperplanesA{\displaystyle {\mathcal {A}}} are now described in a way that involvesonly the vector space operations onH{\displaystyle H}; this results in an interpretation of the vector space operations on the algebraic dual spaceH#{\displaystyle H^{\#}} that is entirely in terms of affine hyperplanes. Fix hyperplanesA,BA.{\displaystyle A,B\in {\mathcal {A}}.} Ifs{\displaystyle s} is a scalar thens^A={hH:shA}.{\displaystyle s{\hat {\cdot }}A=\left\{h\in H:sh\in A\right\}.} Describing the operationA+^B{\displaystyle A{\hat {+}}B} in terms of only the setsA=φ1(1){\displaystyle A=\varphi ^{-1}(1)} andB=ψ1(1){\displaystyle B=\psi ^{-1}(1)} is more complicated because by definition,A+^B=I(φ)+^I(ψ):=I(φ+ψ)=(φ+ψ)1(1).{\displaystyle A{\hat {+}}B=I(\varphi ){\hat {+}}I(\psi ):=I(\varphi +\psi )=(\varphi +\psi )^{-1}(1).} IfA={\displaystyle A=\varnothing } (respectively, ifB={\displaystyle B=\varnothing }) thenA+^B{\displaystyle A{\hat {+}}B} is equal toB{\displaystyle B} (resp. is equal toA{\displaystyle A}) so assumeA{\displaystyle A\neq \varnothing } andB.{\displaystyle B\neq \varnothing .} The hyperplanesA{\displaystyle A} andB{\displaystyle B} are parallel if and only if there exists some scalarr{\displaystyle r} (necessarily non-0) such thatA=rB,{\displaystyle A=rB,} in which caseA+^B={hH:(1+r)hB};{\displaystyle A{\hat {+}}B=\left\{h\in H:(1+r)h\in B\right\};} this can optionally be subdivided into two cases: ifr=1{\displaystyle r=-1} (which happens if and only if the linear functionalsI1(A){\displaystyle I^{-1}(A)} andI1(B){\displaystyle I^{-1}(B)} are negatives of each) thenA+^B={\displaystyle A{\hat {+}}B=\varnothing } while ifr1{\displaystyle r\neq -1} thenA+^B=11+rB=r1+rA.{\displaystyle A{\hat {+}}B={\frac {1}{1+r}}B={\frac {r}{1+r}}A.} Finally, assume now thatkerAkerB.{\displaystyle \ker A\neq \ker B.} ThenA+^B{\displaystyle A{\hat {+}}B} is the unique affine hyperplane containing bothAkerB{\displaystyle A\cap \ker B} andBkerA{\displaystyle B\cap \ker A} as subsets; explicitly,ker(A+^B)=span(AkerBBkerA){\displaystyle \ker \left(A{\hat {+}}B\right)=\operatorname {span} \left(A\cap \ker B-B\cap \ker A\right)} andA+^B=AkerB+ker(A+^B)=BkerA+ker(A+^B).{\displaystyle A{\hat {+}}B=A\cap \ker B+\ker \left(A{\hat {+}}B\right)=B\cap \ker A+\ker \left(A{\hat {+}}B\right).} To see why this formula forA+^B{\displaystyle A{\hat {+}}B} should hold, considerH:=R3,{\displaystyle H:=\mathbb {R} ^{3},}A:=φ1(1),{\displaystyle A:=\varphi ^{-1}(1),} andB:=ψ1(1),{\displaystyle B:=\psi ^{-1}(1),} whereφ(x,y,z):=x{\displaystyle \varphi (x,y,z):=x} andψ(x,y,z):=x+y{\displaystyle \psi (x,y,z):=x+y} (or alternatively,ψ(x,y,z):=y{\displaystyle \psi (x,y,z):=y}). Then by definition,A+^B:=(φ+ψ)1(1){\displaystyle A{\hat {+}}B:=(\varphi +\psi )^{-1}(1)} andker(A+^B):=(φ+ψ)1(0).{\displaystyle \ker \left(A{\hat {+}}B\right):=(\varphi +\psi )^{-1}(0).} NowAkerB = φ1(1)ψ1(0)  (φ+ψ)1(1){\displaystyle A\cap \ker B~=~\varphi ^{-1}(1)\cap \psi ^{-1}(0)~\subseteq ~(\varphi +\psi )^{-1}(1)} is an affine subspace ofcodimension2{\displaystyle 2} inH{\displaystyle H} (it is equal to a translation of thez{\displaystyle z}-axis{(0,0)}×R{\displaystyle \{(0,0)\}\times \mathbb {R} }). The same is true ofBkerA.{\displaystyle B\cap \ker A.} Plotting anx{\displaystyle x}-y{\displaystyle y}-plane cross section (that is, settingz={\displaystyle z=} constant) of the setskerA,kerB,A{\displaystyle \ker A,\ker B,A} andB{\displaystyle B} (each of which will be plotted as a line), the set(φ+ψ)1(1){\displaystyle (\varphi +\psi )^{-1}(1)} will then be plotted as the (unique) line passing through theAkerB{\displaystyle A\cap \ker B} andBkerA{\displaystyle B\cap \ker A} (which will be plotted as two distinct points) while(φ+ψ)1(0)=ker(A+^B){\displaystyle (\varphi +\psi )^{-1}(0)=\ker \left(A{\hat {+}}B\right)} will be plotted the line through the origin that is parallel toA+^B=(φ+ψ)1(1).{\displaystyle A{\hat {+}}B=(\varphi +\psi )^{-1}(1).} The above formulas forker(A+^B):=(φ+ψ)1(0){\displaystyle \ker \left(A{\hat {+}}B\right):=(\varphi +\psi )^{-1}(0)} andA+^B:=(φ+ψ)1(1){\displaystyle A{\hat {+}}B:=(\varphi +\psi )^{-1}(1)} follow naturally from the plot and they also hold in general.
  4. ^Showing that there is a non-zero vectorv{\displaystyle v} inK{\displaystyle K^{\bot }} relies on the continuity ofϕ{\displaystyle \phi } and theCauchy completeness ofH.{\displaystyle H.} This is the only place in the proof in which these properties are used.
  5. ^Technically,H=KK{\displaystyle H=K\oplus K^{\bot }} means that the addition mapK×KH{\displaystyle K\times K^{\bot }\to H} defined by(k,p)k+p{\displaystyle (k,p)\mapsto k+p} is a surjectivelinear isomorphism andhomeomorphism. See the article oncomplemented subspaces for more details.
  6. ^The usual notation for plugging an elementg{\displaystyle g} into a linear mapF{\displaystyle F} isF(g){\displaystyle F(g)} and sometimesFg.{\displaystyle Fg.} ReplacingF{\displaystyle F} withh∣:= Φh{\displaystyle \langle h\mid :=~\Phi h} producesh(g){\displaystyle \langle h\mid (g)} orhg,{\displaystyle \langle h\mid g,} which is unsightly (despite being consistent with the usual notation used with functions). Consequently, the symbol{\displaystyle \,\rangle \,} is appended to the end, so that the notationhg{\displaystyle \langle h\mid g\rangle } is used instead to denote this value(Φh)g.{\displaystyle (\Phi h)g.}
  7. ^abcdeThe notationzA()Z{\displaystyle \left\langle z\mid A(\cdot )\right\rangle _{Z}} denotes the continuous linear functional defined bygzAgZ.{\displaystyle g\mapsto \left\langle z\mid Ag\right\rangle _{Z}.}

Proofs

  1. ^This is becausexK=xx,fφfφ2fφ.{\displaystyle x_{K}=x-{\frac {\left\langle x,f_{\varphi }\right\rangle }{\left\|f_{\varphi }\right\|^{2}}}f_{\varphi }.} Now usefφ2=φ2{\displaystyle \left\|f_{\varphi }\right\|^{2}=\|\varphi \|^{2}} andx,fφ=φ(x){\displaystyle \left\langle x,f_{\varphi }\right\rangle =\varphi (x)} and solve forfφ.{\displaystyle f_{\varphi }.}{\displaystyle \blacksquare }
  2. ^AAzh=AzAhH=ΦAhΦAzH{\displaystyle \left\langle A^{*}Az\mid h\right\rangle =\left\langle \,Az\mid Ah\,\right\rangle _{H}=\left\langle \,\Phi Ah\mid \Phi Az\,\right\rangle _{H^{*}}} whereΦAh:=Ah{\displaystyle \Phi Ah:=\left\langle Ah\mid \cdot \,\right\rangle } andΦAz:=Az.{\displaystyle \Phi Az:=\left\langle Az\mid \cdot \,\right\rangle .} By definition of the adjoint,AhAz=hAAz{\displaystyle \left\langle A^{*}h\mid A^{*}z\,\right\rangle =\left\langle h\mid AA^{*}z\,\right\rangle } so taking the complex conjugate of both sides proves thatAAzh=AzAh.{\displaystyle \left\langle AA^{*}z\mid h\right\rangle =\left\langle A^{*}z\mid A^{*}h\right\rangle .} FromA=Φ1tAΦ,{\displaystyle A^{*}=\Phi ^{-1}\circ {}^{t}A\circ \Phi ,} it follows thatAAz|hH=AzAhH=Φ1tAΦzΦ1tAΦhH=tAΦhtAΦzH{\displaystyle \left\langle AA^{*}z\,|\,h\right\rangle _{H}=\left\langle A^{*}z\mid A^{*}h\right\rangle _{H}=\left\langle \Phi ^{-1}\circ {}^{t}A\circ \Phi z\mid \Phi ^{-1}\circ {}^{t}A\circ \Phi h\right\rangle _{H}=\left\langle \,{}^{t}A\circ \Phi h\mid {}^{t}A\circ \Phi z\right\rangle _{H^{*}}} where(tAΦ)h=h|A(){\displaystyle \left({}^{t}A\circ \Phi \right)h=\langle h\,|\,A(\cdot )\rangle } and(tAΦ)z=z|A().{\displaystyle \left({}^{t}A\circ \Phi \right)z=\langle z\,|\,A(\cdot )\rangle .}{\displaystyle \blacksquare }

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