About this blog post

People call the Banach-Steinhaus theorem the first of the big three, which sits at the foundation of linear functional analysis. None of them can go without the Baire’s category theorem.

This blog post offers the Banach-Steinhaus theorem on different abstract levels. Recall that we have

$\text{TVS} \supset \text{Metrizable TVS} \supset \text{F-space} \supset \text{Fréchet space}\supset\text{Banach space} \supset \text{Hilbert space}$

First, there will be a simple version for Banach spaces, which may be more frequently used, and you will realize why it’s referred to as the uniform boundedness principle. After that, there will be a much more generalized version for TVS. Typically, the metrization of the space will not be considered.

Also, it will be a good chance to get a better view of the first and second space by Baire.

Equicontinuity

For metric spaces, equicontinuity is defined as follows. Let $(X,d_X)$ and $(Y,d_Y)$ be two metric spaces.

Let $\Lambda$ be a collection of functions from $X$ to $Y$. We have three different levels of equicontinuity.
Equicontinuous at a point. For $x_0 \in X$, if for every $\varepsilon>0$, there exists a $\delta>0$ such that $d_Y(Lx_0,Lx)<\varepsilon$ for all $L \in \Lambda$ and $d_X(x_0,x)<\delta$ (that is, the continuity holds for all $L$ in a ball centered at $x_0$ with radius $r$).
Pointwise equicontinuous. $\Lambda$ is equicontinuous at each point of $X$.
Uniformly equicontinuous. For every $\varepsilon>0$, there exists a $\delta>0$ such that $d_Y(Lx,Ly)<\varepsilon$ for all $x \in \Lambda$ and $x,y \in X$ such that $d_X(x,y) < \delta$.

Indeed, if $\Lambda$ contains only one element, namely $L$, then everything goes with the continuity and uniform continuity.

But for Banach-Steinhaus theorem, we need a little more restrictions. In fact, $X$ and $Y$ should be considered Banach spaces, and $\Lambda$ contains linear functions only. In this sense, for $L \in \Lambda$, we have the following three conditions equivalent.

$L$ is bounded.
$L$ is continuous.
$L$ is continuous at one point of $X$.

For topological vector spaces, where only topology and linear structure are taken into consideration, things get different. Since no metrization is considered, we have to state it in the language of topology.

Suppose $X$ and $Y$ are TVS and $\Lambda$ is a collection of linear functions from $X$ to $Y$. $\Lambda$ is equicontinuous if for every neighborhood $N$ of $0$ in $Y$, there corresponds a neighborhood $V$ of $0$ in $X$ such that $L(V) \subset N$ for all $L \in \Lambda$.

Indeed, for TVS, $L \in \Lambda$ has the three conditions equivalent as well. With that being said, equicontinuous collection has the boundedness property in a uniform manner. That’s why the Banach-Steinhaus theorem is always referred to as the uniform boundedness principle.

The Banach-Steinhaus theorem, a sufficient condition for being equicontinuous

Banach space version

Suppose $X$ is a Banach space, $Y$ is a normed linear space, and ${F}$ is a collection of bounded linear transformation of $X$ into $Y$, we have two equivalent statements:
(The Resonance Theorem) If $\sup\limits_{L \in \Lambda}\left\Vert{L}\right\Vert=\infty$, then there exists some $x \in X$ such that $\sup\limits_{L \in {L}}\left\Vert{Lx}\right\Vert=\infty$. (In fact, these $x$ form a dense $G_\delta$.)
(The Uniform Boundedness Principle) If $\sup\limits_{L \in {\Lambda}}\left\Vert{Lx}\right\Vert<\infty$ for all $x \in X$, then we have $ \left\Vert L \right\Vert \leq M$ for all $L \in {\Lambda}$ and some $M<\infty$.
(A summary of 1 and 2) Either there exists an $M<\infty$ such that $\lVert L \rVert \leq M$ for all $L \in {L}$, or $\sup\lVert Lx \rVert = \infty$ for all $x$ belonging to some dense $G_\delta$ in $X$.

Proof

Though it would be easier if we finish the TVS version proof, it’s still a good idea to leave the formal proof without the help of TVS here. The equicontinuity of $\Lambda$ will be shown in the next section.

An elementary proof of the Resonance theorem

First, we offer an elementary proof in which the hardest part is the Cauchy sequence.

(Lemma) For any $x \in X$ and $r >0$, we have
$\sup_{y\in B(x,r)}\lVert Lx \rVert \geq \lVert L \rVert r$
where $B(x,r)={y \in X:\lVert x-y \rVert < r}$.
(Proof of the lemma)
For $t \in X$ we have a simple relation
$\begin{aligned}\max(\lVert{L(x+t)}\rVert,\lVert{L(x-t)}\rVert)&=\frac{1}{2}(\lVert{L(x+t)}\rVert+\lVert{L(x-t)}\rVert)+\frac{1}{2}\left\vert\lVert{L(x+t)}\rVert-\lVert{L(x-t)}\rVert\right\vert \\&\geq \frac{1}{2}(\lVert{L(x+t)}\rVert+\lVert{L(x-t)}\rVert) \\&\geq \frac{1}{2}\lVert{L(2t)}\rVert=\lVert Lt \rVert\end{aligned}$
If we have $t \in B(0,r)$, then $x+t,x-t\in{B(x,r)}$. And the desired inequality follows by taking the supremum over $t \in B(0,r)$. (If you find trouble understanding this, take a look at the definition of $\lVert L \rVert$.)

