# Banach-Steinhaus Theorem

## Theorem

Let $\struct {X, \norm {\,\cdot\,}_X}$ be a Banach space.

Let $\struct {Y, \norm {\,\cdot\,}_Y}$ be a normed vector space.

Let $\family {T_\alpha: X \to Y}_{\alpha \mathop \in A}$ be an $A$-indexed family of bounded linear transformations from $X$ to $Y$.

Suppose that:

$\ds \forall x \in X: \sup_{\alpha \mathop \in A} \norm {T_\alpha x}_Y$ is finite.

Then:

$\ds \sup_{\alpha \mathop \in A} \norm {T_\alpha}$ is finite

where $\norm {T_\alpha}$ denotes the norm of the linear transformation $T_\alpha$.

## Proof

For each $n \in \N$, define:

$F_n = \set {x \in X : \norm {T_\alpha x} \le n \text { for each } \alpha \in A}$

We have:

 $\ds F_n$ $=$ $\ds \bigcap_{\alpha \in A} \set {x \in X : \norm {T x}_Y \le n}$ $\ds$ $=$ $\ds \bigcap_{\alpha \in A} {T_\alpha}^{-1} \sqbrk {\map {\overline {B_Y} } {0, n} }$

where $\map {\overline {B_Y} } {0, n}$ is the closed ball in $\struct {Y, \norm \cdot_Y}$ of radius $n$ centred at $0$.

From Closed Ball is Closed in Normed Vector Space, we have:

$\map {\overline {B_Y} } {0, n}$ is closed.
${T_\alpha}^{-1} \sqbrk {\map {\overline {B_Y} } {0, n} }$ is closed for each $\alpha \in A$.
$F_n$ is closed for each $n \in \N$.

Clearly we have:

$\ds \bigcup_{n \mathop = 1}^\infty F_n \subseteq X$

Recall that by hypothesis, for every $x \in X$, we have:

$\ds \sup_{\alpha \mathop \in A} \norm {T_\alpha x}_Y$ is finite.

So in particular there exists $N \in \N$ such that:

$\ds \norm {T_\alpha x}_Y \le N$

so that:

$x \in F_N$

That is:

$\ds x \in \bigcup_{n \mathop = 1}^\infty F_n$

So:

$\ds X = \bigcup_{n \mathop = 1}^\infty F_n$

Since $\struct {X, \norm \cdot_X}$ is a Banach space, from the Baire Category Theorem we have that:

$\struct {X, \norm \cdot_X}$ is a Baire space.

From Baire Space is Non-Meager, we therefore have:

$X$ is non-meager.

So:

$X$ is not the countable union of nowhere dense subsets of $X$.

So, we must have:

$F_n$ is not nowhere dense for some $n \in \N$.

Fix such an $n$.

Recall that $F_n$ is closed.

From Set is Closed iff Equals Topological Closure, we therefore have:

$F_n^- = F_n$

where $F_n^-$ denotes the closure of $F_n$.

So, since $F_n$ is not nowhere dense, we have that:

there exists a non-empty open $U \subseteq X$ such that $U \subseteq F_n$.

Pick $u \in U$.

Since $U$ is open, there exists $y \in X$ and $r > 0$ such that:

$\map {B_X} {y, r} \subseteq F_n$

where $\map {B_X} {y, r}$ denotes the open ball in $\struct {X, \norm \cdot_X}$ of radius $r$, centre $y$.

Now let $\alpha \in A$ and $x \ne 0$.

Since:

$\ds \norm {\frac {r x} {2 \norm x} }_X = \frac r 2 < r$

we have:

$\ds y + \frac {r x} {2 \norm x_X} \in \map {B_X} {y, r}$

So, we have:

$\ds \norm {\map {T_\alpha} {y + \frac r {2 \norm x_X} } }_Y \le n$

So, from linearity, we have:

$\ds \norm {T_\alpha y + \frac r {2 \norm x_X} T_\alpha x}_Y \le n$

From Reverse Triangle Inequality: Normed Vector Space and positive homogeneity of the norm, we therefore have:

$\ds \size {\norm {T_\alpha y}_Y - \frac r {2 \norm x_X} \norm {T_\alpha x}_Y} \le n$

So that:

$\ds \frac r {2 \norm x_X} \norm {T_\alpha x}_Y \le \norm {T_\alpha y}_Y + n$

Since $y \in F_n$, we have:

$\ds \norm {T_\alpha y}_Y \le n$

So:

$\ds \frac r {2 \norm x_X} \norm {T_\alpha x}_Y \le 2 n$

That is:

$\ds \norm {T_\alpha x}_Y \le \frac {4 n} r \norm x_X$

for all $\alpha \in A$ and $x \ne 0$.

This inequality also clearly holds for $x = 0$.

So:

$\ds \norm {T_\alpha} \le \frac {4 n} r$

for all $\alpha \in A$ from the definition of the norm on bounded linear transformations.

From the Continuum Property, we therefore have:

$\ds \sup_{\alpha \mathop \in A} \norm {T_\alpha}$ exists as a real number

as required.

$\blacksquare$

## Also known as

This theorem is also known as the uniform boundedness principle or uniform bounded principle.

## Source of Name

This entry was named for Stefan Banach and Władysław Hugo Dionizy Steinhaus.

## Historical Note

The Banach-Steinhaus Theorem was first proved by Eduard Helly in around $1912$, some years before Stefan Banach's work, but he failed to obtain recognition for this.