Let $$h^1=\left\{ (u_n)_{n \geq 0}\: \middle| \:u_n \in \mathbb{C}, \,\sum_{n \geq 0}{(n^2+1)|u_n|^2 < \infty} \right\}$$
with the norm
$$\lVert u \rVert_{h^1} = \sum_{n \geq 0}{(n^2+1)|u_n|^2}$$
and let $T:h^1 \rightarrow \ell^2$ (with usual definition of $\ell^2$ space) such that
\begin{split}
(Tu)_0 &= 0 \\
(Tu)_n &= nu_n + u_{n-1}.
\end{split}
I want to find the operator norm $\lVert T \rVert$.
I already showed that $T$ is linear and bounded with
$$ \lVert Tu \rVert_{\ell^2}^2 \leq 2 \lVert u \rVert_{h^1}^2.$$
Moreover I tried to find a sequence in $h^1$ such that $\lVert u \rVert_{h^1} = 1$ and $\lVert Tu \rVert_{\ell^2}^2 = 2$ but it doesn't come to an end.
Best Answer
Define $u$ such that $u_0=1$ and $$ (nu_n + u_{n-1})^2 = 2n^2u_n^2 + 2 u_{n-1}^2 $$ (in the norm estimate one gets $\le$ instead of equality). This is equivalent to $u_n = \frac1n u_{n-1}$ or $u_n=\frac1{n!}$. Then $u\in h^1$ by quotient rule.
This implies $$ \|Tu\|_{l^2} = \sum_{n=1}^\infty (nu_n + u_{n-1})^2 = \sum_{n=1}^\infty 2n^2u_n^2 + 2 u_{n-1}^2 \\ =\sum_{n=0}^\infty 2n^2u_n^2 + 2 u_{n}^2 = 2 \|u\|_{h^1}^2. $$