Infunctional analysis and related areas ofmathematics, acontinuous linear operator orcontinuous linear mapping is acontinuouslinear transformation betweentopological vector spaces.

An operator between twonormed spaces is abounded linear operator if and only if it is a continuous linear operator.

Suppose that${\displaystyle F:X\to Y}$ is alinear operator between twotopological vector spaces (TVSs). The following are equivalent:

1. ${\displaystyle F}$ is continuous.
2. ${\displaystyle F}$ iscontinuous at some point${\displaystyle x\in X.}$
3. ${\displaystyle F}$ is continuous at the origin in${\displaystyle X.}$

If${\displaystyle Y}$ islocally convex then this list may be extended to include:

4. for every continuousseminorm${\displaystyle q}$ on${\displaystyle Y,}$ there exists a continuous seminorm${\displaystyle p}$ on${\displaystyle X}$ such that${\displaystyle q\circ F\le p.}$^[1]

If${\displaystyle X}$ and${\displaystyle Y}$ are bothHausdorff locally convex spaces then this list may be extended to include:

5. ${\displaystyle F}$ isweakly continuous and itstranspose${\displaystyle {}^{t}F:Y^{\mathrm{\prime }}\to X^{\mathrm{\prime }}}$ mapsequicontinuous subsets of${\displaystyle Y^{\mathrm{\prime }}}$ to equicontinuous subsets of${\displaystyle X^{\mathrm{\prime }}.}$

If${\displaystyle X}$ is asequential space (such as apseudometrizable space) then this list may be extended to include:

6. ${\displaystyle F}$ issequentially continuous at some (or equivalently, at every) point of its domain.

If${\displaystyle X}$ ispseudometrizable or metrizable (such as a normed orBanach space) then we may add to this list:

7. ${\displaystyle F}$ is abounded linear operator (that is, it maps bounded subsets of${\displaystyle X}$ to bounded subsets of${\displaystyle Y}$).^[2]

If${\displaystyle Y}$ isseminormable space (such as anormed space) then this list may be extended to include:

8. ${\displaystyle F}$ maps some neighborhood of 0 to a bounded subset of${\displaystyle Y.}$^[3]

If${\displaystyle X}$ and${\displaystyle Y}$ are bothnormed orseminormed spaces (with both seminorms denoted by${\displaystyle \Vert \cdot \Vert }$) then this list may be extended to include:

9. for every${\displaystyle r>0}$ there exists some${\displaystyle \delta >0}$ such that

If${\displaystyle X}$ and${\displaystyle Y}$ are Hausdorff locally convex spaces with${\displaystyle Y}$ finite-dimensional then this list may be extended to include:

10. the graph of${\displaystyle F}$ is closed in${\displaystyle X\times Y.}$^[4]

Throughout,${\displaystyle F:X\to Y}$ is alinear map betweentopological vector spaces (TVSs).

Bounded subset

The notion of a "bounded set" for a topological vector space is that of being avon Neumann bounded set. If the space happens to also be anormed space (or aseminormed space) then a subset${\displaystyle S}$ is von Neumann bounded if and only if it isnorm bounded, meaning that A subset of a normed (or seminormed) space is calledbounded if it is norm-bounded (or equivalently, von Neumann bounded). For example, the scalar field (${\displaystyle \mathbb{R}}$ or${\displaystyle \mathbb{C}}$) with theabsolute value${\displaystyle |\cdot |}$ is a normed space, so a subset${\displaystyle S}$ is bounded if and only if${\displaystyle \underset{s\in S}{sup}|s|}$ is finite, which happens if and only if${\displaystyle S}$ is contained in some open (or closed) ball centered at the origin (zero).

Any translation, scalar multiple, and subset of a bounded set is again bounded.

Function bounded on a set

If${\displaystyle S\subseteq X}$ is a set then${\displaystyle F:X\to Y}$ is said to bebounded on${\displaystyle S}$ if${\displaystyle F(S)}$ is abounded subset of${\displaystyle Y,}$ which if${\displaystyle (Y,\Vert \cdot \Vert )}$ is a normed (or seminormed) space happens if and only if A linear map${\displaystyle F}$ is bounded on a set${\displaystyle S}$ if and only if it is bounded on${\displaystyle x+S:=\{x+s:s\in S\}}$ for every${\displaystyle x\in X}$ (because${\displaystyle F(x+S)=F(x)+F(S)}$ and any translation of a bounded set is again bounded) if and only if it is bounded on${\displaystyle cS:=\{cs:s\in S\}}$ for every non-zero scalar${\displaystyle c\ne 0}$ (because${\displaystyle F(cS)=cF(S)}$ and any scalar multiple of a bounded set is again bounded). Consequently, if${\displaystyle (X,\Vert \cdot \Vert )}$ is a normed or seminormed space, then a linear map${\displaystyle F:X\to Y}$ is bounded on some (equivalently, on every) non-degenerate open or closed ball (not necessarily centered at the origin, and of any radius) if and only if it is bounded on the closed unit ball centered at the origin${\displaystyle \{x\in X:\Vert x\Vert \le 1\}.}$

