uniform boundedness principle

{{math theorem|name=Uniform Boundedness Principle|math_statement=

Let $X$ be a Banach space, $Y$ a normed vector space and $B(X,Y)$ the space of all continuous linear operators from $X$ into $Y$. Suppose that $F$ is a collection of continuous linear operators from $X$ to $Y.$ If, for every $x\; \backslash in\; X$,

then

}}

The first inequality (that is, for all $x$) states that the functionals in $F$ are pointwise bounded while the second states that they are uniformly bounded.

The second supremum always equals

$$\backslash sup\_\{T\; \backslash in\; F\}\; \backslash |T\backslash |\_\{B(X,Y)\}\; =\; \backslash sup\_\{\backslash stackrel\{T\; \backslash in\; F\}\{\backslash |x\backslash |\; \backslash leq\; 1\}\}\; \backslash |T(x)\backslash |\_Y\; =\; \backslash sup\_\{T\; \backslash in\; F\}\; \backslash sup\_\{\backslash |x\backslash |\; \backslash leq\; 1\}\; \backslash |T(x)\backslash |\_Y$$

and if $X$ is not the trivial vector space (or if the supremum is taken over $[0,\; \backslash infty]$ rather than $[-\backslash infty,\; \backslash infty]$) then closed unit ball can be replaced with the unit sphere

$$\backslash sup\_\{T\; \backslash in\; F\}\; \backslash |T\backslash |\_\{B(X,Y)\}\; =\; \backslash sup\_\{\backslash stackrel\{T\; \backslash in\; F,\}\{\backslash |x\backslash |\; =\; 1\}\}\; \backslash |T(x)\backslash |\_Y.$$

The completeness of the Banach space $X$ enables the following short proof, using the Baire category theorem.

{{math proof | proof =

Suppose $X$ is a Banach space and that for every $x\; \backslash in\; X,$

For every integer $n\; \backslash in\; \backslash N,$ let

$$X\_n\; =\; \backslash left\backslash \{x\; \backslash in\; X\; \backslash \; :\; \backslash \; \backslash sup\_\{T\; \backslash in\; F\}\; \backslash |T\; (x)\backslash |\_Y\; \backslash leq\; n\; \backslash right\backslash \}.$$

Each set $X\_n$ is a closed set and by the assumption,

$$\backslash bigcup\_\{n\; \backslash in\; \backslash N\}\; X\_n\; =\; X\; \backslash neq\; \backslash varnothing.$$

By the Baire category theorem for the non-empty complete metric space $X,$ there exists some $m\; \backslash in\; \backslash N$ such that

$X\_m$ has non-empty interior; that is, there exist $x\_0\; \backslash in\; X\_m$ and $\backslash varepsilon\; >\; 0$ such that

$$\backslash overline\{B\_\backslash varepsilon\; (x\_0)\}\; ~:=~\; \backslash left\backslash \{x\; \backslash in\; X\; \backslash ,:\backslash ,\; \backslash |x\; -\; x\_0\backslash |\; \backslash leq\; \backslash varepsilon\; \backslash right\backslash \}\; ~\backslash subseteq~\; X\_m.$$

Let $u\; \backslash in\; X$ with $\backslash |u\backslash |\; \backslash leq\; 1$ and $T\; \backslash in\; F.$

Then:

$$\backslash begin\{align\}$$

\|T(u)\|_Y &= \varepsilon^{-1}\left\|T\left(x_0 + \varepsilon u\right) - T\left(x_0\right)\right\|_Y & [\text{by linearity of } T ] \\

&\leq \varepsilon^{-1}\left(\left\| T (x_0 + \varepsilon u) \right\|_Y + \left\|T(x_0)\right\|_Y \right ) \\

&\leq \varepsilon^{-1}(m + m). & [ \text{since } \ x_0 + \varepsilon u, \ x_0 \in X_m ] \\

\end{align}

Taking the supremum over $u$ in the unit ball of $X$ and over $T\; \backslash in\; F$ it follows that

}}

There are also simple proofs not using the Baire theorem {{harv|Sokal|2011}}.

