# Uniform boundedness principle

(Redirected from Banach–Steinhaus theorem)

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.

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

## Theorem

Theorem (Uniform Boundedness Principle). Let X be a Banach space and Y a normed vector space. Suppose that F is a collection of continuous linear operators from X to Y. If

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

for all x in X, then

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

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

Proof —

Suppose that for every x in the Banach space X, one has:

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

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

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

The set $X_{n}$  is a closed set and by the assumption,

$\bigcup \nolimits _{n\in \mathbf {N} }X_{n}=X\neq \varnothing .$

By the Baire category theorem for the non-empty complete metric space X, there exists m such that $X_{m}$  has non-empty interior, i.e., there exist $x_{0}\in X_{m}$  and ε > 0 such that

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

Let uX with ǁuǁ ≤ 1 and TF. One has that:

{\begin{aligned}\|T(u)\|_{Y}&=\varepsilon ^{-1}\left\|T\left(x_{0}+\varepsilon u\right)-T(x_{0})\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 u in the unit ball of X and over TF it follows that

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

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

## Corollaries

Corollary. If a sequence of bounded operators (Tn) converges pointwise, that is, the limit of {Tn(x)} exists for all x in X, then these pointwise limits define a bounded operator T.

Note it is not claimed above that Tn converges to T in operator norm, i.e. uniformly on bounded sets. (However, since {Tn} is bounded in operator norm, and the limit operator T is continuous, a standard "3-ε" estimate shows that Tn converges to T uniformly on compact sets.)

Corollary. Any weakly bounded subset S 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*, continuous dual 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**, are bounded. But for every s in S, the norm in the second dual coincides with the norm in Y, by a consequence of the Hahn–Banach theorem.

Let L(XY) denote the continuous operators from X to Y, with the operator norm. If the collection F is unbounded in L(XY), then by the uniform boundedness principle, we have:

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

In fact, R is dense in X. The complement of R in X is the countable union of closed sets ∪Xn. By the argument used in proving the theorem, each Xn is nowhere dense, i.e. the subset ∪Xn is 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 residual sets) are dense. Such reasoning leads to the principle of condensation of singularities, which can be formulated as follows:

Theorem. Let X be a Banach space, {Yn} a sequence of normed vector spaces, and Fn an unbounded family in L(X, Yn). Then the set

$R=\left\{x\in X\ :\ \forall n\in \mathbf {N} :\sup \nolimits _{T\in F_{n}}\|Tx\|_{Y_{n}}=\infty \right\}$

is a residual set, and thus dense in X.

Proof —

The complement of R is the countable union

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

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

## Example: pointwise convergence of Fourier series

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

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

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

and the N-th symmetric partial sum is

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

where DN is the N-th Dirichlet kernel. Fix $x\in \mathbb {T}$  and consider the convergence of {SN(f)(x)}. The functional $\varphi _{N,x}:C(\mathbb {T} )\rightarrow \mathbb {C}$  defined by

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

is bounded. The norm of φN,x, in the dual of $C(\mathbb {T} )$ , is the norm of the signed measure (2π)−1DN(xt) dt, namely

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

One can verify that

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

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

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

## Generalisations

The least restrictive setting for the uniform boundedness principle is a barrelled space where the following generalised version of the theorem holds (Bourbaki 1987, Theorem III.2.1):

Theorem. 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 (even uniformly equicontinuous).

Alternatively, the statement also holds whenever X is a Baire space and Y is a locally convex space.

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

Theorem. Let X be a Fréchet space, Y a normed space, and H a set of continuous linear mappings of X into Y. If for every x in X
$\sup \nolimits _{u\in H}\|u(x)\|<\infty ,$

then the family H is equicontinuous.