I want to find the operator norm of
\begin{equation} Ax=(x_1+x_2,x_2+x_3,...) \end{equation}
for x$\epsilon \ell_2$
I can show that this operator is bounded, as
\begin{equation} \sum_{i=0}^\infty|x_i+x_{i+1}|^2<6\sum_{i=0}^\infty|x_i|^2 \end{equation}
Now in order to find the norm of this operator I search for $sup(Ax)$ for |x|=1. Thus I have that such a sequence is given by
\begin{equation} (x_n)_k=\delta_{n,k}\rightarrow||Ax_{n,k}||=\sqrt{\sum_{i=0}^\infty|x_i+x_{i+1}|^2}=\sqrt{2} \end{equation}
Does that allows me to conclude that this is the norm of my operator? I think it does, because we search for the supremum. For instance, if we took, as our sequence to be the normalized Basel sequence, then we have a smaller norm.
or another argument could be that:
\begin{equation} \sum_{i=1}^\infty|x_i+x_{i+1}|^2 \leq 2\sum_{i=1}^\infty|x_i|^2+|x_{i+1}|^2=2\sum_{i=1}^\infty|x_i|^2-|x_1|^2. \end{equation}
The last can be maximized for $|x_0|=0$
Thank you in advance.
First, your upper bounds are wrong: $|x_i+x_{i+1}|^2$ is not $\le|x_i|^2+|x_{i+1}|^2.$
Now, let $S$ be the shift operator, defined by $(S(x))_i=x_{i+1}.$ Then, $$\|A\|=\|I+S\|\le\|I\|+\|S\|=2.$$
Conversely, for any positive integer $n,$ the sequence $x:=\frac1{\sqrt n}\sum_{i=1}^n\delta_i$ has norm $1$ and its image $$A(x)=\frac1{\sqrt n}\left(2\sum_{i=1}^{n-1}\delta_i+\delta_n\right)$$ has norm $\sqrt{\frac{4n-3}n},$ so $$\|A\|\ge\sup_{n\ge1}\sqrt{\frac{4n-3}n}=2.$$ Conclusion: $$\|A\|=2.$$