Suppose now $\sup\limits_{L \in \Lambda}\left\Vert{L}\right\Vert=\infty$. Pick a sequence of linear transformation in $\Lambda$, say $(L_n)_{n=1}^{\infty}$, such that $\lVert L_n \rVert \geq 4^n$. Pick $x_0 \in X$, and for $n \geq 1$, we pick $x_n$ inductively.

Set $r_n=3^{-n}$. With $x_{n-1}$ being picked, $x_n \in B(x_{n-1},r_n)$ is picked in such a way that

$\lVert L_n x_n \rVert \geq \frac{2}{3}\lVert L_n \rVert r_n$

(It’s easy to validate this inequality by reaching a contradiction.) Also, it’s easy to check that $(x_n)_{n=1}^{\infty}$ is Cauchy. Since $X$ is complete, $(x_n)$ converges to some $x \in X$. Further we have

$\begin{aligned}\lVert x-x_n \rVert &\leq \sum_{k=n}^{\infty}\lVert x_k - x_{k+1}\rVert \\&=\frac{1}{2\cdot 3^n}\end{aligned}$

Therefore we have

$\begin{aligned}\lVert L_n x \rVert &=\lVert L_n[x_n-(x_n-x)] \rVert \\&\geq \lVert L_nx_n \rVert - \lVert L_n(x_n-x) \rVert \\&\geq \frac{2}{3}\lVert{L_n}\rVert{3}^{-n}-\lVert{L_n}\rVert\lVert{x_n-x}\rVert\\&\geq \frac{1}{6}\lVert{L_n}\rVert{3}^{-n} \\& \geq \frac{1}{6}\left(\frac{4}{3}\right)^n \to\infty\end{aligned}$

A topology-based proof

The previous proof is easy to understand but it’s not easy to see the topological properties of the set formed by such $x$. Thus we are offering a topology-based proof which enables us to get a topology view.

Put

$\varphi(x)=\sup_{L \in \Lambda}\lVert Lx \rVert$

and let

$V_n=\{x:\varphi(x)>n\}$

we claim that each $V_n$ is open. Indeed, we have to show that $x \mapsto \lVert Lx \rVert$ is continuous. It suffice to show that $\lVert\cdot\rVert$ defined in $Y$ is continuous. This follows immediately from triangle inequality since for $x,y \in Y$ we have

$\lVert x \rVert \leq \lVert x-y \rVert + \lVert y \rVert$

which implies

$\lVert x \rVert - \lVert y \rVert \leq \lVert x-y \rVert$

by interchanging $x$ and $y$, we get

$|\lVert x \rVert - \lVert y \rVert | \leq \lVert x-y \rVert$

Thus $x \mapsto \lVert Lx \rVert$ is continuous since it’s a composition of $\lVert\cdot\rVert$ and $L$. Hence $\varphi$, by the definition, is lower semicontinuous, which forces $V_n$ to be open.

If every $V_n$ is dense in $X$ (consider $\sup\lVert L \rVert=\infty$), then by BCT, $B=\bigcap_{n=1}^{\infty} V_n$ is dense in $X$. Since each $V_n$ is open, $B$ is a dense $G_\delta$. Again by the definition of $B$, we have $\varphi(x)=\infty$ for all $x \in B$.

If one of these sets, namely $V_N$, fails to be dense in $X$, then there exist an $x_0 \in X - V_N$ and an $r>0$ such that for $x \in B(0,r)$ we have $x_0+x \notin V_N$, which is equivalent to

$\varphi(x+x_0) \leq N$

considering the definition of $\varphi$, we also have

$\lVert L(x+x_0) \rVert \leq N$

for all $L \in \Lambda$. Since $x=(x+x_0)-x_0$, we also have

$\lVert Lx \rVert \leq \lVert L(x+x_0) \rVert+\lVert Lx_0 \rVert \leq 2N$

Dividing $r$ on two sides, we got

$\lVert L\frac{x}{r}\rVert \leq \frac{2N}{r}$

therefore $\lVert L \rVert \leq M=\frac{2N}{r}$ as is to be shown. Again, this follows from the definition of $\lVert L \rVert$.

Topological vector space version

Suppose $X$ and $Y$ are topological vector spaces, $\Lambda$ is a collection of continuous linear mapping from $X$ into $Y$, and $B$ is the set of all $x \in X$ whose orbits
$\Lambda(x)=\{Lx:L\in\Lambda\}$
are bounded in $Y$. For this $B$, we have:
If $B$ is of the second category, then $\Lambda$ is equicontinuous.

A proof using properties of TVS

Pick balanced neighborhoods $W$ and $U$ of the origin in $Y$ such that $\overline{U} + \overline{U} \subset W$. The balanced neighborhood exists since every neighborhood of $0$ contains a balanced one.

