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

Question

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.

Answer 1

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 $\lambda\in\sigma(T)$. $\;\;\blacksquare$