Prove that T is self-adjoint

Question

Define the linear operator $T:\ell^2 \to\ell^2 $ by $$T(x_1,x_2,x_3,...)= \left(x_1,\frac{1}{2}x_2,\frac{1}{3}x_3,...\right)$$

I need to show that T is self-adjoint.
I know that T is self-adjoint iff $\langle Tx,x\rangle \in \mathbb{R}$, so I tried to check this. I am not sure I did it correctly: $$\langle Tx,x\rangle= \sum_{k=1}^{\infty}\,(Tx)_k\,\overline{x_k}= \left(x_1,\frac{1}{2}x_2,\frac{1}{3}x_3,\dots\right) \cdot (\,\overline{x_1},\overline{x_2},\overline{x_3},\dots\,) \\ =x_1\overline{x_1}+\frac12 x_2\overline{x_2}+\frac13x_3\overline{x_3} +\dots$$ Since $x_n\overline{x_n} \in \mathbb{R}$ for all $n\in \mathbb{N}$ we conclude that $\langle Tx,x\rangle \in \mathbb{R}$, so T is self-adjoint.

Is this correct?

Answer 1

This would more be a comment, but I am not allowed to do so yet.

First, you should check if your operator $T$ is well-defined. But, I assume you did that.

Overall, you are on the right way, but somehow your summation symbols are not used correctly.

For instance, instead of $$\sum_{k=1}^{\infty}(x_1,\frac{1}{2}x_2,\frac{1}{3}x_3,...) \cdot (\bar{x_1},\bar{x_2},\bar{x_3},...),$$ I guess you mean $$\sum_{k=1}^{\infty}\frac{1}{k}x_k\bar{x}_k.$$

I also wonder why you have the factors $\frac{1}{4}$ and $\frac{1}{9}$ in the last equation line. Maybe this has been a typo on your side?

Answer 2

T is self-adjoint if $T=T^*$. In the space $\ell^2$ the standard inner product is $$(x,y)=\sum_{n\in\mathbb{N}}x_n\,\overline{y_n}\,,\quad x,y\in\ell^2$$

Thus, let $x,y\in\ell^2$, then

$$(Tx,y)=\sum_{n\in\mathbb{N}}\frac{x_n}{n}\overline{y_n}= \sum_{n\in\mathbb{N}}x_n\frac{\overline{y_n}}{n}=(x,Ty)$$

Therefore, $T$ is self-adjoint.

Note that I used the fact that $\overline{n}=n$ for all $n\in\mathbb{N}$.