Uniform boundedness principle

Uniform Boundedness Principle — Let ${\displaystyle X}$ be a Banach space, ${\displaystyle Y}$ a normed vector space and ${\displaystyle B(X,Y)}$ the space of all continuous linear operators from ${\displaystyle X}$ into ${\displaystyle Y}$. Suppose that ${\displaystyle F}$ is a collection of continuous linear operators from ${\displaystyle X}$ to ${\displaystyle Y.}$ If

$${\displaystyle \sup _{T\in F}\|T(x)\|_{Y}<\infty \quad {\text{ for all }}x\in X,}$$

then

$${\displaystyle \sup _{\stackrel {T\in F,}{\|x\|\leq 1}}\|T(x)\|_{Y}=\sup _{T\in F}\|T\|_{B(X,Y)}<\infty .}$$

In the case that ${\displaystyle X}$ is not the trivial vector space, then the semi-inequality used in the supremum of the first term in this last chain of equalities (which has ${\displaystyle x}$ range over the closed unit ball) may be replaced by a proper equality (which has ${\displaystyle x}$ range over the closed unit sphere).

The completeness of ${\displaystyle X}$ enables the following short proof, using the Baire category theorem.

Proof

Let X be a Banach space. Suppose that for every ${\displaystyle x\in X,}$

$${\displaystyle \sup _{T\in F}\|T(x)\|_{Y}<\infty .}$$

For every integer ${\displaystyle n\in \mathbb {N} ,}$ let

$${\displaystyle X_{n}=\left\{x\in X\$$ :\ \sup _{T\in F}\|T(x)\|_{Y}\leq n\right\}.}

Each set ${\displaystyle X_{n}}$ is a closed set and by the assumption,

$${\displaystyle \bigcup _{n\in \mathbb {N} }X_{n}=X\neq \varnothing .}$$

By the Baire category theorem for the non-empty complete metric space ${\displaystyle X,}$ there exists some ${\displaystyle m\in \mathbb {N} }$ such that ${\displaystyle X_{m}}$ has non-empty interior; that is, there exist ${\displaystyle x_{0}\in X_{m}}$ and ${\displaystyle \varepsilon >0}$ such that

$${\displaystyle {\overline {B_{\varepsilon }(x_{0})}}~:=~\left\{x\in X\,:\,\|x-x_{0}\|\leq \varepsilon \right\}~\subseteq ~X_{m}.}$$

Let ${\displaystyle u\in X}$ with ${\displaystyle \|u\|\leq 1}$ and ${\displaystyle T\in F.}$ Then:

$${\displaystyle {\begin{aligned}\|T(u)\|_{Y}&=\varepsilon ^{-1}\left\|T\left(x_{0}+\varepsilon u\right)-T\left(x_{0}\right)\right\|_{Y}&[{\text{ by linearity of }}T]\\&\leq \varepsilon ^{-1}\left(\left\|T(x_{0}+\varepsilon u)\right\|_{Y}+\left\|T(x_{0})\right\|_{Y}\right)\\&\leq \varepsilon ^{-1}(m+m).&[{\text{ since }}\ x_{0}+\varepsilon u,\ x_{0}\in X_{m}]\\\end{aligned}}}$$

Taking the supremum over ${\displaystyle u}$ in the unit ball of ${\displaystyle X}$ and over ${\displaystyle T\in F}$ it follows that

$${\displaystyle \sup _{T\in F}\|T\|_{B(X,Y)}~\leq ~2\varepsilon ^{-1}m~<~\infty .}$$

There are also simple proofs not using the Baire theorem (Sokal 2011).

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

Indeed, the elements of ${\displaystyle S}$ define a pointwise bounded family of continuous linear forms on the Banach space ${\displaystyle X:=Y',}$ which is the continuous dual space of ${\displaystyle Y.}$ By the uniform boundedness principle, the norms of elements of ${\displaystyle S,}$ as functionals on ${\displaystyle X,}$ that is, norms in the second dual ${\displaystyle Y'',}$ are bounded. But for every ${\displaystyle s\in S,}$ the norm in the second dual coincides with the norm in ${\displaystyle Y,}$ by a consequence of the Hahn–Banach theorem.

