# Banach-Steinhaus Theorem/Normed Vector Space

## 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 1

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.
${T_\alpha}^{-1} \sqbrk {\map {\overline {B_Y} } {0, n} }$ 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.

$\blacksquare$

## Proof 2

From Banach Space is F-Space, $\struct {X, \norm {\, \cdot \,}_X}$ can be considered as an $F$-Space.

From Normed Vector Space is Hausdorff Topological Vector Space, $\struct {Y, \norm {\, \cdot \,}_Y}$ can be considered as a topological vector space.

Let $\Gamma = \set {T_\alpha : \alpha \in A}$ and:

$\map \Gamma x = \set {T_\alpha x : \alpha \in A}$

for each $x \in X$.

By hypothesis, we have:

$\ds \sup_{\alpha \in A} \norm {T_\alpha x}_Y < \infty$ for each $x \in X$.

From Characterization of von Neumann-Boundedness in Normed Vector Space, we have that $\map \Gamma x$ is von Neumann-bounded in $\struct {Y, \norm {\, \cdot \,}_Y}$ for each $x \in X$.

From Banach-Steinhaus Theorem: $F$-Space, $\Gamma$ is equicontinuous.

So there exists an open neighborhood $U$ of ${\mathbf 0}_X$ such that:

$T_\alpha \sqbrk U \subseteq \map {B_1^Y} { {\mathbf 0}_Y} \subseteq B^-_Y$ for each $\alpha \in A$.

where:

$\map {B_1^Y} { {\mathbf 0}_Y}$ is the open ball of center ${\mathbf 0}_Y$ and radius $1$ in $\struct {Y, \norm {\, \cdot \,}_Y}$
$B_Y^-$ is the closed unit ball of $\struct {Y, \norm {\, \cdot \,}_Y}$.

From the definition of an open set in a normed vector space, there exists $\delta > 0$ such that:

$\map {B_\delta^X} { {\mathbf 0}_X} \subseteq U$

and so:

$T_\alpha \sqbrk {\map {B_\epsilon^X} { {\mathbf 0}_X} } \subseteq \map {B_1^Y} { {\mathbf 0}_Y}$ for each $\alpha \in A$.

We then have:

$\paren {\delta/2} B_X^- \subseteq \map {B_\delta^X} { {\mathbf 0}_X}$

and:

$\map {B_1^Y} { {\mathbf 0}_Y} \subseteq B_Y^-$

where $B_X^-$ denotes the closed unit ball of $\struct {X, \norm {\, \cdot \,}_X}$.

$T_\alpha \sqbrk {B_X^-} \subseteq \paren {2/\delta} B_Y^-$ for each $\alpha \in A$.
$\ds \norm {T_\alpha} \le \frac 2 \delta$ for each $\alpha \in A$.

That is:

$\ds \sup_{\alpha \in A} \norm {T_\alpha} \le \frac 2 \delta$

as required.

$\blacksquare$

## Proof 3

It suffices to show that there exist an $x_0 \in X$ and an $r \in \R_{>0}$ such that:

$\ds K : = \sup_{x \mathop \in \map {B_r} {x_0} } \sup_{\alpha \mathop \in A} \norm {T_\alpha x}_Y$ is finite

where $\map {B_r} {x_0}$ is the open $r$-ball of $x_0$.

Indeed, then we have for all $x \in X \setminus \set 0$:

 $\ds \frac{r}{2} \frac {\norm {T_\alpha x}_Y}{\norm x_X}$ $=$ $\ds \norm {\map {T_\alpha} {x_0 + \frac{r}{2 \norm x_X} x } - T_\alpha x_0 }_Y$ $\ds$ $\le$ $\ds \norm {\map {T_\alpha} {\underbrace {x_0 + \frac{r}{2 \norm x_X} x}_{\in \map {B_r} {x_0} } } }_Y + \norm {\T_\alpha \underbrace {x_0}_{\in \map {B_r} {x_0} } }_Y$ $\ds$ $\le$ $\ds 2 K$

That is:

$\ds \norm {T_\alpha} = \sup_{x \mathop \in X \setminus \set 0} \frac {\norm {T_\alpha x}_Y}{\norm x_X} \le \frac {4 K} r$

$\Box$

Aiming for a contradiction, suppose that for all $x_0 \in X$ and $r \in \R_{>0}$:

$\ds \sup_{x \mathop \in \map {B_r} {x_0} } \sup_{\alpha \mathop \in A} \norm {T_\alpha x}_Y = + \infty$

Then we can choose:

$\map {B_{r_1} } {x_1} \supseteq \map {B_{r_2} } {x_2} \supseteq \cdots$

and:

$\alpha_1, \alpha_2, \ldots \in A$

such that:

$\ds \inf _{x \mathop \in \map {B_{r_k} } {x_k} } \norm {T_{\alpha_k} x}_Y \ge k$

and:

$0 < r_k < \frac 1 k$

Indeed, let $r_0 \in \R_{>0}$ and $x_0 \in X$ be fixed.

Then, for $k = 1, 2, \ldots$, we have:

$\exists x_k \in \map {B_{r_{k-1} } } {x_{k-1} } \exists \alpha_k \in A : \norm {T_{\alpha_k} {x_k} }_Y > k$

Furthermore, as $T_{\alpha_k}$ is continuous, $\exists r_k \in \openint 0 {\frac 1 k}$ such that:

$\map {B_{r_{k-1} } } {x_{k-1} } \supseteq \map {B_{r_k} } {x_k}$

and:

$\ds \inf _{x \mathop \in \map {B_{r_k} } {x_k} } \norm {T_{\alpha_k} x}_Y \ge k$

$\Box$

Then $\sequence {x_k}$ is a Cauchy sequence in $X$, since for each $N \in \N_{>0}$:

$k, l \ge N \implies \norm {x_k - x_l}_X \le \norm {\underbrace {x_k - x_N}_{\in \map {B_{r_N} } {x_N} } }_X + \norm {\underbrace {x_l - x_N}_{\in \map {B_{r_N} } {x_N} } }_X\le 2 \, r_N \le \frac 2 N$

As $X$ is a Banach space, there exists:

$\ds z := \lim_{k \mathop \to +\infty} x_k$

Observe that:

$\ds z \in \bigcap_{k \in \N_{>0} } \map {\overline {B_{r_k} } } {x_k}$

where $\map {\overline {B_{r_k} } } {x_k}$ is the $r_k$-closed ball of $x_k$.

In particular:

$\forall k \in \N_{>0} : \norm {T_{\alpha_k} z}_Y \ge k$

Thus:

$\ds \sup_{\alpha \mathop \in A} \norm {T_\alpha z}_Y = + \infty$

$\blacksquare$

## Also known as

This form of the Banach-Steinhaus Theorem is also known as the Uniform Boundedness Principle.

Some sources give it as the Uniform Bounded Principle, but it is possible that this is a mistake.

## Source of Name

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

## Historical Note

The Banach-Steinhaus Theorem was first proved, in the context of normed vector spaces, by Eduard Helly in around $1912$.

This was some years before Stefan Banach's work, but Helly failed to obtain recognition for this.