# $T:\ell^{2}\rightarrow\ell^{2}$ defines a bounded self-adjoint linear operator but has an unbounded self-adjoint inverse.

I'm tryint to show the next statement:

$Tx=(x_{n}/n)_n$ defines a bounded self-adjoint linear operator $T:\ell^{2}\rightarrow\ell^{2}$ which has an unbounded self-adjoint inverse.

I've proved that $T$ is linear and bounded. $T$ is injective. I'd like to prove that $R(T)^{\bot}=\{0\}$ to prove that $R(T)$ is dense in $\ell^{2}.$ So $T^*$ is injective and $(T^{-1})^{*}=(T^{*})^{-1},$ but I'm stuck. Proving that we can conclude that $T^{*}=T.$

How could I prove this?

Any kind of help is thanked in advanced.

• The image contains $e_j$ for all $j$, so contains all finite linear combinations of $e_j$'s and thus are dense. – user99914 Dec 14 '17 at 7:12
• Wow! Many thanks @JohnMa. Very clever observation. – Squird37 Dec 14 '17 at 7:15

## 1 Answer

You can prove that $T$ is selfadjoint directly: $$\langle T^*x,y\rangle=\langle x,Ty\rangle=\sum_n x_n\,\overline{\left(\frac1n\,y_n\right)}=\sum_n\frac1n\,x_n\overline{y_n}=\langle Tx,y\rangle.$$ As this works for all $x,y$ we conclude that $T^*=T$.

Now you have $$R(T)^\perp=N(T^*)=N(T)=\{0\},$$ and so $R(T)$ is dense.

• Thanks @Martin Argerami! Your way is easier. – Squird37 Dec 14 '17 at 16:57