Let ${\displaystyle L(X,Y)}$ denote the continuous operators from ${\displaystyle X}$ to ${\displaystyle Y,}$ endowed with the operator norm. If the collection ${\displaystyle F}$ is unbounded in ${\displaystyle L(X,Y),}$ then the uniform boundedness principle implies:

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

In fact, ${\displaystyle R}$ is dense in ${\displaystyle X.}$ The complement of ${\displaystyle R}$ in ${\displaystyle X}$ is the countable union of closed sets ${\textstyle \bigcup X_{n}.}$ By the argument used in proving the theorem, each ${\displaystyle X_{n}}$ is nowhere dense, i.e. the subset ${\textstyle \bigcup X_{n}}$ is of first category. Therefore ${\displaystyle R}$ is the complement of a subset of first category in a Baire space. By definition of a Baire space, such sets (called comeagre or residual sets) are dense. Such reasoning leads to the principle of condensation of singularities, which can be formulated as follows:

Theorem — Let ${\displaystyle X}$ be a Banach space, ${\displaystyle \left(Y_{n}\right)}$ a sequence of normed vector spaces, and for every ${\displaystyle n,}$ let ${\displaystyle F_{n}}$ an unbounded family in ${\displaystyle L\left(X,Y_{n}\right).}$ Then the set

$${\displaystyle R:=\left\{x\in X\$$ :\ {\text{ for all }}n\in \mathbb {N} ,\sup _{T\in F_{n}}\|Tx\|_{Y_{n}}=\infty \right\}}

is a residual set, and thus dense in ${\displaystyle X.}$

Proof

The complement of ${\displaystyle R}$ is the countable union

$${\displaystyle \bigcup _{n,m}\left\{x\in X\$$ :\ \sup _{T\in F_{n}}\|Tx\|_{Y_{n}}\leq m\right\}}

of sets of first category. Therefore, its residual set ${\displaystyle R}$ is dense.

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

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

$${\displaystyle \sum _{k\in \mathbb {Z} }{\hat {f}}(k)e^{ikx}=\sum _{k\in \mathbb {Z} }{\frac {1}{2\pi }}\left(\int _{0}^{2\pi }f(t)e^{-ikt}dt\right)e^{ikx},}$$

and the N-th symmetric partial sum is

$${\displaystyle S_{N}(f)(x)=\sum _{k=-N}^{N}{\hat {f}}(k)e^{ikx}={\frac {1}{2\pi }}\int _{0}^{2\pi }f(t)D_{N}(x-t)\,dt,}$$

where ${\displaystyle D_{N}}$ is the ${\displaystyle N}$-th Dirichlet kernel. Fix ${\displaystyle x\in \mathbb {T} }$ and consider the convergence of ${\displaystyle \left\{S_{N}(f)(x)\right\}.}$ The functional ${\displaystyle \varphi _{N,x}:C(\mathbb {T} )\to \mathbb {C} }$ defined by

$${\displaystyle \varphi _{N,x}(f)=S_{N}(f)(x),\qquad f\in C(\mathbb {T} ),}$$

is bounded. The norm of ${\displaystyle \varphi _{N,x},}$ in the dual of ${\displaystyle C(\mathbb {T} ),}$ is the norm of the signed measure ${\displaystyle (2(2\pi )^{-1}D_{N}(x-t)dt,}$ namely

$${\displaystyle \left\|\varphi _{N,x}\right\|={\frac {1}{2\pi }}\int _{0}^{2\pi }\left|D_{N}(x-t)\right|\,dt={\frac {1}{2\pi }}\int _{0}^{2\pi }\left|D_{N}(s)\right|\,ds=\left\|D_{N}\right\|_{L^{1}(\mathbb {T} )}.}$$

It can be verified that

$${\displaystyle {\frac {1}{2\pi }}\int _{0}^{2\pi }|D_{N}(t)|\,dt\geq {\frac {1}{2\pi }}\int _{0}^{2\pi }{\frac {\left|\sin \left((N+{\tfrac {1}{2}})t\right)\right|}{t/2}}\,dt\to \infty .}$$

So the collection ${\displaystyle \left(\varphi _{N,x}\right)}$ is unbounded in ${\displaystyle C(\mathbb {T} )^{\ast },}$ the dual of ${\displaystyle C(\mathbb {T} ).}$ Therefore, by the uniform boundedness principle, for any ${\displaystyle x\in \mathbb {T} ,}$ the set of continuous functions whose Fourier series diverges at ${\displaystyle x}$ is dense in ${\displaystyle C(\mathbb {T} ).}$

