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.


1 Answer 1


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 $\sup \{\|A\|:A \in \mathcal A\}<\infty$.

  • $\begingroup$ Why $\sup ||g \circ A|| < \infty \implies \sup ||A|| < \infty$ ? $\endgroup$
    – dnes
    Aug 16, 2021 at 12:28
  • $\begingroup$ $\sup \{ \|g\circ A\|: \|g\|\leq 1\}=\|A\|$. @dnes $\endgroup$ Aug 16, 2021 at 12:31
  • $\begingroup$ Could you elaborate on $\sup \{ ||g \circ A|| : ||g|| \leq 1 \} = ||A||$? That doesn't look familiar to me. $\endgroup$
    – dnes
    Aug 16, 2021 at 13:19
  • $\begingroup$ @dnes $\sup\{ \|g\circ A\|: \|g\|\leq 1\}=\sup \{|g(A(x))|: \|g\|\leq 1, \|x\|\leq 1\}=\sup \|A(x)\|: \|x\|\leq 1\}=\|A\|$. $\endgroup$ Aug 16, 2021 at 23:12

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