Put

$E=\bigcap_{L \in \Lambda}L^{-1}(\overline{U}).$

If $x \in B$, then $\Lambda(x)$ is bounded, which means that to $U$, there exists some $n$ such that $\Lambda(x) \subset nU$ (Be aware, no metric is introduced, this is the definition of boundedness in topological space). Therefore we have $x \in nE$. Consequently,

$B\subset \bigcup_{n=1}^{\infty}nE.$

If no $nE$ is of the second category, then $B$ is of the first category. Therefore, there exists at least one $n$ such that $nE$ is of the second category. Since $x \mapsto nx$ is a homeomorphism of $X$ onto $X$, $E$ is of the second category as well. But $E$ is closed since each $L$ is continuous. Therefore $E$ has an interior point $x$. In this case, $x-E$ contains a neighborhood $V$ of $0$ in $X$, and

$L(V) \subset Lx-L(E) \subset \overline{U} - \overline{U} \subset W$

This proves that $\Lambda$ is equicontinuous.

Equicontinuity and uniform boundedness

We’ll show that $B=X$. But before that, we need another lemma, which states the connection between equicontinuity and uniform boundedness

(Lemma) Suppose $X$ and $Y$ are TVS, $\Gamma$ is an equicontinuous collection of linear mappings from $X$ to $Y$, and $E$ is a bounded subset of $X$. Then $Y$ has a bounded subset $F$ such that $T(E) \subset F$ for every $T \in \Gamma$.
(Proof of the lemma) We’ll show that, the set
$F=\bigcup_{T \in \Gamma}T(E)$
is bounded. By the definition of equicontinuity, there is an neighborhood $V$ of the origin in $X$ such that $T(V) \subset W$ for all $T \in \Gamma$. Since $E$ is bounded, there exists some $t$ such that $E \subset tV$. For these $t$, by the definition of linear functions, we have
$T(E) \subset T(tV)=tT(V) \subset tW$
Therefore $F \subset tW$. $F$ is bounded.

Thus $\Lambda$ is uniformly bounded. Picking $E=\{x\}$ in the lemma, we also see $\Lambda(x)$ is bounded in $Y$ for every $x$. Thus $B=X$.

A special case when $X$ is a $F$-space or Banach space

$X$ is a $F$-space if its topology $\tau$ is induced by a complete invariant metric $d$. By BCT, $X$ is of the second category. If we already have $B=X$, in which case $B$ is of the second category, then by Banach-Steinhaus theorem, $\Lambda$ is equicontinuous. Formally speaking, we have:

If $\Lambda$ is a collection of continuous linear mappings from an $F$-space $X$ into a topological vector space $Y$, and if the sets
$\Lambda(x)=\{Lx:L\in\Lambda\}$
are bounded in $Y$ for every $x \in X$, then $\Lambda$ is equicontinuous.

Notice that all Banach spaces are $F$-spaces. Therefore we can restate the Uniform Boundedness Principle in Banach space with equicontinuity.

Suppose $X$ is a Banach space, $Y$ is a normed linear space, and ${F}$ is a collection of bounded linear transformation of $X$ into $Y$, we have:
(The Uniform Boundedness Principle) If $\sup\limits_{L \in {\Lambda}}\left\Vert{Lx}\right\Vert<\infty$ for all $x \in X$, then we have $ \left\Vert L \right\Vert \leq M$ for all $L \in {\Lambda}$ and some $M<\infty$. Further, $\Lambda$ is equicontinuous.

Application

Surprisingly enough, the Banach-Steinhaus theorem can be used to do Fourier analysis. An important example follows.

There is a periodic continuous function $f$ on $[0,1]$ such that the Fourier series
$\sum_{n\in\mathbb{Z}}\hat{f}(n)e^{2\pi inx}$
of $f$ diverges at $0$. $\hat{f}(n)$ is defined by
$\hat{f}(n)=\int_{0}^{1}e^{-2\pi inx}f(x)dx$

Notice that $f \mapsto \hat{f}$ is linear, and the divergence of the series at $0$ can be considered by

$\sum_{n\in\mathbb{Z}}\hat{f}(n)e^{2\pi in\cdot0}=\sum_{n\in\mathbb{Z}}\hat{f}(n)$

To invoke Banach-Steinhaus theorem, the family of linear functionals are defined by

$\lambda_N(f)=\sum_{|n| \leq N}\hat{f}(n)$

It can be proved that

$\lVert \lambda_N \rVert=\int_0^1\left\vert\sum_{|n| \leq N}e^{-2\pi inx}\right\vert dx$

which goes to infinity as $N \to \infty$. The existence of such $f$ that

$\sup_{N}|\lambda_N(f)|=+\infty$

follows from the resonance theorem. Further, we also know that these $f$ are in a dense $G_\delta$ subset of the vector space generated by all periodic continuous functions on $[0,1]$.

References / Further readings

arXiv:1005.1585v2
W. Rudin, Real and Complex Analysis
W. Rudin, Functional Analysiss
Applications to Fourier series