More can be concluded by applying the principle of condensation of singularities. Let ${\displaystyle \left(x_{m}\right)}$ be a dense sequence in ${\displaystyle \mathbb {T} .}$ Define ${\displaystyle \varphi _{N,x_{m}}}$ 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 ${\displaystyle x_{m}}$ is dense in ${\displaystyle C(\mathbb {T} )}$ (however, the Fourier series of a continuous function ${\displaystyle f}$ converges to ${\displaystyle f(x)}$ for almost every ${\displaystyle x\in \mathbb {T} ,}$ by Carleson's theorem).

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

Uniform boundedness in topological vector spaces

Main article: Uniformly bounded sets (topological vector space)

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

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

which happens if and only if

$${\displaystyle \bigcup _{B\in {\mathcal {B}}}B}$$

is a bounded subset of ${\displaystyle Y}$; if ${\displaystyle Y}$ is a normed space then this happens if and only if there exists some real ${\displaystyle M\geq 0}$ such that ${\textstyle \sup _{\stackrel {b\in B}{B\in {\mathcal {B}}}}\|b\|\leq M.}$ In particular, if ${\displaystyle H}$ is a family of maps from ${\displaystyle X}$ to ${\displaystyle Y}$ and if ${\displaystyle C\subseteq X}$ then the family ${\displaystyle \{h(C):h\in H\}}$ is uniformly bounded in ${\displaystyle Y}$ if and only if there exists some bounded subset ${\displaystyle D}$ of ${\displaystyle Y}$ such that ${\displaystyle h(C)\subseteq D{\text{ for all }}h\in H,}$ which happens if and only if ${\textstyle H(C):=\bigcup _{h\in H}h(C)}$ is a bounded subset of ${\displaystyle Y.}$

Proposition[2] — Let ${\displaystyle H\subseteq L(X,Y)}$ be a set of continuous linear operators between two topological vector spaces ${\displaystyle X}$ and ${\displaystyle Y}$ and let ${\displaystyle C\subseteq X}$ be any bounded subset of ${\displaystyle X.}$ Then the family of sets ${\displaystyle \{h(C):h\in H\}}$ is uniformly bounded in ${\displaystyle Y}$ if any of the following conditions are satisfied:

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

Generalizations involving nonmeager subsets

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

Theorem[2] — Let ${\displaystyle H\subseteq L(X,Y)}$ be a set of continuous linear operators between two topological vector spaces ${\displaystyle X}$ and ${\displaystyle Y}$ (not necessarily Hausdorff or locally convex). For every ${\displaystyle x\in X,}$ denote the orbit of ${\displaystyle x}$ by

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

and let ${\displaystyle B}$ denote the set of all ${\displaystyle x\in X}$ whose orbit ${\displaystyle H(x)}$ is a bounded subset of ${\displaystyle Y.}$ If ${\displaystyle B}$ is of the second category (that is, nonmeager) in ${\displaystyle X}$ then ${\displaystyle B=X}$ and ${\displaystyle H}$ is equicontinuous.

Every proper vector subspace of a TVS ${\displaystyle X}$ has an empty interior in ${\displaystyle X.}$[3] So in particular, every proper vector subspace that is closed is nowhere dense in ${\displaystyle X}$ and thus of the first category (meager) in ${\displaystyle X}$ (and the same is thus also true of all its subsets). Consequently, any vector subspace of a TVS ${\displaystyle X}$ that is of the second category (nonmeager) in ${\displaystyle X}$ must be a dense subset of ${\displaystyle X}$ (since otherwise its closure in ${\displaystyle X}$ would a closed proper vector subspace of ${\displaystyle X}$ and thus of the first category).[3]

Proof[2]

Proof that ${\displaystyle H}$ is equicontinuous:

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

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

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

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

(as will be shown, the set ${\displaystyle C-C}$ is in fact a neighborhood of the origin in ${\displaystyle X}$ because the topological interior of ${\displaystyle C}$ in ${\displaystyle X}$ is not empty). If ${\displaystyle b\in B}$ then ${\displaystyle H(b)}$ being bounded in ${\displaystyle Y}$ implies that there exists some integer ${\displaystyle n\in \mathbb {N} }$ such that ${\displaystyle H(b)\subseteq nV}$ so if ${\displaystyle h\in H,}$ then ${\displaystyle b~\in ~h^{-1}\left(nV\right)~=~nh^{-1}(V).}$ Since ${\displaystyle h\in H}$ was arbitrary,