Bounded linear maps

By definition, a linear map${\displaystyle F:X\to Y}$ betweenTVSs is said to bebounded and is called abounded linear operator if for every(von Neumann) bounded subset${\displaystyle B\subseteq X}$ of its domain,${\displaystyle F(B)}$ is a bounded subset of it codomain; or said more briefly, if it is bounded on every bounded subset of its domain. When the domain${\displaystyle X}$ is a normed (or seminormed) space then it suffices to check this condition for the open or closed unit ball centered at the origin. Explicitly, if${\displaystyle B_1}$ denotes this ball then${\displaystyle F:X\to Y}$ is a bounded linear operator if and only if${\displaystyle F\left(B_1\right)}$ is a bounded subset of${\displaystyle Y;}$ if${\displaystyle Y}$ is also a (semi)normed space then this happens if and only if theoperator norm is finite. Everysequentially continuous linear operator is bounded.^[5]

Function bounded on a neighborhood and local boundedness

In contrast, a map${\displaystyle F:X\to Y}$ is said to bebounded on a neighborhood of a point${\displaystyle x\in X}$ orlocally bounded at${\displaystyle x}$ if there exists aneighborhood${\displaystyle U}$ of this point in${\displaystyle X}$ such that${\displaystyle F(U)}$ is abounded subset of${\displaystyle Y.}$ It is "bounded on a neighborhood" (of some point) if there existssome point${\displaystyle x}$ in its domain at which it is locally bounded, in which case this linear map${\displaystyle F}$ is necessarily locally bounded atevery point of its domain. The term "locally bounded" is sometimes used to refer to a map that is locally bounded at every point of its domain, but some functional analysis authors define "locally bounded" to instead be a synonym of "bounded linear operator", which are related butnot equivalent concepts. For this reason, this article will avoid the term "locally bounded" and instead say "locally bounded at every point" (there is no disagreement about the definition of "locally boundedat a point").

A linear map is "bounded on a neighborhood" (of some point) if and only if it is locally bounded at every point of its domain, in which case it is necessarilycontinuous^[2] (even if its domain is not anormed space) and thus alsobounded (because a continuous linear operator is always abounded linear operator).^[6]

For any linear map, if it isbounded on a neighborhood then it is continuous,^[2][7] and if it is continuous then it isbounded.^[6] The converse statements are not true in general but they are both true when the linear map's domain is anormed space. Examples and additional details are now given below.

The next example shows that it is possible for a linear map to becontinuous (and thus also bounded) but not bounded on any neighborhood. In particular, it demonstrates that being "bounded on a neighborhood" isnot always synonymous with being "bounded".

Example: A continuous and bounded linear map that is not bounded on any neighborhood: If${\displaystyle \mathrm{Id}:X\to X}$ is the identity map on somelocally convex topological vector space then this linear map is always continuous (indeed, even aTVS-isomorphism) andbounded, but${\displaystyle \mathrm{Id}}$ is bounded on a neighborhood if and only if there exists a bounded neighborhood of the origin in${\displaystyle X,}$ whichis equivalent to${\displaystyle X}$ being aseminormable space (which if${\displaystyle X}$ is Hausdorff, is the same as being anormable space). This shows that it is possible for a linear map to be continuous butnot bounded on any neighborhood. Indeed, this example shows that everylocally convex space that is not seminormable has a linear TVS-automorphism that is not bounded on any neighborhood of any point. Thus although every linear map that is bounded on a neighborhood is necessarily continuous, the converse is not guaranteed in general.

