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Sublinear function

From Wikipedia, the free encyclopedia
Type of function in linear algebra

Inlinear algebra, asublinear function (orfunctional as is more often used infunctional analysis), also called aquasi-seminorm or aBanach functional, on avector spaceX{\displaystyle X} is areal-valuedfunction with only some of the properties of aseminorm. Unlike seminorms, a sublinear function does not have to benonnegative-valued and also does not have to beabsolutely homogeneous. Seminorms are themselves abstractions of the more well known notion ofnorms, where a seminorm has all the defining properties of a normexcept that it is not required to map non-zero vectors to non-zero values.

Infunctional analysis the nameBanach functional is sometimes used, reflecting that they are most commonly used when applying a general formulation of theHahn–Banach theorem. The notion of a sublinear function was introduced byStefan Banach when he proved his version of theHahn-Banach theorem.[1]

There is also a different notion incomputer science, described below, that also goes by the name "sublinear function."

Definitions

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LetX{\displaystyle X} be avector space over a fieldK,{\displaystyle \mathbb {K} ,} whereK{\displaystyle \mathbb {K} } is either thereal numbersR{\displaystyle \mathbb {R} } orcomplex numbersC.{\displaystyle \mathbb {C} .} A real-valued functionp:XR{\displaystyle p:X\to \mathbb {R} } onX{\displaystyle X} is called asublinear function (or asublinearfunctional ifK=R{\displaystyle \mathbb {K} =\mathbb {R} }), and also sometimes called aquasi-seminorm or aBanach functional, if it has these two properties:[1]

  1. Positive homogeneity/Nonnegative homogeneity:[2]p(rx)=rp(x){\displaystyle p(rx)=rp(x)} for all realr0{\displaystyle r\geq 0} and allxX.{\displaystyle x\in X.}
  2. Subadditivity/Triangle inequality:[2]p(x+y)p(x)+p(y){\displaystyle p(x+y)\leq p(x)+p(y)} for allx,yX.{\displaystyle x,y\in X.}

A functionp:XR{\displaystyle p:X\to \mathbb {R} } is calledpositive[3] ornonnegative ifp(x)0{\displaystyle p(x)\geq 0} for allxX,{\displaystyle x\in X,} although some authors[4] definepositive to instead mean thatp(x)0{\displaystyle p(x)\neq 0} wheneverx0;{\displaystyle x\neq 0;} these definitions are not equivalent. It is asymmetric function ifp(x)=p(x){\displaystyle p(-x)=p(x)} for allxX.{\displaystyle x\in X.} Every subadditive symmetric function is necessarily nonnegative.[proof 1] A sublinear function on a real vector space issymmetric if and only if it is aseminorm. A sublinear function on a real or complex vector space is a seminorm if and only if it is abalanced function or equivalently, if and only ifp(ux)p(x){\displaystyle p(ux)\leq p(x)} for everyunit length scalaru{\displaystyle u} (satisfying|u|=1{\displaystyle |u|=1}) and everyxX.{\displaystyle x\in X.}

The set of all sublinear functions onX,{\displaystyle X,} denoted byX#,{\displaystyle X^{\#},} can bepartially ordered by declaringpq{\displaystyle p\leq q} if and only ifp(x)q(x){\displaystyle p(x)\leq q(x)} for allxX.{\displaystyle x\in X.} A sublinear function is calledminimal if it is aminimal element ofX#{\displaystyle X^{\#}} under this order. A sublinear function is minimal if and only if it is a reallinear functional.[1]

Examples and sufficient conditions

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Everynorm,seminorm, and real linear functional is a sublinear function. Theidentity functionRR{\displaystyle \mathbb {R} \to \mathbb {R} } onX:=R{\displaystyle X:=\mathbb {R} } is an example of a sublinear function (in fact, it is even a linear functional) that is neither positive nor a seminorm; the same is true of this map's negationxx.{\displaystyle x\mapsto -x.}[5] More generally, for any realab,{\displaystyle a\leq b,} the mapSa,b:RRx{ax if x0bx if x0{\displaystyle {\begin{alignedat}{4}S_{a,b}:\;&&\mathbb {R} &&\;\to \;&\mathbb {R} \\[0.3ex]&&x&&\;\mapsto \;&{\begin{cases}ax&{\text{ if }}x\leq 0\\bx&{\text{ if }}x\geq 0\\\end{cases}}\\\end{alignedat}}}is a sublinear function onX:=R{\displaystyle X:=\mathbb {R} } and moreover, every sublinear functionp:RR{\displaystyle p:\mathbb {R} \to \mathbb {R} } is of this form; specifically, ifa:=p(1){\displaystyle a:=-p(-1)} andb:=p(1){\displaystyle b:=p(1)} thenab{\displaystyle a\leq b} andp=Sa,b.{\displaystyle p=S_{a,b}.}