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

This proves that

$${\displaystyle B~\subseteq ~\bigcup _{n\in \mathbb {N} }nC.}$$

Because ${\displaystyle B}$ is of the second category in ${\displaystyle X,}$ the same must be true of at least one of the sets ${\displaystyle nC}$ for some ${\displaystyle n\in \mathbb {N} .}$ The map ${\displaystyle X\to X}$ defined by ${\textstyle x\mapsto {\frac {1}{n}}x}$ is a (surjective) homeomorphism, so the set ${\textstyle {\frac {1}{n}}(nC)=C}$ is necessarily of the second category in ${\displaystyle X.}$ Because ${\displaystyle C}$ is closed and of the second category in ${\displaystyle X,}$ its topological interior in ${\displaystyle X}$ is not empty. Pick ${\displaystyle c\in \operatorname {Int} _{X}C.}$ Because the map ${\displaystyle X\to X}$ defined by ${\displaystyle x\mapsto c-x}$ is a homeomorphism, the set

$${\displaystyle N~:=~c-\operatorname {Int} _{X}C~=~\operatorname {Int} _{X}(c-C)}$$

is a neighborhood of ${\displaystyle 0=c-c}$ in ${\displaystyle X,}$ which implies that the same is true of its superset ${\displaystyle C-C.}$ And so for every ${\displaystyle h\in H,}$

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

This proves that ${\displaystyle H}$ is equicontinuous. Q.E.D.

---

Proof that ${\displaystyle B=X}$:

Because ${\displaystyle H}$ is equicontinuous, if ${\displaystyle S\subseteq X}$ is bounded in ${\displaystyle X}$ then ${\displaystyle H(S)}$ is uniformly bounded in ${\displaystyle Y.}$ In particular, for any ${\displaystyle x\in X,}$ because ${\displaystyle S:=\{x\}}$ is a bounded subset of ${\displaystyle X,}$ ${\displaystyle H(\{x\})=H(x)}$ is a uniformly bounded subset of ${\displaystyle Y.}$ Thus ${\displaystyle B=X.}$ Q.E.D.

Sequences of continuous linear maps

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

Theorem[4] — Suppose that ${\displaystyle h_{1},h_{2},\ldots }$ is a sequence of continuous linear maps between two topological vector spaces ${\displaystyle X}$ and ${\displaystyle Y.}$

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

Theorem[3] — If ${\displaystyle h_{1},h_{2},\ldots }$ is a sequence of continuous linear maps from an F-space ${\displaystyle X}$ into a Hausdorff topological vector space ${\displaystyle Y}$ such that for every ${\displaystyle x\in X,}$ the limit

$${\displaystyle h(x)~:=~\lim _{n\to \infty }h_{n}(x)}$$

exists in ${\displaystyle Y,}$ then ${\displaystyle h:X\to Y}$ is a continuous linear map and the maps ${\displaystyle h,h_{1},h_{2},\ldots }$ are equicontinuous.

If in addition the domain is a Banach space and the codomain is a normed space then ${\displaystyle \|h\|\leq \liminf _{n\to \infty }\left\|h_{n}\right\|<\infty .}$

Complete metrizable domain

Dieudonné (1970) proves a weaker form of this theorem with Fréchet spaces rather than the usual Banach spaces.

Theorem[2] — Let ${\displaystyle H\subseteq L(X,Y)}$ be a set of continuous linear operators from a complete metrizable topological vector space ${\displaystyle X}$ (such as a Fréchet space or an F-space) into a Hausdorff topological vector space ${\displaystyle Y.}$ If for every ${\displaystyle x\in X,}$ the orbit

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

is a bounded subset of ${\displaystyle Y}$ then ${\displaystyle H}$ is equicontinuous.

So in particular, if ${\displaystyle Y}$ is also a normed space and if

$${\displaystyle \sup _{h\in H}\|h(x)\|<\infty \quad {\text{ for every }}x\in X,}$$

then ${\displaystyle H}$ is equicontinuous.

• Barrelled space – Type of topological vector space
• Ursescu theorem – Generalization of closed graph, open mapping, and uniform boundedness theorem

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