Show that operator is continuous and selfadjoint (or not) In this thread
Show compactness/ noncompactness of an operator by approximation
I came to the conclusion that the operator
$$
T\colon\ell^2\to\ell^2, (a_n)_{n\in\mathbb{N}}\mapsto\left(\frac{a_n+a_{n+1}}{2}\right)_{n\in\mathbb{N}}
$$
is not compact.
Now my task is to show whether this operator is (i) continuous (or not) and (ii) selfadjoint (or not).

(i) Concerning continuity I wrote down
$$
\lVert Tx\rVert_{\ell^2}^2=\frac{1}{4}\sum\limits_{n\geq 1}\lvert x_n+x_{n+1}\rvert^2\leq\sum\limits_{n\geq 1}\left(\lvert x_n\rvert+\lvert x_{n+1}\rvert\right)^2\leq\sum\limits_{n\geq 1}(2\lvert x_n\rvert)^2=4\lVert x\rVert_{\ell^2}^2,
$$
so
$$
\lVert Tx\rVert_{\ell^2}\leq 2\lVert x\rVert_{\ell^2}
$$
and so $T$ is continuous, because it is linear and bounded.

(ii) But I do not know exactly how to show if $T$ is selfadjoint. I try to put it more precisely: $\ell^2$ is a Hilbert space with scalar product
$$
\langle (x_n),(y_n)\rangle=\sum\limits_{n\geq 1}x_n\overline{y_n}.
$$
Therefore I started with
$$
\langle Tx,y\rangle=\sum\limits_{n\geq 1}\frac{x_n+x_{n+1}}{2}\overline{y_n}.
$$
Now I have to work with this in order to get an expression with is $\langle x,T^*y\rangle$. But I do not know how to continue.



Could you please say me if (i) is correct and help me with (ii)?

With regards

math12
The operator is not self-adjoint. Consider $x=(1,2,0,\dots,0,\dots)$ and $y=(3,4,0,\dots,0,\dots)$.
It is linear and bounded, and you argued correctly.
This is not an answer for the question but I want to comment on part (i). It seems to me that your estimate 
$$\lVert Tx\rVert_{\ell^2}\leqslant 2\lVert x\rVert_{\ell^2}$$is too rough. My calculation shows that
$$\lVert Tx\rVert_{\ell^2}\leqslant \lVert x\rVert_{\ell^2}$$and it turns out that $\|T\|=1$ since $\|T\| \geqslant \sqrt{1-\frac 3{4n}}$ for any $n$ if you choose the point $(1,\dots,1,0,\dots)$ to test $\|T\|$.