This is a theorem from Rudin Functional Analysis.

$T$ is a self adjoint operator in $H$ (a Hilbert space). $T$ is not necessarily bounded and $\mathscr D(T)$ denotes the domain of $T$. We are to show that -

If $\langle Tx,x \rangle \ge 0$ for all $x\in \mathscr D(T)$ then $\sigma (T)\subset [0,\infty )$.

Infact the converse is also true. It is written in book that the proof works similar to the one for bounded case. I tried it the following way- Let $\lambda\ge 0$ observe, $$0\le\lambda\|x\|^2= \langle \lambda x,x \rangle \le \langle Tx+\lambda x,x\rangle\le \|Tx+\lambda x\|\cdot \|x\|$$ $\implies\lambda\|x\|\le \|(T+\lambda I) x\|$ for all $x\in \mathscr D(T)$.

Therefore $(T+\lambda I)$ is one-one on $\mathscr D(T)$.

I want to show that $T+\lambda I$ has bounded inverse, for that I must have range of $(T+\lambda I)$ is all of $H$. How do I show that ? Please help.


1 Answer 1


Since $T$ is self-adjoint its spectrunm is contained in $\mathbb R$. We only have to show that if $\lambda >0$ then $T+\lambda I$ is invertible. We have $ \langle Tx+\lambda x,x\rangle \geq \lambda \|x\|^{2}$. This gives $\|Tx+\lambda x\| \geq \lambda \|x\|$. From this you conclude that $T+\lambda I$ is one-to-one and its range is closed. [See details below]. To show that the range is dense take any $y$ orthogonal to the range. You get $ \langle Tx+\lambda x, y \rangle=0$ for all $x$. In particular, $ \langle Ty+\lambda y, y \rangle=0$. But LHS is a sum of two non-negative numbers. so each of them must be $0$. In particular $\|y\|^2=0$, so $y=0$.

To show that the range of $T+\lambda I$ is closed let $Tx_n+\lambda x_n \to y$. Then the inequality proved above shows that $\|x_n-x_m\| \to 0$ as $n, m \to \infty$. Hence, $x_n \to x$ for some $x \in H$. Now $Tx_n\to y-\lambda x$. The sequence $(x_,Tx_n)$ in the graph of $T$ converges to $(x,y-\lambda x)$. But the graph of $T$ is closed [Theorem 13.9 in Rudin's book]. It follows that $x \in D(T)$ and $y-\lambda x=Tx$. Thus $y=(T+\lambda I)x$.

  • $\begingroup$ How do I show the range is closed? @KaviRamMurthy Sir $\endgroup$ Mar 21, 2021 at 6:02
  • $\begingroup$ @Noobmathematician I have added some details,. Theorem number in my edition may differ from the one in your copy. $\endgroup$ Mar 21, 2021 at 7:34
  • $\begingroup$ Thank you very much sir this works fine for me . $\endgroup$ Mar 21, 2021 at 8:33
  • $\begingroup$ @KaviRamaMurthy I think your proof contains a flaw. When you say $\langle Ty + \lambda y , y \rangle = 0$ you implicitly assume that $y$ is in the domain of $T$. To fix this, consult Rudin's functional analysis book theorem 13.11 (b). $\endgroup$
    – J. De Ro
    Feb 27 at 12:41

You must log in to answer this question.

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