Ifp{\displaystyle p} andq{\displaystyle q} are sublinear functions on a real vector spaceX{\displaystyle X} then so is the mapxmax{p(x),q(x)}.{\displaystyle x\mapsto \max\{p(x),q(x)\}.} More generally, ifP{\displaystyle {\mathcal {P}}} is any non-empty collection of sublinear functionals on a real vector spaceX{\displaystyle X} and if for allxX,{\displaystyle x\in X,}q(x):=sup{p(x):pP},{\displaystyle q(x):=\sup\{p(x):p\in {\mathcal {P}}\},} thenq{\displaystyle q} is a sublinear functional onX.{\displaystyle X.}[5]


A functionp:XR{\displaystyle p:X\to \mathbb {R} } which issubadditive,convex, and satisfiesp(0)0{\displaystyle p(0)\leq 0} is also positively homogeneous (the latter conditionp(0)0{\displaystyle p(0)\leq 0} is necessary as the example ofp(x):=x2+1{\displaystyle p(x):={\sqrt {x^{2}+1}}} onX:=R{\displaystyle X:=\mathbb {R} } shows). Ifp{\displaystyle p} is positively homogeneous, it is convex if and only if it is subadditive. Therefore, assumingp(0)0{\displaystyle p(0)\leq 0}, any two properties among subadditivity, convexity, and positive homogeneity implies the third.

Properties

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Every sublinear function is aconvex function: For0t1,{\displaystyle 0\leq t\leq 1,}p(tx+(1t)y)p(tx)+p((1t)y) subadditivity=tp(x)+(1t)p(y) nonnegative homogeneity{\displaystyle {\begin{alignedat}{3}p(tx+(1-t)y)&\leq p(tx)+p((1-t)y)&&\quad {\text{ subadditivity}}\\&=tp(x)+(1-t)p(y)&&\quad {\text{ nonnegative homogeneity}}\\\end{alignedat}}}

Ifp:XR{\displaystyle p:X\to \mathbb {R} } is a sublinear function on a vector spaceX{\displaystyle X} then[proof 2][3]p(0) = 0  p(x)+p(x),{\displaystyle p(0)~=~0~\leq ~p(x)+p(-x),} for everyxX,{\displaystyle x\in X,} which implies that at least one ofp(x){\displaystyle p(x)} andp(x){\displaystyle p(-x)} must be nonnegative; that is, for everyxX,{\displaystyle x\in X,}[3]0  max{p(x),p(x)}.{\displaystyle 0~\leq ~\max\{p(x),p(-x)\}.}Moreover, whenp:XR{\displaystyle p:X\to \mathbb {R} } is a sublinear function on a real vector space then the mapq:XR{\displaystyle q:X\to \mathbb {R} } defined byq(x) =def max{p(x),p(x)}{\displaystyle q(x)~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\max\{p(x),p(-x)\}} is a seminorm.[3]

Subadditivity ofp:XR{\displaystyle p:X\to \mathbb {R} } guarantees that for all vectorsx,yX,{\displaystyle x,y\in X,}[1][proof 3]p(x)p(y)  p(xy),{\displaystyle p(x)-p(y)~\leq ~p(x-y),}p(x)  p(x),{\displaystyle -p(x)~\leq ~p(-x),}so ifp{\displaystyle p} is alsosymmetric then thereverse triangle inequality will hold for all vectorsx,yX,{\displaystyle x,y\in X,}|p(x)p(y)|  p(xy).{\displaystyle |p(x)-p(y)|~\leq ~p(x-y).}

