Is T self adjoint and unitary?

Question

Consider the Hilbert space $H=l^2 $over $\mathbb C$ .If $x\in l^2$,then $\displaystyle{ \sum_{i=1}^\infty}|x_k|^2<\infty$.If $x,y\in l^2$, the inner product is defined by $$\langle x,y\rangle=\displaystyle\sum_{k=1}^{\infty}x_k\bar y_k$$.

Let $T:l^2\to l^2$ is defined by $T(x_1,x_2...)=(x_1,\frac{x_2}{2},\frac{x_3}{3}....)$.

Is T self- adjoint and unitary ?

My thought:

I am not sure which criteria's are necessary to show self-adjointness.I know T is self adjoint if $T=T^\star$

Again I found $\langle T(x),y\rangle=\langle (x_1,\frac{x_2}{2},\frac{x_3}{3}....), (y_1,y_2...)\rangle= \displaystyle{ \sum_{k=1}^\infty}\frac{x_k}{k}.\bar y_k$, I can't further proceed.

For unitary, $TT^\star=T^\star T=I$.

Please someone explain and give a brief idea how to solve this types of problems.

Thanks.

Answer 1

$T$, Self-adjoint?

$\langle x,y\rangle=\displaystyle\sum_{k=1}^{\infty}x_k\bar y_k$

$\langle x,Tx\rangle=\displaystyle\sum_{k=1}^{\infty}x_k\cdot\bar {\frac{x_k}{k}}=\displaystyle\sum_{k=1}^{\infty}\frac{|x_k|^2}{k}$

$\langle Tx,x\rangle=\displaystyle\sum_{k=1}^{\infty} {\frac{\bar x_k}{k}\cdot x_k}=\displaystyle\sum_{k=1}^{\infty}\frac{|x_k|^2}{k}$

Thus T is self-adjoint.

$T$, unitary?

Since $T$ is self-adjoint, if $T$ were unitary then $T$ it is its own inverse so that $T^2=I$

$T^2(x)=T(T(x))=T(x_1,\frac{x_2}{2},\frac{x_3}{3},.....)=(x_1,\frac{x_2}{4},\frac{x_3}{9},......)\ne (x_1,x_2,x_3,.....)$

Thus $T$ is not unitary.