To summarize the discussion below, for a linear map on a normed (or seminormed) space, being continuous, beingbounded, and being bounded on a neighborhood are allequivalent. A linear map whose domainor codomain is normable (or seminormable) is continuous if and only if it bounded on a neighborhood. And abounded linear operator valued in alocally convex space will be continuous if its domain is(pseudo)metrizable^[2] orbornological.^[6]

Guaranteeing that "continuous" implies "bounded on a neighborhood"

A TVS is said to belocally bounded if there exists a neighborhood that is also abounded set.^[8] For example, everynormed orseminormed space is a locally bounded TVS since the unit ball centered at the origin is a bounded neighborhood of the origin. If${\displaystyle B}$ is a bounded neighborhood of the origin in a (locally bounded) TVS then its image under any continuous linear map will be a bounded set (so this map is thus bounded on this neighborhood${\displaystyle B}$). Consequently, a linear map from a locally bounded TVS into any other TVS is continuous if and only if it isbounded on a neighborhood. Moreover, any TVS with this property must be a locally bounded TVS. Explicitly, if${\displaystyle X}$ is a TVS such that every continuous linear map (into any TVS) whose domain is${\displaystyle X}$ is necessarily bounded on a neighborhood, then${\displaystyle X}$ must be a locally bounded TVS (because theidentity function${\displaystyle X\to X}$ is always a continuous linear map).

Any linear map from a TVS into a locally bounded TVS (such as any linear functional) is continuous if and only if it is bounded on a neighborhood.^[8] Conversely, if${\displaystyle Y}$ is a TVS such that every continuous linear map (from any TVS) with codomain${\displaystyle Y}$ is necessarilybounded on a neighborhood, then${\displaystyle Y}$ must be a locally bounded TVS.^[8] In particular, a linear functional on an arbitrary TVS is continuous if and only if it is bounded on a neighborhood.^[8]

Thus when the domainor the codomain of a linear map is normable or seminormable, then continuity will beequivalent to being bounded on a neighborhood.

Guaranteeing that "bounded" implies "continuous"

A continuous linear operator is always abounded linear operator.^[6] But importantly, in the most general setting of a linear operator between arbitrary topological vector spaces, it is possible for a linear operator to bebounded but tonot be continuous.

A linear map whose domain ispseudometrizable (such as anynormed space) isbounded if and only if it is continuous.^[2] The same is true of a linear map from abornological space into alocally convex space.^[6]

Guaranteeing that "bounded" implies "bounded on a neighborhood"

In general, without additional information about either the linear map or its domain or codomain, the map being "bounded" is not equivalent to it being "bounded on a neighborhood". If${\displaystyle F:X\to Y}$ is a bounded linear operator from anormed space${\displaystyle X}$ into some TVS then${\displaystyle F:X\to Y}$ is necessarily continuous; this is because any open ball${\displaystyle B}$ centered at the origin in${\displaystyle X}$ is both a bounded subset (which implies that${\displaystyle F(B)}$ is bounded since${\displaystyle F}$ is a bounded linear map) and a neighborhood of the origin in${\displaystyle X,}$ so that${\displaystyle F}$ is thus bounded on this neighborhood${\displaystyle B}$ of the origin, which (as mentioned above) guarantees continuity.

Every linear functional on atopological vector space (TVS) is a linear operator so all of the properties described above for continuous linear operators apply to them. However, because of their specialized nature, we can say even more about continuous linear functionals than we can about more general continuous linear operators.

Let${\displaystyle X}$ be atopological vector space (TVS) over the field${\displaystyle \mathbb{F}}$ (${\displaystyle X}$ need not beHausdorff orlocally convex) and let${\displaystyle f:X\to \mathbb{F}}$ be alinear functional on${\displaystyle X.}$ The following are equivalent:^[1]