{{math theorem|name=Corollary | math_statement=

If a sequence of bounded operators $\backslash left(T\_n\backslash right)$ converges pointwise, that is, the limit of $\backslash left(T\_n(x)\backslash right)$ exists for all $x\; \backslash in\; X,$ then these pointwise limits define a bounded linear operator $T.$

}}

The above corollary does {{em|not}} claim that $T\_n$ converges to $T$ in operator norm, that is, uniformly on bounded sets. However, since $\backslash left\backslash \{T\_n\backslash right\backslash \}$ is bounded in operator norm, and the limit operator $T$ is continuous, a standard "$3\backslash varepsilon$" estimate shows that $T\_n$ converges to $T$ uniformly on {{em|compact}} sets.

{{math proof | proof =

Essentially the same as that of the proof that a pointwise convergent sequence of equicontinuous functions on a compact set converges to a continuous function.

By uniform boundedness principle, let $M\; =\; \backslash max\backslash \{\backslash sup\_n\; \backslash |T\_n\backslash |,\; \backslash |T\backslash |\backslash \}$ be a uniform upper bound on the operator norms.

Fix any compact $K\backslash subset\; X$. Then for any $\backslash epsilon\; >\; 0$, finitely cover (use compactness) $K$ by a finite set of open balls $\backslash \{B(x\_i,\; r)\backslash \}\_\{i=1,\; ...,\; N\}$ of radius $r\; =\; \backslash frac\{\backslash epsilon\}\{M\}$

Since $T\_n\; \backslash to\; T$ pointwise on each of $x\_1,\; ...,\; x\_N$, for all large $n$, $\backslash |T\_n(x\_i)\; -\; T(x\_i)\backslash |\backslash leq\; \backslash epsilon$ for all $i=\; 1,...,\; N$.

Then by triangle inequality, we find for all large $n$, $\backslash forall\; x\backslash in\; K,\; \backslash |T\_n(x)\; -\; T(x)\backslash |\backslash leq\; 3\backslash epsilon$.

}}

{{math theorem | name=Corollary | math_statement=

Any weakly bounded subset $S\; \backslash subseteq\; Y$ in a normed space $Y$ is bounded.

}}

Indeed, the elements of $S$ define a pointwise bounded family of continuous linear forms on the Banach space $X\; :=\; Y\text{'},$ which is the continuous dual space of $Y.$

By the uniform boundedness principle, the norms of elements of $S,$ as functionals on $X,$ that is, norms in the second dual $Y\text{'}\text{'},$ are bounded.

But for every $s\; \backslash in\; S,$ the norm in the second dual coincides with the norm in $Y,$ by a consequence of the Hahn–Banach theorem.

Let $L(X,\; Y)$ denote the continuous operators from $X$ to $Y,$ endowed with the operator norm.

If the collection $F$ is unbounded in $L(X,\; Y),$ then the uniform boundedness principle implies:

$$R\; =\; \backslash left\; \backslash \{x\; \backslash in\; X\; \backslash \; :\; \backslash \; \backslash sup\backslash nolimits\_\{T\; \backslash in\; F\}\; \backslash |Tx\backslash |\_Y\; =\; \backslash infty\; \backslash right\backslash \}\; \backslash neq\; \backslash varnothing.$$

In fact, $R$ is dense in $X.$ The complement of $R$ in $X$ is the countable union of closed sets $\backslash bigcup\; X\_n.$

By the argument used in proving the theorem, each $X\_n$ is nowhere dense, i.e. the subset $\backslash bigcup\; X\_n$ is {{em|of first category}}.

Therefore $R$ is the complement of a subset of first category in a Baire space. By definition of a Baire space, such sets (called {{em|comeagre}} or {{em|residual sets}}) are dense.

Such reasoning leads to the {{em|principle of condensation of singularities}}, which can be formulated as follows:

{{math theorem | math_statement=

Let $X$ be a Banach space, $\backslash left(Y\_n\backslash right)$ a sequence of normed vector spaces, and for every $n,$ let $F\_n$ an unbounded family in $L\backslash left(X,\; Y\_n\backslash right).$ Then the set

$$R\; :=\; \backslash left\backslash \{x\; \backslash in\; X\; \backslash \; :\; \backslash \; \backslash text\{\; for\; all\; \}\; n\; \backslash in\; \backslash N\; ,\; \backslash sup\_\{T\; \backslash in\; F\_n\}\; \backslash |Tx\backslash |\_\{Y\_n\}\; =\; \backslash infty\backslash right\backslash \}$$

is a residual set, and thus dense in $X.$

}}

{{math proof| proof =

The complement of $R$ is the countable union

$$\backslash bigcup\_\{n,m\}\; \backslash left\backslash \{x\; \backslash in\; X\; \backslash \; :\; \backslash \; \backslash sup\_\{T\; \backslash in\; F\_n\}\; \backslash |Tx\backslash |\_\{Y\_n\}\; \backslash leq\; m\backslash right\backslash \}$$

of sets of first category. Therefore, its residual set $R$ is dense.

