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.
 A: 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 $y=(T+\lambda I)x$.
