Bounded operators versus bounded sequences

Question

Let $X$ be a Banach space, $Y$ normed, $A_n\in B(X,Y)$. Prove that $(\|A_n\|)$ is bounded if and only if for all $x\in X$ and for all $f\in Y^*$: $(|f(A_n(x))|)$ is bounded.

Any hints on that? $A_n$ is bounded, so there is $M_n\geq 0$ such that $\|A_n(x)\|\leq M_n\|x\|$. From there it follows that $\|A_n\|\leq M_n$ but it is not clear that the sequence $(M_n)$ is bounded? So I need to show that using that $(|f(A_n(x))|)$ is bounded? How do I do that? And what about the other direction?

Answer 1

One direction is very straightforward. If $(\|A_n\|)$ is bounded then $$|f(A_nx)| \leq \|f\| \|A_n\| \|x\|$$ so $(|f(A_nx)|)$ is bounded.

The converse follows from an application of the Uniform Boundedness Theorem.

Uniform Boundedness Theorem: Let $X$ and $Y$ be Banach spaces and let $F \subseteq B(X,Y)$. Suppose that for every $x \in X$ $$\sup_{T \in F} \|Tx\| < \infty.$$ Then $\sup_{F} \|T\| < \infty$.

Applying the U.B.T. to the family of operators $\{T_n=f \circ A_n: X \to \mathbb{R} \mid n \in \mathbb{N}\}$ gives us that $\sup_n \|T_n\| < \infty$. Now, notice that $T_n = A_n^*f$. So, since $\sup_n \|T_n\| < \infty$, we can apply the U.B.T. to the family of maps $\{A_n^*:Y^* \to X^* \mid n \in \mathbb{N}\}$ to conclude that $(\|A_n^*\|)$ is a bounded sequence. The result then follows by recalling that $\|A_n\| = \|A_n^*\|$.