}}

Let $\backslash mathbb\{T\}$ be the circle, and let $C(\backslash mathbb\{T\})$ be the Banach space of continuous functions on $\backslash mathbb\{T\},$ with the uniform norm. Using the uniform boundedness principle, one can show that there exists an element in $C(\backslash mathbb\{T\})$ for which the Fourier series does not converge pointwise.

For $f\; \backslash in\; C(\backslash mathbb\{T\}),$ its Fourier series is defined by

$$\backslash sum\_\{k\; \backslash in\; \backslash Z\}\; \backslash hat\{f\}(k)\; e^\{ikx\}\; =\; \backslash sum\_\{k\; \backslash in\; \backslash Z\}\; \backslash frac\{1\}\{2\backslash pi\}\; \backslash left\; (\backslash int\_0\; ^\{2\; \backslash pi\}\; f(t)\; e^\{-ikt\}\; dt\; \backslash right)\; e^\{ikx\},$$

and the N-th symmetric partial sum is

$$S\_N(f)(x)\; =\; \backslash sum\_\{k=-N\}^N\; \backslash hat\{f\}(k)\; e^\{ikx\}\; =\; \backslash frac\{1\}\{2\; \backslash pi\}\; \backslash int\_0^\{2\; \backslash pi\}\; f(t)\; D\_N(x\; -\; t)\; \backslash ,\; dt,$$

where $D\_N$ is the $N$-th Dirichlet kernel. Fix $x\; \backslash in\; \backslash mathbb\{T\}$ and consider the convergence of $\backslash left\backslash \{S\_N(f)(x)\backslash right\backslash \}.$

The functional $\backslash varphi\_\{N,x\}\; :\; C(\backslash mathbb\{T\})\; \backslash to\; \backslash Complex$ defined by

$$\backslash varphi\_\{N,\; x\}(f)\; =\; S\_N(f)(x),\; \backslash qquad\; f\; \backslash in\; C(\backslash mathbb\{T\}),$$

is bounded.

The norm of $\backslash varphi\_\{N,x\},$ in the dual of $C(\backslash mathbb\{T\}),$ is the norm of the signed measure $(2(2\; \backslash pi)^\{-1\}\; D\_N(x\; -\; t)\; d\; t,$ namely

$$\backslash left\backslash |\backslash varphi\_\{N,x\}\backslash right\backslash |\; =\; \backslash frac\{1\}\{2\; \backslash pi\}\; \backslash int\_0^\{2\; \backslash pi\}\; \backslash left|D\_N(x-t)\backslash right|\; \backslash ,\; dt\; =\; \backslash frac\{1\}\{2\; \backslash pi\}\; \backslash int\_0^\{2\; \backslash pi\}\; \backslash left|D\_N(s)\backslash right|\; \backslash ,\; ds\; =\; \backslash left\backslash |D\_N\backslash right\backslash |\_\{L^1(\backslash mathbb\{T\})\}.$$

It can be verified that

$$\backslash frac\{1\}\{2\; \backslash pi\}\; \backslash int\_0\; ^\{2\; \backslash pi\}\; |D\_N(t)|\; \backslash ,\; dt\; \backslash geq\; \backslash frac\{1\}\{2\backslash pi\}\backslash int\_0^\{2\backslash pi\}\; \backslash frac\{\backslash left|\backslash sin\backslash left(\; (N\; +\; \backslash tfrac\{1\}\{2\})t\; \backslash right)\backslash right|\}\{t/2\}\; \backslash ,\; dt\; \backslash to\; \backslash infty.$$

So the collection $\backslash left(\backslash varphi\_\{N,\; x\}\backslash right)$ is unbounded in $C(\backslash mathbb\{T\})^\{\backslash ast\},$ the dual of $C(\backslash mathbb\{T\}).$

Therefore, by the uniform boundedness principle, for any $x\; \backslash in\; \backslash mathbb\{T\},$ the set of continuous functions whose Fourier series diverges at $x$ is dense in $C(\backslash mathbb\{T\}).$

More can be concluded by applying the principle of condensation of singularities.

