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.

  • 4
    $\begingroup$ The image contains $e_j$ for all $j$, so contains all finite linear combinations of $e_j$'s and thus are dense. $\endgroup$ – user99914 Dec 14 '17 at 7:12
  • $\begingroup$ Wow! Many thanks @JohnMa. Very clever observation. $\endgroup$ – Squird37 Dec 14 '17 at 7:15

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.

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

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