$\sup_{n}\|A_n\|<+\infty$ using Banach - Steinhaus theorem

Question

Let's consider $l^2$ with standard dot product and $A_n:l^2 \rightarrow l^2$ linear and continuous maps. Additionaly assume that:

$$\sum_{n=1}^\infty|<A_n(x), y>| < + \infty,\;\;\;\;\;\;\; \text{for}\; x,y \in l^2$$

I want to show that $\sup_{n}\|A_n\| < + \infty$ but there can be situation that $\sum_{n=1}^\infty\|A_n\| = + \infty$.

My work so far

Since $l^2$ is Banach space we can use Banach - Steinhaus theorem. So to show $\sup\|A_n\| < +\infty$ we jut want to show $\sup\|A_n(x)\|<+\infty$.

Rewriting sum $\sum_{n=1}^\infty|<A_n(x), y>|$ and using Cauchy Schwarz inequality:

$$\sum_{n=1}^\infty|<A_n(x), y>| = |<A_1(x), y>| + |<A_2(x), y>| +....\le$$ $$\le \|A_1(x)\| \|y\| + \|A_2(x)\|\|y\| +... = \|y\| \cdot \sum_{n=1}^\infty\|A_n(x)\|$$

We know that $\|y\| \cdot \sum_{n=1}^\infty\|A_n(x)\| < + \infty$. Since $y \in l^2$ we have that $\sum_{n=1}^\infty\|A_n(x)\| < +\infty$ . Finally:

$$ + \infty > \sum_{n=1}^\infty \|A_n(x)\| \ge \sup\|A_n(x)\|$$

So now by using Banach Steinhaus theorem we have that $\sup\|A_n\| < +\infty$.

Questions

1. Is my justification that $\sup_{n}\|A_n\| < + \infty$ correct?

2. I tried to find some examples that $\sum_{n=1}^\infty \|A_n\| = + \infty$ but I couldn't find any good example. Could you please give me a hint how can I construct such a example?

Answer 1

Answer for the second part: Take $A_nx= \langle x, e_n \rangle e_n$. Then the hypothesis holds by Cauchy-Schwarz inequality but $\|A_n\|=1$ for all $n$.

Hint for the first part: Apply Banach Steinhaus Theorem twice. Since $ \langle A_nx, y \rangle \to 0$ it follows that $(\langle A_nx, y \rangle)$ is bounded for each $y$ and one application of the theorem shows that $\sup_n \|A_n x\| <\infty$ for each $x$. Another applicatin of the theorem finishes the proof.