Resolvent operator exists and is bounded

Question

Let T denote a, not necessarily bounded, positive self adjoint operator. Let $\lambda>0$, then $(T+\lambda I)^{-1}$ exists and is a bounded positive self adjoint operator. My attempt is that for $x\in D(T)$ we have $$||(T+\lambda I)x||^{2}=\lambda^{2}||x||^{2}+2\lambda (Tx, x)+||Tx||^{2}\geq \lambda^{2}||x||^{2}$$ so that $T+\lambda I$ is injective. Then I am stuck. Can someone help?

Answer 1

Let $y$ be orthogonal to the range of $T+\lambda I$. Then $\langle y, Tx \rangle + \lambda \langle y, x \rangle =0$ for all $x$. This implies that $y$ is in domain of $T^{*}$ which is same as domain of $T$. Also $\langle Ty, x \rangle +\lambda \langle y, x \rangle =0$ for all $x$. Hence $Ty=-\lambda y$. But then $\langle Ty, y \rangle =-\lambda \langle y, y \rangle $. Since $T$ is positive and $\lambda >0$ it follows that $y=0$. Hence the range of $T+\lambda I$ is dense. From the inequality you have obtained it is easy to see that the range is also closed. Hence $T+\lambda I$ is injective and surjective. Its inverse is bounded with norm not exceeding $\frac 1 {\lambda}$ by your inequality.