Banach-Steinhaus Theorem/Normed Vector Space/Proof 1
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:
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.
$\blacksquare$