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.