Let $\backslash left(x\_m\backslash right)$ be a dense sequence in $\backslash mathbb\{T\}.$

Define $\backslash varphi\_\{N,\; x\_m\}$ in the similar way as above. The principle of condensation of singularities then says that the set of continuous functions whose Fourier series diverges at each $x\_m$ is dense in $C(\backslash mathbb\{T\})$ (however, the Fourier series of a continuous function $f$ converges to $f(x)$ for almost every $x\; \backslash in\; \backslash mathbb\{T\},$ by Carleson's theorem).

In a topological vector space (TVS) $X,$ "bounded subset" refers specifically to the notion of a von Neumann bounded subset. If $X$ happens to also be a normed or seminormed space, say with (semi)norm $\backslash |\backslash cdot\backslash |,$ then a subset $B$ is (von Neumann) bounded if and only if it is {{em|norm bounded}}, which by definition means

{{Main|Barrelled space}}

Attempts to find classes of locally convex topological vector spaces on which the uniform boundedness principle holds eventually led to barrelled spaces.

That is, the least restrictive setting for the uniform boundedness principle is a barrelled space, where the following generalized version of the theorem holds {{harv|Bourbaki|1987|loc=Theorem III.2.1}}:

{{math theorem |math_statement=

Given a barrelled space $X$ and a locally convex space $Y,$ then any family of pointwise bounded continuous linear mappings from $X$ to $Y$ is equicontinuous (and even uniformly equicontinuous).

Alternatively, the statement also holds whenever $X$ is a Baire space and $Y$ is a locally convex space.{{sfn|Shtern|2001}}

}}

{{Main|Uniformly bounded sets (topological vector space)}}

A family $\backslash mathcal\{B\}$ of subsets of a topological vector space $Y$ is said to be {{em|uniformly bounded}} in $Y,$ if there exists some bounded subset $D$ of $Y$ such that

$$B\; \backslash subseteq\; D\; \backslash quad\; \backslash text\{\; for\; every\; \}\; B\; \backslash in\; \backslash mathcal\{B\},$$

which happens if and only if

$$\backslash bigcup\_\{B\; \backslash in\; \backslash mathcal\{B\}\}\; B$$

is a bounded subset of $Y$;

if $Y$ is a normed space then this happens if and only if there exists some real $M\; \backslash geq\; 0$ such that $\backslash sup\_\{\backslash stackrel\{b\; \backslash in\; B\}\{B\; \backslash in\; \backslash mathcal\{B\}\}\}\; \backslash |b\backslash |\; \backslash leq\; M.$

In particular, if $H$ is a family of maps from $X$ to $Y$ and if $C\; \backslash subseteq\; X$ then the family $\backslash \{h(C)\; :\; h\; \backslash in\; H\backslash \}$ is uniformly bounded in $Y$ if and only if there exists some bounded subset $D$ of $Y$ such that $h(C)\; \backslash subseteq\; D\; \backslash text\{\; for\; all\; \}\; h\; \backslash in\; H,$ which happens if and only if $H(C)\; :=\; \backslash bigcup\_\{h\; \backslash in\; H\}\; h(C)$ is a bounded subset of $Y.$

{{math theorem | name = Proposition{{sfn|Rudin|1991|pp=42−47}} | math_statement=

Let $H\; \backslash subseteq\; L(X,\; Y)$ be a set of continuous linear operators between two topological vector spaces $X$ and $Y$ and let $C\; \backslash subseteq\; X$ be any bounded subset of $X.$

Then the family of sets $\backslash \{h(C)\; :\; h\; \backslash in\; H\backslash \}$ is uniformly bounded in $Y$ if any of the following conditions are satisfied:

1. $H$ is equicontinuous.
2. $C$ is a convex compact Hausdorff subspace of $X$ and for every $c\; \backslash in\; C,$ the orbit $H(c)\; :=\; \backslash \{h(c)\; :\; h\; \backslash in\; H\backslash \}$ is a bounded subset of $Y.$

}}

Although the notion of a nonmeager set is used in the following version of the uniform bounded principle, the domain $X$ is {{em|not}} assumed to be a Baire space.

