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Consider a bounded, self-adjoint operator $T:H\rightarrow H$, $H$ is a Hilbert space, such that \begin{align*} \langle Tx,x\rangle \geq \beta ||x||^2 \end{align*} where $\beta>0$ is a constant. I know that the null space of this operator is simply $\{0\}$. However, I am having trouble showing that $\mathcal{R}(T)$ is closed. How would I go about doing this?

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$\beta \|x\|^{2} \leq \|Tx\| \|x\|$ so $\beta \|x\| \leq \|Tx\| $. If $Tx_n \to y$ then $T(x_n-x_m) \to 0$ so $\beta \|x_n-x_m\| \to 0$. Hence $\{x_n\}$ is Cauchy. if $x_n \to x$ then $y =\lim Tx_n =Tx$.

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  • $\begingroup$ I see. It seems that your argument is using $T$ is linear. However, I don't think that $T$ is necessarily linear in this problem. Please correct me if I'm wrong. $\endgroup$ – Rebecca Hardenbrook Aug 15 '18 at 13:13
  • $\begingroup$ @RebeccaHardenbrook I have never seen anybody talking about about a bounded self adjoint operator that is not linear. How do you define adjoint of a non-linear operator and prove its existence? $\endgroup$ – Kavi Rama Murthy Aug 15 '18 at 23:21

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