Definingkerp =def p1(0),{\displaystyle \ker p~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~p^{-1}(0),} then subadditivity also guarantees that for allxX,{\displaystyle x\in X,} the value ofp{\displaystyle p} on the setx+(kerpkerp)={x+k:p(k)=0=p(k)}{\displaystyle x+(\ker p\cap -\ker p)=\{x+k:p(k)=0=p(-k)\}} is constant and equal top(x).{\displaystyle p(x).}[proof 4] In particular, ifkerp=p1(0){\displaystyle \ker p=p^{-1}(0)} is a vector subspace ofX{\displaystyle X} thenkerp=kerp{\displaystyle -\ker p=\ker p} and the assignmentx+kerpp(x),{\displaystyle x+\ker p\mapsto p(x),} which will be denoted byp^,{\displaystyle {\hat {p}},} is a well-defined real-valued sublinear function on thequotient spaceX/kerp{\displaystyle X\,/\,\ker p} that satisfiesp^1(0)=kerp.{\displaystyle {\hat {p}}^{-1}(0)=\ker p.} Ifp{\displaystyle p} is a seminorm thenp^{\displaystyle {\hat {p}}} is just the usual canonical norm on the quotient spaceX/kerp.{\displaystyle X\,/\,\ker p.}

Pryce's sublinearity lemma[2]Supposep:XR{\displaystyle p:X\to \mathbb {R} } is a sublinear functional on a vector spaceX{\displaystyle X} and thatKX{\displaystyle K\subseteq X} is a non-empty convex subset. IfxX{\displaystyle x\in X} is a vector anda,c>0{\displaystyle a,c>0} are positive real numbers such thatp(x)+ac < infkKp(x+ak){\displaystyle p(x)+ac~<~\inf _{k\in K}p(x+ak)}then for every positive realb>0{\displaystyle b>0} there exists somezK{\displaystyle \mathbf {z} \in K} such thatp(x+az)+bc < infkKp(x+az+bk).{\displaystyle p(x+a\mathbf {z} )+bc~<~\inf _{k\in K}p(x+a\mathbf {z} +bk).}

Addingbc{\displaystyle bc} to both sides of the hypothesisp(x)+ac<infp(x+aK){\textstyle p(x)+ac\,<\,\inf _{}p(x+aK)} (wherep(x+aK) =def {p(x+ak):kK}{\displaystyle p(x+aK)~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\{p(x+ak):k\in K\}}) and combining that with the conclusion givesp(x)+ac+bc < infp(x+aK)+bc  p(x+az)+bc < infp(x+az+bK){\displaystyle p(x)+ac+bc~<~\inf _{}p(x+aK)+bc~\leq ~p(x+a\mathbf {z} )+bc~<~\inf _{}p(x+a\mathbf {z} +bK)} which yields many more inequalities, including, for instance,p(x)+ac+bc < p(x+az)+bc < p(x+az+bz){\displaystyle p(x)+ac+bc~<~p(x+a\mathbf {z} )+bc~<~p(x+a\mathbf {z} +b\mathbf {z} )} in which an expression on one side of a strict inequality<{\displaystyle \,<\,} can be obtained from the other by replacing the symbolc{\displaystyle c} withz{\displaystyle \mathbf {z} } (or vice versa) and moving the closing parenthesis to the right (or left) of an adjacent summand (all other symbols remain fixed and unchanged).

Associated seminorm

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Ifp:XR{\displaystyle p:X\to \mathbb {R} } is a real-valued sublinear function on a real vector spaceX{\displaystyle X} (or ifX{\displaystyle X} is complex, then when it is considered as a real vector space) then the mapq(x) =def max{p(x),p(x)}{\displaystyle q(x)~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\max\{p(x),p(-x)\}} defines aseminorm on the real vector spaceX{\displaystyle X} called theseminorm associated withp.{\displaystyle p.}[3] A sublinear functionp{\displaystyle p} on a real or complex vector space is asymmetric function if and only ifp=q{\displaystyle p=q} whereq(x) =def max{p(x),p(x)}{\displaystyle q(x)~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\max\{p(x),p(-x)\}} as before.

More generally, ifp:XR{\displaystyle p:X\to \mathbb {R} } is a real-valued sublinear function on a (real or complex) vector spaceX{\displaystyle X} thenq(x) =def sup|u|=1p(ux) = sup{p(ux):u is a unit scalar }{\displaystyle q(x)~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\sup _{|u|=1}p(ux)~=~\sup\{p(ux):u{\text{ is a unit scalar }}\}} will define aseminorm onX{\displaystyle X} if this supremum is always a real number (that is, never equal to{\displaystyle \infty }).

