Banach-Steinhaus Theorem

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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:

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

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.

From Mapping is Continuous iff Inverse Images of Open Sets are Open: Corollary, we have:

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

From Intersection of Closed Sets is Closed in Normed Vector Space, we have:

$F_n$ 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.


Sources

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