{{math theorem | name = Theorem{{sfn|Rudin|1991|pp=42−47}} | math_statement=

Let $H\; \backslash subseteq\; L(X,\; Y)$ be a set of continuous linear operators between two topological vector spaces $X$ and $Y$ (not necessarily Hausdorff or locally convex).

For every $x\; \backslash in\; X,$ denote the orbit of $x$ by

$$H(x)\; :=\; \backslash \{h(x)\; :\; h\; \backslash in\; H\backslash \}$$

and let $B$ denote the set of all $x\; \backslash in\; X$ whose orbit $H(x)$ is a bounded subset of $Y.$

If $B$ is of the second category (that is, nonmeager) in $X$ then $B\; =\; X$ and $H$ is equicontinuous.

}}

Every proper vector subspace of a TVS $X$ has an empty interior in $X.${{sfn|Rudin|1991|p=46}} So in particular, every proper vector subspace that is closed is nowhere dense in $X$ and thus of the first category (meager) in $X$ (and the same is thus also true of all its subsets).

Consequently, any vector subspace of a TVS $X$ that is of the second category (nonmeager) in $X$ must be a dense subset of $X$ (since otherwise its closure in $X$ would a closed proper vector subspace of $X$ and thus of the first category).{{sfn|Rudin|1991|p=46}}

{{math proof | title = Proof{{sfn|Rudin|1991|pp=42−47}} | proof =

{{em|Proof that $H$ is equicontinuous:}}

Let $W,\; V\; \backslash subseteq\; Y$ be balanced neighborhoods of the origin in $Y$ satisfying $\backslash overline\{V\}\; +\; \backslash overline\{V\}\; \backslash subseteq\; W.$ It must be shown that there exists a neighborhood $N\; \backslash subseteq\; X$ of the origin in $X$ such that $h(N)\; \backslash subseteq\; W$ for every $h\; \backslash in\; H.$

Let

$$C\; ~:=~\; \backslash bigcap\_\{h\; \backslash in\; H\}\; h^\{-1\}\backslash left(\backslash overline\{V\}\backslash right),$$

which is a closed subset of $X$ (because it is an intersection of closed subsets) that for every $h\; \backslash in\; H,$ also satisfies $h(C)\; \backslash subseteq\; \backslash overline\{V\}$ and

$$h(C\; -\; C)\; ~=~\; h(C)\; -\; h(C)\; ~\backslash subseteq~\; \backslash overline\{V\}\; -\; \backslash overline\{V\}\; ~=~\; \backslash overline\{V\}\; +\; \backslash overline\{V\}\; ~\backslash subseteq~\; W$$

(as will be shown, the set $C\; -\; C$ is in fact a neighborhood of the origin in $X$ because the topological interior of $C$ in $X$ is not empty).

If $b\; \backslash in\; B$ then $H(b)$ being bounded in $Y$ implies that there exists some integer $n\; \backslash in\; \backslash N$ such that $H(b)\; \backslash subseteq\; n\; V$ so if $h\; \backslash in\; H,$ then $b\; ~\backslash in~\; h^\{-1\}\backslash left(n\; V\backslash right)\; ~=~\; n\; h^\{-1\}(V).$

Since $h\; \backslash in\; H$ was arbitrary,

$$b\; ~\backslash in~\; \backslash bigcap\_\{h\; \backslash in\; H\}\; nh^\{-1\}(V)\; ~=~\; n\; \backslash bigcap\_\{h\; \backslash in\; H\}\; h^\{-1\}(V)\; ~\backslash subseteq~\; n\; C.$$

This proves that

$$B\; ~\backslash subseteq~\; \backslash bigcup\_\{n\; \backslash in\; \backslash N\}\; n\; C.$$

Because $B$ is of the second category in $X,$ the same must be true of at least one of the sets $n\; C$ for some $n\; \backslash in\; \backslash N.$

The map $X\; \backslash to\; X$ defined by $x\; \backslash mapsto\; \backslash frac\{1\}\{n\}\; x$ is a (surjective) homeomorphism, so the set $\backslash frac\{1\}\{n\}\; (n\; C)\; =\; C$ is necessarily of the second category in $X.$

Because $C$ is closed and of the second category in $X,$ its topological interior in $X$ is not empty.

