Range of unbounded operator is closed

Question

Suppose I have a densely defined symmetric positive operator $T$. I want to show that the range of $T+I$ is closed if $T$ is closed. I've been able to show that $\| x\|^2 + \| Tx\|^2 \leq \| (T+I)x\|^2$ and understand that we need to consider a convergent sequence $(T+I)x_n \to y$ and show that $y = (T+I)x$, where $x$ is the limit of $x_n$.

The inequality just doesn't come out.

Answer 1

Consider $(x_n)_{n \in \mathbb{N}} \in Dom(T)$ a converging sequence, with limit point $x$, such that $(T+I)x_n$ converges to some $y$. Note that: \begin{equation} \|Tx_n -(y-x)\| \leq \|Tx_n + x_n - y \|+\|x-x_n\| = \|(T+I)x_n -y \| +\|x-x_n\| \end{equation}

So $Tx_n$ converges to $y-x$. Because $T$ is closed, $x \in Dom(T)$ and $Tx = y -x \Rightarrow (T+I)x = y$.

Answer 2

Consider the map $L : \mathcal{G}(T)\rightarrow\mathcal{R}(T+I)$ defined by $$ L(x,Tx) = Tx+x. $$

Your inequality, combined with the triangle inequality gives $$ \|Tx\|^2+ \|x\|^2 \le \|(T+I)x\|^2 \le 2(\|Tx\|^2+\|x\|^2), $$ which proves that $L$ is a bicontinuous bijection. Because $\mathcal{G}(T)$ is complete, then the range of $T+I$ must be complete, which proves that $\mathcal{R}(T+I)$ is closed.