1. ${\displaystyle f}$ is continuous.
2. ${\displaystyle f}$ is uniformly continuous on${\displaystyle X.}$
3. ${\displaystyle f}$ iscontinuous at some point of${\displaystyle X.}$
4. ${\displaystyle f}$ is continuous at the origin.
   • By definition,${\displaystyle f}$ said to be continuous at the origin if for every open (or closed) ball${\displaystyle B_r}$ of radius${\displaystyle r>0}$ centered at${\displaystyle 0}$ in the codomain${\displaystyle \mathbb{F},}$ there exists someneighborhood${\displaystyle U}$ of the origin in${\displaystyle X}$ such that${\displaystyle f(U)\subseteq B_r.}$
   • If${\displaystyle B_r}$ is a closed ball then the condition${\displaystyle f(U)\subseteq B_r}$ holds if and only if${\displaystyle \underset{u\in U}{sup}|f(u)|\le r.}$
     • It is important that${\displaystyle B_r}$ be a closed ball in thissupremum characterization. Assuming that${\displaystyle B_r}$ is instead an open ball, then is a sufficient butnot necessary condition for${\displaystyle f(U)\subseteq B_r}$ to be true (consider for example when${\displaystyle f=\mathrm{Id}}$ is the identity map on${\displaystyle X=\mathbb{F}}$ and${\displaystyle U=B_r}$), whereas the non-strict inequality${\displaystyle \underset{u\in U}{sup}|f(u)|\le r}$ is instead a necessary butnot sufficient condition for${\displaystyle f(U)\subseteq B_r}$ to be true (consider for example${\displaystyle X=\mathbb{R},f=\mathrm{Id},}$ and the closed neighborhood${\displaystyle U=[-r,r]}$). This is one of several reasons why many definitions involving linear functionals, such aspolar sets for example, involve closed (rather than open) neighborhoods and non-strict${\displaystyle \le }$ (rather than strict) inequalities.
5. ${\displaystyle f}$ isbounded on a neighborhood (of some point). Said differently,${\displaystyle f}$ is alocally bounded at some point of its domain.
   • Explicitly, this means that there exists some neighborhood${\displaystyle U}$ of some point${\displaystyle x\in X}$ such that${\displaystyle f(U)}$ is abounded subset of${\displaystyle \mathbb{F};}$^[2] that is, such that This supremum over the neighborhood${\displaystyle U}$ is equal to${\displaystyle 0}$ if and only if${\displaystyle f=0.}$
   • Importantly, a linear functional being "bounded on a neighborhood" is in generalnot equivalent to being a "bounded linear functional" because (as described above) it is possible for a linear map to bebounded butnot continuous. However, continuity andboundedness are equivalent if the domain is anormed orseminormed space; that is, for a linear functional on a normed space, being "bounded" is equivalent to being "bounded on a neighborhood".
6. ${\displaystyle f}$ isbounded on a neighborhood of the origin. Said differently,${\displaystyle f}$ is alocally bounded at the origin.
   • The equality${\displaystyle \underset{x\in sU}{sup}|f(x)|=|s|\underset{u\in U}{sup}|f(u)|}$ holds for all scalars${\displaystyle s}$ and when${\displaystyle s\ne 0}$ then${\displaystyle sU}$ will be neighborhood of the origin. So in particular, if$R:={\displaystyle \underset{u\in U}{sup}|f(u)|}$ is a positive real number then for every positive real${\displaystyle r>0,}$ the set${\displaystyle N_r:=\frac{r}{R}U}$ is a neighborhood of the origin and${\displaystyle {\displaystyle \underset{n\in N_r}{sup}|f(n)|=r.}}$ Using${\displaystyle r:=1}$ proves the next statement when${\displaystyle R\ne 0.}$
7. There exists some neighborhood${\displaystyle U}$ of the origin such that${\displaystyle \underset{u\in U}{sup}|f(u)|\le 1}$
   • This inequality holds if and only if${\displaystyle \underset{x\in rU}{sup}|f(x)|\le r}$ for every real${\displaystyle r>0,}$ which shows that the positive scalar multiples${\displaystyle \{rU:r>0\}}$ of this single neighborhood${\displaystyle U}$ will satisfy the definition ofcontinuity at the origin given in (4) above.
   • By definition of the set${\displaystyle U^{\circ },}$ which is called the(absolute) polar of${\displaystyle U,}$ the inequality${\displaystyle \underset{u\in U}{sup}|f(u)|\le 1}$ holds if and only if${\displaystyle f\in U^{\circ }.}$ Polar sets, and so also this particular inequality, play important roles induality theory.
8. ${\displaystyle f}$ is alocally bounded at every point of its domain.
9. The kernel of${\displaystyle f}$ is closed in${\displaystyle X.}$^[2]
10. Either${\displaystyle f=0}$ or else the kernel of${\displaystyle f}$ isnot dense in${\displaystyle X.}$^[2]
11. There exists a continuous seminorm${\displaystyle p}$ on${\displaystyle X}$ such that${\displaystyle |f|\le p.}$
    • In particular,${\displaystyle f}$ is continuous if and only if the seminorm${\displaystyle p:=|f|}$ is a continuous.
12. The graph of${\displaystyle f}$ is closed.^[9]
13. ${\displaystyle \mathrm{Re}f}$ is continuous, where${\displaystyle \mathrm{Re}f}$ denotes thereal part of${\displaystyle f.}$

If${\displaystyle X}$ and${\displaystyle Y}$ are complex vector spaces then this list may be extended to include:

14. The imaginary part${\displaystyle \mathrm{Im}f}$ of${\displaystyle f}$ is continuous.

If the domain${\displaystyle X}$ is asequential space then this list may be extended to include:

15. ${\displaystyle f}$ issequentially continuous at some (or equivalently, at every) point of its domain.^[2]

If the domain${\displaystyle X}$ ismetrizable or pseudometrizable (for example, aFréchet space or anormed space) then this list may be extended to include:

16. ${\displaystyle f}$ is abounded linear operator (that is, it maps bounded subsets of its domain to bounded subsets of its codomain).^[2]

