Characterizations of continuity

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

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

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

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

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

5. ${\displaystyle F}$ is weakly continuous and its transpose ${\displaystyle {}^{t}F:Y^{\prime }\to X^{\prime }}$ maps equicontinuous subsets of ${\displaystyle Y^{\prime }}$ to equicontinuous subsets of ${\displaystyle X^{\prime }.}$

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

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

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

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

If ${\displaystyle Y}$ is seminormable space (such as a normed 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 both normed or seminormed spaces (with both seminorms denoted by ${\displaystyle \|\cdot \|}$) then this list may be extended to include:

9. for every ${\displaystyle r>0}$ there exists some ${\displaystyle \delta >0}$ such that
   $${\displaystyle {\text{ for all }}x,y\in X,{\text{ if }}\|x-y\|<\delta {\text{ then }}\|Fx-Fy\|<r.}$$

   

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]

Bounded on a neighborhood implies continuous implies bounded

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 necessarily continuous^[2] (even if its domain is not a normed space) and thus also bounded (because a continuous linear operator is always a bounded linear operator).^[6]

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

Guaranteeing converses

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

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

A TVS is said to be locally bounded if there exists a neighborhood that is also a bounded set.^[8] For example, every normed or seminormed 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 is bounded 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 the identity 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 necessarily bounded on a neighborhood, then ${\displaystyle Y}$ must be a locally bounded TVS.^[8] In particular, a linear functional on a arbitrary TVS is continuous if and only if it is bounded on a neighborhood.^[8]

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

Guaranteeing that "bounded" implies "continuous"

A continuous linear operator is always a bounded 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 be bounded but to not be continuous.

A linear map whose domain is pseudometrizable (such as any normed space) is bounded if and only if it is continuous.^[2] The same is true of a linear map from a bornological space into a locally 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 a normed 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.

Characterizing continuous linear functionals

Let ${\displaystyle X}$ be a topological vector space (TVS) over the field ${\displaystyle \mathbb {F} }$ (${\displaystyle X}$ need not be Hausdorff or locally convex) and let ${\displaystyle f:X\to \mathbb {F} }$ be a linear functional on ${\displaystyle X.}$ The following are equivalent:^[1]