Relation to linear functionals

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Ifp{\displaystyle p} is a sublinear function on a real vector spaceX{\displaystyle X} then the following are equivalent:[1]

  1. p{\displaystyle p} is alinear functional.
  2. for everyxX,{\displaystyle x\in X,}p(x)+p(x)0.{\displaystyle p(x)+p(-x)\leq 0.}
  3. for everyxX,{\displaystyle x\in X,}p(x)+p(x)=0.{\displaystyle p(x)+p(-x)=0.}
  4. p{\displaystyle p} is a minimal sublinear function.

Ifp{\displaystyle p} is a sublinear function on a real vector spaceX{\displaystyle X} then there exists a linear functionalf{\displaystyle f} onX{\displaystyle X} such thatfp.{\displaystyle f\leq p.}[1]

IfX{\displaystyle X} is a real vector space,f{\displaystyle f} is a linear functional onX,{\displaystyle X,} andp{\displaystyle p} is a positive sublinear function onX,{\displaystyle X,} thenfp{\displaystyle f\leq p} onX{\displaystyle X} if and only iff1(1){xX:p(x)<1}=.{\displaystyle f^{-1}(1)\cap \{x\in X:p(x)<1\}=\varnothing .}[1]

Dominating a linear functional

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A real-valued functionf{\displaystyle f} defined on a subset of a real or complex vector spaceX{\displaystyle X} is said to bedominated by a sublinear functionp{\displaystyle p} iff(x)p(x){\displaystyle f(x)\leq p(x)} for everyx{\displaystyle x} that belongs to the domain off.{\displaystyle f.} Iff:XR{\displaystyle f:X\to \mathbb {R} } is a reallinear functional onX{\displaystyle X} then[6][1]f{\displaystyle f} is dominated byp{\displaystyle p} (that is,fp{\displaystyle f\leq p}) if and only ifp(x)f(x)p(x) for every xX.{\displaystyle -p(-x)\leq f(x)\leq p(x)\quad {\text{ for every }}x\in X.} Moreover, ifp{\displaystyle p} is a seminorm or some othersymmetric map (which by definition means thatp(x)=p(x){\displaystyle p(-x)=p(x)} holds for allx{\displaystyle x}) thenfp{\displaystyle f\leq p} if and only if|f|p.{\displaystyle |f|\leq p.}

Theorem[1]Ifp:XR{\displaystyle p:X\to \mathbb {R} } be a sublinear function on a real vector spaceX{\displaystyle X} and ifzX{\displaystyle z\in X} then there exists a linear functionalf{\displaystyle f} onX{\displaystyle X} that is dominated byp{\displaystyle p} (that is,fp{\displaystyle f\leq p}) and satisfiesf(z)=p(z).{\displaystyle f(z)=p(z).} Moreover, ifX{\displaystyle X} is atopological vector space andp{\displaystyle p} is continuous at the origin thenf{\displaystyle f} is continuous.

Continuity

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Theorem[7]Supposef:XR{\displaystyle f:X\to \mathbb {R} } is a subadditive function (that is,f(x+y)f(x)+f(y){\displaystyle f(x+y)\leq f(x)+f(y)} for allx,yX{\displaystyle x,y\in X}). Thenf{\displaystyle f} is continuous at the origin if and only iff{\displaystyle f} is uniformly continuous onX.{\displaystyle X.} Iff{\displaystyle f} satisfiesf(0)=0{\displaystyle f(0)=0} thenf{\displaystyle f} is continuous if and only if its absolute value|f|:X[0,){\displaystyle |f|:X\to [0,\infty )} is continuous. Iff{\displaystyle f} is non-negative thenf{\displaystyle f} is continuous if and only if{xX:f(x)<1}{\displaystyle \{x\in X:f(x)<1\}} is open inX.{\displaystyle X.}

SupposeX{\displaystyle X} is atopological vector space (TVS) over the real or complex numbers andp{\displaystyle p} is a sublinear function onX.{\displaystyle X.} Then the following are equivalent:[7]

  1. p{\displaystyle p} is continuous;
  2. p{\displaystyle p} is continuous at 0;
  3. p{\displaystyle p} is uniformly continuous onX{\displaystyle X};

and ifp{\displaystyle p} is positive then this list may be extended to include:

