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let $H$ be a complex Hilbert Space and Let $T \in B(H)$ such that $|\langle Tx,x \rangle| \geq \langle x,x \rangle $ for all $x \in X$. Show that $0 \notin \sigma(T)$. Here $\sigma(T)$ denotes the spectrum of $T$.

Now what I have deduced is that from Cauchy-Schwarz inequality we have $||T(x)|| ||x|| \geq |\langle Tx,x \rangle| \geq ||x||^{2}$.Hence T is bounded below. Therefore T is injective. Hence $0 \notin \sigma_{p}(T)$ is fine, where $\sigma_{p}(T)$ denotes the point spectrum. If T was self adjoint then obviously $0 \notin \sigma(T)$. This is because we have a theorem for bounded self adjoint operator which says that $\lambda \in \rho(T)$ iff $T-\lambda I$ is bounded below, where $\rho(T)$ is the resolvent of $T$, but T is not given to be self-adjoint. So, I don't know how to proceed now. Also range of T is closed. Will that yield anything??

Thanks in advance!!

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As you noted, $$ \|x\|^2 \le \langle Tx,x\rangle \le \|Tx\|\|x\| \\ \|x\| \le \|Tx\|. $$ So $T$ is injective. The range of $T$ is dense because, if $y \perp \mathcal{R}(T)$, then $$ 0=\langle Ty,y\rangle \ge \|y\|^2 \implies y = 0. $$ The range of $T$ is closed because, if $\{ Tx_n \}$ converges to some $y$, the sequence $\{ Tx_n \}$ is Cauchy, which forces $\{ x_n \}$ to be Cauchy and, hence, to converge to some $x$. By the continuity of $T$, $Tx=\lim_n Tx_n =y$, which proves that every $y \in X$ is in the range of $T$. The inequality $\|x\| \le \|Tx\|$, which holds for all $x$, establishes the continuity of the inverse $T^{-1}$. So $0$ is in the resolvent set of $T$.

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  • $\begingroup$ I just missed that dense part... Thanks very much!! $\endgroup$ – Riju May 9 '17 at 12:45

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