# Uniform boundedness principle

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In mathematics, the uniform boundedness principle or Banach–Steinhaus theorem is one of the fundamental results in functional analysis. Together with the Hahn–Banach theorem and the open mapping theorem, it is considered one of the cornerstones of the field. In its basic form, it asserts that for a family of continuous linear operators (and thus bounded operators) whose domain is a Banach space, pointwise boundedness is equivalent to uniform boundedness in operator norm.

## Contents

The theorem was first published in 1927 by Stefan Banach and Hugo Steinhaus, but it was also proven independently by Hans Hahn.

## Theorem

Uniform Boundedness Principle  Let ${\displaystyle X}$ be a Banach space and ${\displaystyle Y}$ a normed vector space. 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

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

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 ).

## Corollaries

Corollary  If a sequence of bounded operators ${\displaystyle \left(T_{n}\right)}$ converges pointwise, that is, the limit of ${\displaystyle \left(T_{n}(x)\right)}$ exists for all ${\displaystyle x\in X,}$ then these pointwise limits define a bounded linear operator ${\displaystyle T.}$

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.

Corollary  Any weakly bounded subset ${\displaystyle S\subseteq Y}$ in a normed space ${\displaystyle Y}$ is bounded.

Indeed, the elements of ${\displaystyle S}$ define a pointwise bounded family of continuous linear forms on the Banach space ${\displaystyle X:=Y^{\prime },}$ 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 ${\displaystyle Y^{\prime \prime },}$ 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 ${\displaystyle \cup X_{n}.}$ By the argument used in proving the theorem, each ${\displaystyle X_{n}}$ is nowhere dense, i.e. the subset ${\displaystyle \cup X_{n}}$ 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 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 ${\displaystyle \left(F_{n}\right)}$ 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 \nolimits _{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 \nolimits _{n,m}\left\{x\in X\$ :\ \sup \nolimits _{T\in F_{n}}\|Tx\|_{Y_{n}}\leq m\right\}}

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

## Example: pointwise convergence of Fourier series

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 ${\displaystyle X_{N}}$ 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).

## Generalizations

### Barrelled spaces

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 ( Bourbaki 1987 , Theorem III.2.1):

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

Alternatively, the statement also holds whenever ${\displaystyle X}$ is a Baire space and ${\displaystyle Y}$ is a locally convex space. [1]

### Uniform boundedness in general topological vector spaces

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 ${\displaystyle \,\sup _{\stackrel {b\in B}{B\in {\mathcal {B}}}}\|b\|\leq M.}$

The below theorem's conclusion that the set ${\displaystyle B}$ is necessarily equal to all to ${\displaystyle X}$ can be deduced, with the help of the first part of the following theorem, from the equicontinuity of ${\displaystyle H}$ and the fact that every single subset of ${\displaystyle X}$ is also a bounded subset.

Theorem [2]   Let ${\displaystyle H\subseteq L(X,Y)}$ be a set of continuous linear operators between two topological vector spaces ${\displaystyle X}$ and ${\displaystyle Y}$ and let ${\displaystyle C\subseteq X}$ be any bounded subset of ${\displaystyle X.}$ If ${\displaystyle H}$ is equicontinuous then the family of sets ${\displaystyle \{h(C):h\in H\}}$ is uniformly bounded in ${\displaystyle Y,}$ [proof 1] meaning that 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 ${\displaystyle \bigcup _{h\in H}h(C)}$ is a bounded subset of ${\displaystyle Y.}$

Also, if ${\displaystyle C}$ is a convex compact Hausdorff subspace and if for every ${\displaystyle c\in C,}$ the orbit ${\displaystyle H(c):=\{h(c):h\in H\}}$ is a bounded subset of ${\displaystyle Y,}$ then the family ${\displaystyle \{h(C):h\in H\}}$ is uniformly bounded in ${\displaystyle Y}$ (for this conclusion, ${\displaystyle H}$ was not assumed to be equicontinuous).

### Generalizations involving nonmeager subsets

In the following version of the theorem, 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.

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 non-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 ${\displaystyle x\mapsto {\frac {1}{n}}x}$ is a (surjective) homeomorphism, so the set ${\displaystyle C={\frac {1}{n}}(nC)}$ 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 non-empty. Pick ${\displaystyle c\in \operatorname {Int} _{X}C.}$ The map ${\displaystyle X\to X}$ defined by ${\displaystyle x\mapsto c-x}$ being a homeomorphism implies that

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

is a neighborhood of ${\displaystyle 0=c-c}$ in ${\displaystyle X}$ (which implies that the same is true of the 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. ${\displaystyle \blacksquare }$

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,}$${\displaystyle H(\{x\})=H(x)}$ is a uniformly bounded subset of ${\displaystyle Y.}$ Thus ${\displaystyle B=X.}$${\displaystyle \blacksquare }$

#### 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) 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   A topological vector space with near minimum requirements for the Banach–Steinhaus theorem to hold.
• Ursescu theorem   Theorem that simultaneously generalizes the closed graph, open mapping, and Banach–Steinhaus theorems

## Notes

1. Let ${\displaystyle V}$ be a neighborhood of the origin in ${\displaystyle Y.}$ Since ${\displaystyle H}$ is equicontinuous, there exists a neighborhood ${\displaystyle U}$ of the origin in ${\displaystyle X}$ such that ${\displaystyle h(U)\subseteq V}$ for every ${\displaystyle h\in H.}$ Because ${\displaystyle C}$ is bounded in ${\displaystyle X,}$ there exists some real ${\displaystyle r>0}$ such that if ${\displaystyle t\geq r}$ then ${\displaystyle C\subseteq tU.}$ So for every ${\displaystyle h\in H}$ and every ${\displaystyle t\geq r,}$${\displaystyle h(C)\subseteq h(tU)=th(U)\subseteq tV,}$ which implies that ${\displaystyle \bigcup _{h\in H}h(C)\subseteq tV.}$ Thus ${\displaystyle \bigcup _{h\in H}h(C)}$ is bounded in ${\displaystyle Y.}$${\displaystyle \blacksquare }$

## Citations

1. Rudin 1991, pp. 42−47.

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