  1. {xX:p(x)<1}{\displaystyle \{x\in X:p(x)<1\}} is open inX.{\displaystyle X.}

IfX{\displaystyle X} is a real TVS,f{\displaystyle f} is a linear functional onX,{\displaystyle X,} andp{\displaystyle p} is a continuous sublinear function onX,{\displaystyle X,} thenfp{\displaystyle f\leq p} onX{\displaystyle X} implies thatf{\displaystyle f} is continuous.[7]

Relation to Minkowski functions and open convex sets

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Theorem[7]IfU{\displaystyle U} is a convex open neighborhood of the origin in atopological vector spaceX{\displaystyle X} then theMinkowski functional ofU,{\displaystyle U,}pU:X[0,),{\displaystyle p_{U}:X\to [0,\infty ),} is a continuous non-negative sublinear function onX{\displaystyle X} such thatU={xX:pU(x)<1};{\displaystyle U=\left\{x\in X:p_{U}(x)<1\right\};} if in additionU{\displaystyle U} is abalanced set thenpU{\displaystyle p_{U}} is aseminorm onX.{\displaystyle X.}

Relation to open convex sets

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Theorem[7]Suppose thatX{\displaystyle X} is atopological vector space (not necessarilylocally convex orHausdorff) over the real or complex numbers. Then the open convex subsets ofX{\displaystyle X} are exactly those that are of the formz+{xX:p(x)<1}={xX:p(xz)<1}{\displaystyle z+\{x\in X:p(x)<1\}=\{x\in X:p(x-z)<1\}} for somezX{\displaystyle z\in X} and some positive continuous sublinear functionp{\displaystyle p} onX.{\displaystyle X.}

Proof

LetV{\displaystyle V} be an open convex subset ofX.{\displaystyle X.} If0V{\displaystyle 0\in V} then letz:=0{\displaystyle z:=0} and otherwise letzV{\displaystyle z\in V} be arbitrary. Letp:X[0,){\displaystyle p:X\to [0,\infty )} be theMinkowski functional ofVz,{\displaystyle V-z,} which is a continuous sublinear function onX{\displaystyle X} sinceVz{\displaystyle V-z} is convex,absorbing, and open (p{\displaystyle p} however is not necessarily a seminorm sinceV{\displaystyle V} was not assumed to bebalanced). FromX=Xz,{\displaystyle X=X-z,} it follows thatz+{xX:p(x)<1}={xX:p(xz)<1}.{\displaystyle z+\{x\in X:p(x)<1\}=\{x\in X:p(x-z)<1\}.} It will be shown thatV=z+{xX:p(x)<1},{\displaystyle V=z+\{x\in X:p(x)<1\},} which will complete the proof.One of the knownproperties of Minkowski functionals guarantees{xX:p(x)<1}=(0,1)(Vz),{\textstyle \{x\in X:p(x)<1\}=(0,1)(V-z),} where(0,1)(Vz)=def{tx:0<t<1,xVz}=Vz{\displaystyle (0,1)(V-z)\;{\stackrel {\scriptscriptstyle {\text{def}}}{=}}\;\{tx:0<t<1,x\in V-z\}=V-z} sinceVz{\displaystyle V-z} is convex and contains the origin. ThusVz={xX:p(x)<1},{\displaystyle V-z=\{x\in X:p(x)<1\},} as desired.{\displaystyle \blacksquare }

Operators

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The concept can be extended to operators that are homogeneous and subadditive. This requires only that thecodomain be, say, anordered vector space to make sense of the conditions.

Computer science definition

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Incomputer science, a functionf:Z+R{\displaystyle f:\mathbb {Z} ^{+}\to \mathbb {R} } is calledsublinear iflimnf(n)n=0,{\displaystyle \lim _{n\to \infty }{\frac {f(n)}{n}}=0,} orf(n)o(n){\displaystyle f(n)\in o(n)} inasymptotic notation (notice the smallo{\displaystyle o}). Formally,f(n)o(n){\displaystyle f(n)\in o(n)} if and only if, for any givenc>0,{\displaystyle c>0,} there exists anN{\displaystyle N} such thatf(n)<cn{\displaystyle f(n)<cn} fornN.{\displaystyle n\geq N.}[8]That is,f{\displaystyle f} grows slower than any linear function.The two meanings should not be confused: while a Banach functional isconvex, almost the opposite is true for functions of sublinear growth: every functionf(n)o(n){\displaystyle f(n)\in o(n)} can be upper-bounded by aconcave function of sublinear growth.[9]

