0
$\begingroup$

Let $T : H → H$ be a bounded self–adjoint linear operator on a Hilbert space $H$. Suppose the range $R(T)$ is dense in $H$. Prove that $T$ is injective.

$\endgroup$
1
  • $\begingroup$ I want to show that the null space only contains zero, but I am confused in where to go. $\endgroup$ Dec 5, 2019 at 20:01

1 Answer 1

0
$\begingroup$

We have $R(T)^{\perp}=\overline{R(H)}^{\perp}=H^{\perp}=\{0\}$, but $\ker T^{\ast}=(R(T))^{\perp}$.

$\endgroup$
3
  • $\begingroup$ So how do we know that the range is closed? We only know that the range of T is dense. $\endgroup$ Dec 12, 2019 at 18:25
  • $\begingroup$ Okay, just made a mistake, we don't need that. Just keep in mind that $A^{\perp}=\overline{A}^{\perp}$. $\endgroup$
    – user284331
    Dec 12, 2019 at 18:30
  • $\begingroup$ Isn't $R(T)$ always dense for all bounded self-adjoint operators on $H$? $\endgroup$
    – ric.san
    Jun 14, 2021 at 14:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .