# On positive self adjoint unbounded operator

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.

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 $$y=(T+\lambda I)x$$.
• @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). Feb 27 at 12:41