# Spectrum of a positive operator in $B(H)$.

We know that for $$T\in B(H)$$. If $$T$$ is positive, then $$T$$ is self-adjoint and $$\sigma(T)\subset R^{+}$$. Do we have the inverse ie: if $$T$$ is self-adjoint and $$\sigma(T)\subset R^{+}$$. $$T$$ is positive.

where $$T$$ is said to be positive in case $$\langle Tx,x\rangle\geq 0$$ for all $$x\in H$$, $$H$$ is a Hilbert space.

Thank you very much for your help.

• Do you know about the continuous functional calculus for a normal element of a C*-algebra? One can use this to immediately get that a normal element (in any C*-algebra) is positive if and only if it has non-negative spectrum. Sep 14, 2020 at 17:03
• yes, I know a bit of the functional calculus. Sep 14, 2020 at 17:35

Theorem: Let $$T$$ be a bounded self-adjoint linear operator on a Hilbert space. Then $$\lambda=\inf_{\|x\|=1}\langle Tx,x\rangle$$ is in the spectrum of $$T$$. Therefore, if $$\sigma(T)\subset[0,\infty)$$, then $$\inf_{\|x\|=1}\langle Tx,x\rangle \ge 0$$.
Proof: To prove what you want, let $$\lambda=\inf_{\|x\|=1}\langle Tx,x\rangle$$, and note that $$\lambda$$ satisfies $$0 \le \langle (T-\lambda I)x,x\rangle.$$ Therefore $$[x,y] = \langle (T-\lambda I)x,x\rangle$$ has all the properties of an inner product, except possibly strict positivity ($$[x,x] \ge 0$$ is always non-negative.) So the Cauchy-Schwarz inequality holds: $$|[x,y]|^2 \le [x,x][y,y] \\ |\langle (T-\lambda I)x,y\rangle|^2 \le \langle (T-\lambda I)x,x\rangle\langle(T-\lambda I)y,y\rangle$$ Now set $$y=(T-\lambda I)x$$ in order to obtain \begin{align} \|(T-\lambda I)x\|^4 &\le \langle (T-\lambda I)x,x\rangle\cdot\langle(T-\lambda I)^2x,(T-\lambda I)x\rangle \\ &\le \langle(T-\lambda I)x,x\rangle\|(T-\lambda I)\|\|(T-\lambda I)x\|\|(T-\lambda I)x\| \\ \|(T-\lambda I)x\|^2 &\le\|(T-\lambda I)\|\langle(T-\lambda I)x,x\rangle \end{align} If you choose a sequence of unit vectors $$\{ x_n\}$$ so that $$\langle (T-\lambda I)x_n,x_n\rangle\rightarrow 0$$, it follows that $$(T-\lambda I)x_n\rightarrow 0$$, which forces $$\lambda\in\sigma(T)$$. $$\;\;\blacksquare$$