Norm of the operator $T:\ell^2 \to \ell^2$ defined as $(Tx)_1=0, (Tx)_n=-x_n+\alpha x_{n+1}$ Consider the operator $T: \ell^2 \to \ell^2$ defined as
$$\begin{cases}
  (Tx)_1 = 0, \\
  (Tx)_n = -x_n + \alpha x_{n+1}, \quad n\ge 2
\end{cases} $$
where $\alpha \in \mathbb{C}$.
I want to find the norm $\|T\|$.

The best constraint I found is
$$ \sqrt{1+|\alpha|^2} \le \|T\| \le 1+|\alpha|. $$
To do this I considered that


*

*$$\|T\|^2=\sum_{n=2}^\infty |-x_n + \alpha x_{n+1}|^2
\le 2 \sum_{n=2}^\infty (|x_n|^2+|\alpha|^2 |x_n|^2) \le 2(1+|\alpha|^2) \|x\|^2$$
hence $\|T\| \le \sqrt{2(1+|\alpha|^2)}$.
This is actually worst that the bound found below so not very useful.

*$$x=(0,0,1+\alpha,0,...) \Longrightarrow Tx=(1+\alpha)(0,\alpha,-1,0,...)$$
hence $\|T\| \ge \sqrt{1+|\alpha|^2}$. 

*Thinking of $T$ as sum of $T_1$ and $T_2$ defined respectively as
$$\begin{cases}
  (T_1 x)_1 = 0, \\
  (T_1 x)_n = -x_n, \quad n \ge 2,
\end{cases}$$
and
$$\begin{cases}
  (T_2 x)_1 = 0, \\
  (T_2 x)_n = \alpha x_n, \quad n \ge 2,
\end{cases}$$
noting that $\|T_1\|=1$ and $\|T_2\|=|\alpha|$, we conclude from the triangle inequality that
$$\|T\| \le 1 + |\alpha|.$$
This constraint is better than the above as is easily seen in the plot below (where $x=|\alpha|$):

What can I use to obtain the exact result?
 A: The "geometric" sequence $x_1=0, x_2=1$, $x_{n+1}=qx_n, n\ge2$, is in $\ell^2$ iff $|q|<1$. For this choice we get $(y_n)=T(x_n)$ where for all $n\ge2$ we have
$$
y_n=-x_n+\alpha x_{n+1}=x_n(-1+q\alpha).
$$
In other words $(x_n)$ is an eigensequence of $T$ belonging to the eigenvalue $\lambda=\lambda(q,\alpha):=-1+q\alpha$. This implies that we can make $|\lambda|$ arbitrarily close to $1+|\alpha|$, which then has to be the norm.
A: Write $\alpha = re^{i\varphi}$ with $r \geqslant 0$ and $\varphi \in [0,2\pi)$. Let $\lambda = -e^{-i\varphi}$. Then consider
$$x_n = \sum_{k=2}^{n+1} \lambda^k e_k.$$
We have $\lVert x_n\rVert = \sqrt{n}$, and
$$T(x_n) = \sum_{k=2}^n (\alpha \lambda^{k+1} - \lambda^k )e_k - \lambda^{n+1}e_{n+1} = - \sum_{k=2}^n (1+r)\lambda^k e_k - \lambda^{n+1}e_{n+1},$$
whence
$$\lVert T(x_n)\rVert > \sqrt{n-1}(1+r).$$
It follows that
$$\lvert T\rVert \geqslant \frac{\lvert T(x_n)\rVert}{\lVert x_n\rVert} > \sqrt{1-\frac{1}{n}}(1+r)$$
for all $n \geqslant 2$, so $\lVert T\rVert = 1+r = 1 + \lvert \alpha\rvert$.