Pick $c\; \backslash in\; \backslash operatorname\{Int\}\_X\; C.$

Because the map $X\; \backslash to\; X$ defined by $x\; \backslash mapsto\; c\; -\; x$ is a homeomorphism, the set

$$N\; ~:=~\; c\; -\; \backslash operatorname\{Int\}\_X\; C\; ~=~\; \backslash operatorname\{Int\}\_X\; (c\; -\; C)$$

is a neighborhood of $0\; =\; c\; -\; c$ in $X,$ which implies that the same is true of its superset $C\; -\; C.$

And so for every $h\; \backslash in\; H,$

$$h(N)\; ~\backslash subseteq~\; h(c\; -\; C)\; ~=~\; h(c)\; -\; h(C)\; ~\backslash subseteq~\; \backslash overline\{V\}\; -\; \backslash overline\{V\}\; ~\backslash subseteq~\; W.$$

This proves that $H$ is equicontinuous.

Q.E.D.

{{hr|1}}

{{em|Proof that $B\; =\; X$:}}

Because $H$ is equicontinuous, if $S\; \backslash subseteq\; X$ is bounded in $X$ then $H(S)$ is uniformly bounded in $Y.$

In particular, for any $x\; \backslash in\; X,$ because $S\; :=\; \backslash \{x\backslash \}$ is a bounded subset of $X,$ $H(\backslash \{x\backslash \})\; =\; H(x)$ is a uniformly bounded subset of $Y.$ Thus $B\; =\; X.$

Q.E.D.

}}

The following theorem establishes conditions for the pointwise limit of a sequence of continuous linear maps to be itself continuous.

{{math theorem | name = Theorem{{sfn|Rudin|1991|pp=45−46}} | math_statement=

Suppose that $h\_1,\; h\_2,\; \backslash ldots$ is a sequence of continuous linear maps between two topological vector spaces $X$ and $Y.$

1. If the set $C$ of all $x\; \backslash in\; X$ for which $h\_1(x),\; h\_2(x),\; \backslash ldots$ is a Cauchy sequence in $Y$ is of the second category in $X,$ then $C\; =\; X.$
2. If the set $L$ of all $x\; \backslash in\; X$ at which the limit $h(x)\; :=\; \backslash lim\_\{n\; \backslash to\; \backslash infty\}\; h\_n(x)$ exists in $Y$ is of the second category in $X$ and if $Y$ is a complete metrizable topological vector space (such as a Fréchet space or an F-space), then $L\; =\; X$ and $h\; :\; X\; \backslash to\; Y$ is a continuous linear map.

}}

{{math theorem | name = Theorem{{sfn|Rudin|1991|p=46}} | math_statement=

If $h\_1,\; h\_2,\; \backslash ldots$ is a sequence of continuous linear maps from an F-space $X$ into a Hausdorff topological vector space $Y$ such that for every $x\; \backslash in\; X,$ the limit

$$h(x)\; ~:=~\; \backslash lim\_\{n\; \backslash to\; \backslash infty\}\; h\_n(x)$$

exists in $Y,$ then $h\; :\; X\; \backslash to\; Y$ is a continuous linear map and the maps $h,\; h\_1,\; h\_2,\; \backslash ldots$ are equicontinuous.

}}

If in addition the domain is a Banach space and the codomain is a normed space then

{{harvtxt|Dieudonné|1970}} proves a weaker form of this theorem with Fréchet spaces rather than the usual Banach spaces.

{{math theorem| name = Theorem{{sfn|Rudin|1991|pp=42−47}} | math_statement=

Let $H\; \backslash subseteq\; L(X,\; Y)$ be a set of continuous linear operators from a complete metrizable topological vector space $X$ (such as a Fréchet space or an F-space) into a Hausdorff topological vector space $Y.$

If for every $x\; \backslash in\; X,$ the orbit

$$H(x)\; :=\; \backslash \{h(x)\; :\; h\; \backslash in\; H\backslash \}$$

is a bounded subset of $Y$ then $H$ is equicontinuous.

So in particular, if $Y$ is also a normed space and if

then $H$ is equicontinuous.

}}

• {{annotated link|Barrelled space}}
• {{annotated link|Ursescu theorem}}

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{{reflist|group=proof}}

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{{Functional analysis}}

{{Topological vector spaces}}

{{Boundedness and bornology}}

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