See also

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Notes

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Proofs

  1. ^LetxX.{\displaystyle x\in X.} The triangle inequality and symmetry implyp(0)=p(x+(x))p(x)+p(x)=p(x)+p(x)=2p(x).{\displaystyle p(0)=p(x+(-x))\leq p(x)+p(-x)=p(x)+p(x)=2p(x).} Substituting0{\displaystyle 0} forx{\displaystyle x} and then subtractingp(0){\displaystyle p(0)} from both sides proves that0p(0).{\displaystyle 0\leq p(0).} Thus0p(0)2p(x){\displaystyle 0\leq p(0)\leq 2p(x)} which implies0p(x).{\displaystyle 0\leq p(x).}{\displaystyle \blacksquare }
  2. ^IfxX{\displaystyle x\in X} andr:=0{\displaystyle r:=0} then nonnegative homogeneity implies thatp(0)=p(rx)=rp(x)=0p(x)=0.{\displaystyle p(0)=p(rx)=rp(x)=0p(x)=0.} Consequently,0=p(0)=p(x+(x))p(x)+p(x),{\displaystyle 0=p(0)=p(x+(-x))\leq p(x)+p(-x),} which is only possible if0max{p(x),p(x)}.{\displaystyle 0\leq \max\{p(x),p(-x)\}.}{\displaystyle \blacksquare }
  3. ^p(x)=p(y+(xy))p(y)+p(xy),{\displaystyle p(x)=p(y+(x-y))\leq p(y)+p(x-y),} which happens if and only ifp(x)p(y)p(xy).{\displaystyle p(x)-p(y)\leq p(x-y).}{\displaystyle \blacksquare } Substitutingy:=x{\displaystyle y:=-x} and givesp(x)p(x)p(x(x))=p(x+x)p(x)+p(x),{\displaystyle p(x)-p(-x)\leq p(x-(-x))=p(x+x)\leq p(x)+p(x),} which impliesp(x)p(x){\displaystyle -p(-x)\leq p(x)} (positive homogeneity is not needed; the triangle inequality suffices).{\displaystyle \blacksquare }
  4. ^LetxX{\displaystyle x\in X} andkp1(0)(p1(0)).{\displaystyle k\in p^{-1}(0)\cap (-p^{-1}(0)).} It remains to show thatp(x+k)=p(x).{\displaystyle p(x+k)=p(x).} The triangle inequality impliesp(x+k)p(x)+p(k)=p(x)+0=p(x).{\displaystyle p(x+k)\leq p(x)+p(k)=p(x)+0=p(x).} Sincep(k)=0,{\displaystyle p(-k)=0,}p(x)=p(x)p(k)p(x(k))=p(x+k),{\displaystyle p(x)=p(x)-p(-k)\leq p(x-(-k))=p(x+k),} as desired.{\displaystyle \blacksquare }

References

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  1. ^abcdefghiNarici & Beckenstein 2011, pp. 177–220.
  2. ^abcSchechter 1996, pp. 313–315.
  3. ^abcdeNarici & Beckenstein 2011, pp. 120–121.
  4. ^Kubrusly 2011, p. 200.
  5. ^abNarici & Beckenstein 2011, pp. 177–221.
  6. ^Rudin 1991, pp. 56–62.
  7. ^abcdeNarici & Beckenstein 2011, pp. 192–193.
  8. ^Thomas H. Cormen,Charles E. Leiserson,Ronald L. Rivest, andClifford Stein (2001) [1990]. "3.1".Introduction to Algorithms (2nd ed.). MIT Press and McGraw-Hill. pp. 47–48.ISBN 0-262-03293-7.{{cite book}}: CS1 maint: multiple names: authors list (link)
  9. ^Ceccherini-Silberstein, Tullio; Salvatori, Maura; Sava-Huss, Ecaterina (2017-06-29).Groups, graphs, and random walks. Cambridge. Lemma 5.17.ISBN 9781316604403.OCLC 948670194.{{cite book}}: CS1 maint: location missing publisher (link)

Bibliography

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