Continuity and norm of operator on $l^2$ I need help with this:
Let $A$ be an operator on $l^2$, defined by
$$A(x)=y, \;x=(x_n)_{n \in N}, \; y = (\alpha_n x_n)_{n \in N}.$$ When is it continuous? Find its norm.
This is what I have done for now:
$$||A|| =  \underset{||x|| \leq 1}{\sup}\sqrt {\sum_{n=1}^{\infty} \alpha_n^2 x_n^2}\leq \underset{||x|| \leq 1}\sup{\sqrt {\underset{n}{\sup}\alpha_n^2\sum_{n=1}^{\infty} x_n^2}} < C,$$
when $(\alpha_n)$ is bounded, that is $\underset{n}{\sup}|\alpha_n|<C<\infty$. So, $A$ is bounded, therefore continuous, because it's linear. Is it enough for the first part?
For the norm, I know that $||A|| = \underset{||x|| = 1}{\sup}\sqrt {\sum_{n=1}^{\infty} \alpha_n^2 x_n^2}$, and $||x||^2=\sum_{n=1}^{\infty} x_n^2$, but I don't know what to do next.
Thank you!
 A: If $(\alpha_n)$ is bounded you have already shown that $A$ is continous. So suppose that $(\alpha_n)$ is not bounded, i.e. $\forall C > 0 : \exists n \in \mathbb{N} : \vert\alpha_n\vert > C$. By definition $\Vert A \Vert = \sup_{\Vert x\Vert =1} \Vert Ax\Vert$. Take any $C>0$ and pick $n\in \mathbb{N}$ with $\vert \alpha_n\vert > C$. Now let $e_n\in \ell^2$ denote the element which is one at its $n$-th entry and zero otherwise. Clearly $\Vert e_n \Vert =1$ and $Ae_n =\alpha_n > C$, so the norm is unbounded.
Suppose that $(\alpha_n)$ is bounded. By a very similar trick we can show $\Vert A\Vert = \Vert (\alpha_n)\Vert_\infty = \sup_n \vert\alpha_n\vert,$ where $\Vert \cdot\Vert_\infty$ denotes the supremum norm in $\mathbb{R}^\mathbb{N}$. We have seen:
$$\Vert Ax\Vert^2 = \sum_i \alpha_i^2 x_i^2 \le \sup_n \vert \alpha_n \vert^2 \Vert x\Vert^2 \implies \Vert Ax\Vert \le \sqrt{\sup_n \vert \alpha_n \vert^2} \Vert x\Vert\le \sqrt{\sup_n \vert \alpha_n \vert^2}, \quad \forall x\in \ell²,\Vert x\Vert\le 1.$$
I claim furthermore $\sqrt{\sup_n \vert \alpha_n \vert^2} = \sup_n \vert \alpha_n\vert$. Clearly for each $k$:
$$
\vert \alpha_k\vert^2 \le \sup_n \vert \alpha_n\vert^2 \implies \sqrt{\vert \alpha_k\vert^2}=\vert \alpha_k \vert \le \sqrt{\sup_n \vert \alpha_n \vert^2} \implies \sqrt{\sup_n \vert \alpha_n \vert^2} \ge \sup_n \vert \alpha_n\vert.
$$
Now pick a subsequence $\alpha_{n_k}$ s.t. $\vert\alpha_{n_k}\vert^2 \to \sup_n \vert \alpha_n\vert^2$ (exists as $(\alpha_n)$ is bounded). Then $$\vert \alpha_{n_k} \vert \to \sqrt{\sup_n \vert \alpha_n \vert^2}$$ and
$$\sup_n \vert \alpha_n \vert \ge \vert \alpha_{n_k}\vert,\forall k\in \mathbb{N} \implies \sup_n \vert \alpha_n\vert \ge \sqrt{\sup_n \vert \alpha_n\vert^2},
$$
so the desired equality holds. We have thus seen $\Vert A\Vert \le \sup_n \vert \alpha_n \vert$. Now pick a subsequence $\alpha_{n_k}$ s.t. $\vert \alpha_{n_k} \vert \to \sup_n \vert \alpha_n\vert$ as $k\to \infty$. Then let $e_{n_k}$ be the $n_k$-th unit vector in $\mathbb{R}^\mathbb{N}$. Clearly $\Vert e_{n_k}\Vert =1$ for each $k$, so
$$
\Vert A\Vert \ge \Vert Ae_{n_k}\Vert = \vert \alpha_{n_k}\vert, \forall k \implies \Vert A\Vert \ge \sup_n \vert \alpha_n\vert
$$
by which the result follows.
A: Here is another piece for a complete solution: If the operator $A$ maps all of $\ell^2$ to $\ell^2$, then $\alpha$ has to be bounded.
Indeed, if $\alpha$ would not be bounded, there would be a subsequence with $|\alpha_{k_n}| \ge n$. Thus, we can define a sequence $x \in \ell^2$ via
$$
x_\ell =
\begin{cases}
1/\alpha_{k_n} & \text{if } \ell = k_n \text{ for some } n \in \mathbb N, \\
0 & \text{else}.
\end{cases}
$$
Then, $x \in \ell^2$, but the sequence $A x$ contains infinitely many $1$s, i.e., $A x \not\in \ell^2$.
