Corollary of Uniform boundedness principle

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.

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 $$\sup \{\|A\|:A \in \mathcal A\}<\infty$$.
• Why $\sup ||g \circ A|| < \infty \implies \sup ||A|| < \infty$ ?
• $\sup \{ \|g\circ A\|: \|g\|\leq 1\}=\|A\|$. @dnes Aug 16, 2021 at 12:31
• Could you elaborate on $\sup \{ ||g \circ A|| : ||g|| \leq 1 \} = ||A||$? That doesn't look familiar to me.
• @dnes $\sup\{ \|g\circ A\|: \|g\|\leq 1\}=\sup \{|g(A(x))|: \|g\|\leq 1, \|x\|\leq 1\}=\sup \|A(x)\|: \|x\|\leq 1\}=\|A\|$. Aug 16, 2021 at 23:12