If the domain${\displaystyle X}$ is abornological space (for example, apseudometrizable TVS) and${\displaystyle Y}$ islocally convex then this list may be extended to include:

17. ${\displaystyle f}$ is abounded linear operator.^[2]
18. ${\displaystyle f}$ issequentially continuous at some (or equivalently, at every) point of its domain.^[10]
19. ${\displaystyle f}$ is sequentially continuous at the origin.

and if in addition${\displaystyle X}$ is a vector space over thereal numbers (which in particular, implies that${\displaystyle f}$ is real-valued) then this list may be extended to include:

20. There exists a continuous seminorm${\displaystyle p}$ on${\displaystyle X}$ such that${\displaystyle f\le p.}$^[1]
21. For some real${\displaystyle r,}$ the half-space${\displaystyle \{x\in X:f(x)\le r\}}$ is closed.
22. For any real${\displaystyle r,}$ the half-space${\displaystyle \{x\in X:f(x)\le r\}}$ is closed.^[11]

If${\displaystyle X}$ is complex then either all three of${\displaystyle f,}$${\displaystyle \mathrm{Re}f,}$ and${\displaystyle \mathrm{Im}f}$ arecontinuous (respectively,bounded), or else all three arediscontinuous (respectively, unbounded).

Every linear map whose domain is a finite-dimensional Hausdorfftopological vector space (TVS) is continuous. This is not true if the finite-dimensional TVS is not Hausdorff.

Every (constant) map${\displaystyle X\to Y}$ between TVS that is identically equal to zero is a linear map that is continuous, bounded, and bounded on the neighborhood${\displaystyle X}$ of the origin. In particular, every TVS has a non-emptycontinuous dual space (although it is possible for the constant zero map to be its only continuous linear functional).

Suppose${\displaystyle X}$ is any Hausdorff TVS. Theneverylinear functional on${\displaystyle X}$ is necessarily continuous if and only if every vector subspace of${\displaystyle X}$ is closed.^[12] Every linear functional on${\displaystyle X}$ is necessarily a bounded linear functional if and only if everybounded subset of${\displaystyle X}$ is contained in a finite-dimensional vector subspace.^[13]

Alocally convexmetrizable topological vector space isnormable if and only if every bounded linear functional on it is continuous.

A continuous linear operator mapsbounded sets into bounded sets.

The proof uses the facts that the translation of an open set in a linear topological space is again an open set, and the equality$${\displaystyle F^{-1}(D)+x=F^{-1}(D+F(x))}$$for any subset${\displaystyle D}$ of${\displaystyle Y}$ and any${\displaystyle x\in X,}$ which is true due to theadditivity of${\displaystyle F.}$

If${\displaystyle X}$ is a complexnormed space and${\displaystyle f}$ is a linear functional on${\displaystyle X,}$ then${\displaystyle \Vert f\Vert =\Vert \mathrm{Re}f\Vert }$^[14] (where in particular, one side is infinite if and only if the other side is infinite).

Every non-trivial continuous linear functional on a TVS${\displaystyle X}$ is anopen map.^[1] If${\displaystyle f}$ is a linear functional on a real vector space${\displaystyle X}$ and if${\displaystyle p}$ is a seminorm on${\displaystyle X,}$ then${\displaystyle |f|\le p}$ if and only if${\displaystyle f\le p.}$^[1]

If${\displaystyle f:X\to \mathbb{F}}$ is a linear functional and${\displaystyle U\subseteq X}$ is a non-empty subset, then by defining the sets$${\displaystyle f(U):=\{f(u):u\in U\}\text{ and }|f(U)|:=\{|f(u)|:u\in U\},}$$the supremum${\displaystyle \underset{u\in U}{sup}|f(u)|}$ can be written more succinctly as${\displaystyle sup|f(U)|}$ because$${\displaystyle sup|f(U)|=sup\{|f(u)|:u\in U\}=\underset{u\in U}{sup}|f(u)|.}$$If${\displaystyle s}$ is a scalar then$${\displaystyle sup|f(sU)|=|s|sup|f(U)|}$$so that if${\displaystyle r>0}$ is a real number and${\displaystyle B_{\le r}:=\{c\in \mathbb{F}:|c|\le r\}}$ is the closed ball of radius${\displaystyle r}$ centered at the origin then the following are equivalent:

1. $f(U)\subseteq B_{\le 1}$
2. $sup|f(U)|\le 1$
3. $sup|f(rU)|\le r$
4. $f(rU)\subseteq B_{\le r}.$

• Bounded linear operator – Kind of linear transformationPages displaying short descriptions of redirect targets
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