Corollary of Uniform boundedness principle

Question

Let $X$ be banach space, $Y$ be normed space, $\mathcal A \subset \mathcal B(X, Y)$ be some set of continuous linear operators $X \to Y$. I need to prove that if $$\forall x \in X, g \in Y^* \;\; \sup_{A \in \mathcal A} |g(Ax)| < \infty,$$ then $$\sup_{A \in \mathcal A} ||A|| < \infty.$$

I have managed to use uniform boundedness principle to deduce $$\forall g \in Y^* \sup_{A \in \mathcal A} ||g \circ A|| <\infty.$$

I have no idea how to proceed.

score 5 · Accepted Answer · answered Aug 16 '21 at 12:22

5

Define $T_A: Y^{*} \to X^{*}$ by $T_A (g)=g\circ A$. For each $g$, $(T_Ag)_{A\ \in \mathcal A}$ is norm bounded (from what you have already observed). By Uniform Boundeness Principle we get $\sup_{A \in \mathcal A, \|g\|\leq 1} \|g\circ A\| <\infty$. This means $\sup \{\|A\|:A \in \mathcal A\}<\infty$.

answered Aug 16 '21 at 12:22

Kavi Rama Murthy

311,013

Why $\sup ||g \circ A|| < \infty \implies \sup ||A|| < \infty$ ? – dnes Aug 16 '21 at 12:28
$\sup { |g\circ A|: |g|\leq 1}=|A|$. @dnes – Kavi Rama Murthy Aug 16 '21 at 12:31
Could you elaborate on $\sup { ||g \circ A|| : ||g|| \leq 1 } = ||A||$? That doesn't look familiar to me. – dnes Aug 16 '21 at 13:19
@dnes $\sup{ |g\circ A|: |g|\leq 1}=\sup {|g(A(x))|: |g|\leq 1, |x|\leq 1}=\sup |A(x)|: |x|\leq 1}=|A|$. – Kavi Rama Murthy Aug 16 '21 at 23:12

Corollary of Uniform boundedness principle

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