1. ${\displaystyle f}$ is continuous.
2. ${\displaystyle f}$ is uniformly continuous on ${\displaystyle X.}$
3. ${\displaystyle f}$ is continuous 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 some neighborhood ${\displaystyle U}$ 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 ${\displaystyle \sup _{u\in U}|f(u)|\leq r.}$
     • It is important that ${\displaystyle B_{r}}$ be a closed ball in this supremum characterization. Assuming that ${\displaystyle B_{r}}$ is instead an open ball, then ${\displaystyle \sup _{u\in U}|f(u)|<r}$ is a sufficient but not necessary condition for ${\displaystyle f(U)\subseteq B_{r}}$ to be true (consider for example when ${\displaystyle f=\operatorname {Id} }$ is the identity map on ${\displaystyle X=\mathbb {F} }$ and ${\displaystyle U=B_{r}}$), whereas the non-strict inequality ${\displaystyle \sup _{u\in U}|f(u)|\leq r}$ is instead a necessary but not sufficient condition for ${\displaystyle f(U)\subseteq B_{r}}$ to be true (consider for example ${\displaystyle X=\mathbb {R} ,f=\operatorname {Id} ,}$ and the closed neighborhood ${\displaystyle U=[-r,r]}$). This is one of several reasons why many definitions involving linear functionals, such as polar sets for example, involve closed (rather than open) neighborhoods and non-strict ${\displaystyle \,\leq \,}$ (rather than strict${\displaystyle \,<\,}$) inequalities.
5. ${\displaystyle f}$ is bounded on a neighborhood (of some point). Said differently, ${\displaystyle f}$ is a locally 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 a bounded subset of ${\displaystyle \mathbb {F}$ ;} ^[2] that is, such that ${\textstyle \displaystyle \sup _{u\in U}|f(u)|<\infty .}$ 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 general not equivalent to being a "bounded linear functional" because (as described above) it is possible for a linear map to be bounded but not continuous. However, continuity and boundedness are equivalent if the domain is a normed or seminormed space; that is, for a linear functional on a normed space, being "bounded" is equivalent to being "bounded on a neighborhood".
6. ${\displaystyle f}$ is bounded on a neighborhood of the origin. Said differently, ${\displaystyle f}$ is a locally bounded at the origin.
   • The equality ${\displaystyle \sup _{x\in sU}|f(x)|=|s|\sup _{u\in U}|f(u)|}$ holds for all scalars ${\displaystyle s}$ and when ${\displaystyle s\neq 0}$ then ${\displaystyle sU}$ will be neighborhood of the origin. So in particular, if ${\textstyle R:=\displaystyle \sup _{u\in U}|f(u)|}$ is a positive real number then for every positive real ${\displaystyle r>0,}$ the set ${\displaystyle N_{r}:={\tfrac {r}{R}}U}$ is a neighborhood of the origin and ${\displaystyle \displaystyle \sup _{n\in N_{r}}|f(n)|=r.}$ Using ${\displaystyle r:=1}$ proves the next statement when ${\displaystyle R\neq 0.}$
7. There exists some neighborhood ${\displaystyle U}$ of the origin such that ${\displaystyle \sup _{u\in U}|f(u)|\leq 1}$
   • This inequality holds if and only if ${\displaystyle \sup _{x\in rU}|f(x)|\leq r}$ 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 of continuity 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 ${\displaystyle \sup _{u\in U}|f(u)|\leq 1}$ holds if and only if ${\displaystyle f\in U^{\circ }.}$ Polar sets, and so also this particular inequality, play important roles in duality theory.
8. ${\displaystyle f}$ is a locally 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 ${\displaystyle f}$ is not dense in ${\displaystyle X.}$^[2]
11. There exists a continuous seminorm ${\displaystyle p}$ on ${\displaystyle X}$ such that ${\displaystyle |f|\leq p.}$
    • 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. ${\displaystyle \operatorname {Re} f}$ is continuous, where ${\displaystyle \operatorname {Re} f}$ denotes the real 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 ${\displaystyle \operatorname {Im} f}$ of ${\displaystyle f}$ is continuous.

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

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

If the domain ${\displaystyle X}$ is metrizable or pseudometrizable (for example, a Fréchet space or a normed space) then this list may be extended to include:

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

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

17. ${\displaystyle f}$ is a bounded linear operator.^[2]
18. ${\displaystyle f}$ is sequentially 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 the real 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 ${\displaystyle f\leq p.}$^[1]
21. For some real ${\displaystyle r,}$ the half-space ${\displaystyle \{x\in X:f(x)\leq r\}}$ is closed.
22. For any real ${\displaystyle r,}$ the half-space ${\displaystyle \{x\in X:f(x)\leq r\}}$ is closed.^[11]

If ${\displaystyle X}$ is complex then either all three of ${\displaystyle f,}$ ${\displaystyle \operatorname {Re} f,}$ and ${\displaystyle \operatorname {Im} f}$ are continuous (respectively, bounded), or else all three are discontinuous (respectively, unbounded).

Properties of continuous linear functionals

If ${\displaystyle X}$ is a complex normed space and ${\displaystyle f}$ is a linear functional on ${\displaystyle X,}$ then ${\displaystyle \|f\|=\|\operatorname {Re} f\|}$^[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 an open 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 ${\displaystyle |f|\leq p}$ if and only if ${\displaystyle f\leq p.}$^[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

$${\displaystyle f(U):=\{f(u):u\in U\}\quad {\text{ and }}\quad |f(U)|:=\{|f(u)|:u\in U\},}$$

the supremum ${\displaystyle \,\sup _{u\in U}|f(u)|\,}$ can be written more succinctly as ${\displaystyle \,\sup |f(U)|\,}$ because

$${\displaystyle \sup |f(U)|~=~\sup\{|f(u)|:u\in U\}~=~\sup _{u\in U}|f(u)|.}$$

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 ${\displaystyle B_{\leq r}:=\{c\in \mathbb {F}$ :|c|\leq r\}} is the closed ball of radius ${\displaystyle r}$ centered at the origin then the following are equivalent:

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