Prove that the spectrum of a bounded, self-adjoint linear operator is real and non-negative iff the operator is positive

Question

Let $H$ be a Hilbert space over the scalar field $\mathbb{C}$. Furthermore let $T$ be a bounded, self-adjoint linear operator on $H$. Now show that $$ T \text{ positive} \iff \sigma(T) \subseteq [0, \infty) $$ I know that if $T$ is positive, the eigenvalues are non-negative, so that proves this theorem for a part of the spectrum of $T$. I am sure I have to apply the spectral theorem in this situation, but I don't find a good approach to this problem.

So how can I approach this proof? I am struggling to understand this theorem

Answer 1

Suppose that $T$ is positive in the sense $\langle Tx,x\rangle\geq0$ for all $x\in H$. Obviously $T$ is self-adjoint because $\langle Tx,x\rangle\in\mathbb{R}$. This shows that $\sigma(T)\subset\mathbb{R}$ (why?) Now if $t<0$ we have $$|\langle (T-tI)x,x\rangle|=|\langle Tx,x\rangle-t\|x\|^2|\geq\langle Tx,x\rangle+|t|\|x\|^2\geq|t|\|x\|^2$$ But by Cauchy-Schwarz $|\langle (T-tI)x,x\rangle|\leq\|(T-tI)x\|\|x\|$, so combining these we get $$|t|\|x\|\leq\|(T-tI)x\|.$$ This shows that $T-tI$ is bounded below and also that $T-tI$ is 1-1. It is an easy result that an operator is invertible if and only if it has dense range and is bounded below. Since the range of $T-tI$ has orthogonal complement equal to $\ker((T-tI^*)=\ker(T-tI)=0$, [i used the well known $Range(S)^\bot=\ker(S^*)$] we have that $T-tI$ is invertible, i.e. $t\not\in\sigma(T)$. As $t$ was arbitrary, this shows $\sigma(T)\subset[0,\infty)$.

Conversely: If $\sigma(T)\subset[0,\infty)$, then $T$ is a positive element in the $C^*$-algebraic sense. So we may write $T=S^*S$ for some operator $S\in B(H)$. Then $\langle Tx,x\rangle=\langle S^*Sx,x\rangle=\langle Sx,Sx\rangle=\|Sx\|^2\geq0$.

Answer 2

The result that you want follows from this theorem, whose proof is a bit tedious, but follows from first principles.

Theorem: Let $T$ be a bounded self-adjoint operator on a Hilbert space $\mathcal{H}$. Then $\lambda=\inf_{\|x\|=1}\langle Tx,x\rangle$ and $\Lambda=\sup_{\|x\|=1}\langle Tx,x\rangle$ are in the spectrum of $T$. Furthermore, $\lambda$ and $\Lambda$ are the minimum and maximum elements of the spectrum of $T$, respectively.

Proof: It is enough to prove the case for $\inf$. For this case, let $\lambda$ be as defined in the theorem, and define $$ [x,y]_{\epsilon}=\langle (T-\lambda I+\epsilon I)x,y\rangle. $$ This is an inner product for all $\epsilon > 0$. So the Cauchy-Schwarz inequality holds: $$ |[x,y]_{\epsilon}|^2 \le [x,x]_{\epsilon}[y,y]_{\epsilon} $$ This continues to hold in the limit as $\epsilon\downarrow 0$, which gives $$ |\langle (T-\lambda I)x,y\rangle|^2 \le \langle (T-\lambda I)x,x\rangle\langle (T-\lambda I)y,y\rangle \\ \le \langle (T-\lambda I)x,x\rangle\|T-\lambda I\|\|y\|^2 $$ Let $y=(T-\lambda I)x$ in order to obtain $$ \|(T-\lambda I)x\|^4 \le \langle (T-\lambda I)x,x\rangle\|T-\lambda I\|\|(T-\lambda I)x\|^2 \\ \|(T-\lambda I)x\|^2 \le \|T-\lambda I\|\langle (T-\lambda I)x,x\rangle. $$ Now let $\{ x_n \}$ be a sequence of unit vectors such that $\langle (T-\lambda)x_n,x_n\rangle\rightarrow 0$ as $n\rightarrow \infty$, which exists because of the way that $\lambda$ is defined. Then, by the last inequality, $(T-\lambda I)x_n\rightarrow 0$ as $n\rightarrow\infty$, which either implies that $\lambda$ is an eigenvalue or an approximate eigenvalue of $T$; in either case $\lambda\in\sigma(T)$.

To see that $\lambda$ is the minimum element of the spectrum of $T$, we argue that $T-\mu I$ is invertible for $\mu < \lambda$. By what has been shown, the following holds for $\epsilon > 0$ and $x\in\mathcal{H}$: $$ \epsilon\|x\|^2 \le \langle (T-\lambda I+\epsilon I)x,x\rangle \le \|(T-\lambda I+\epsilon I)x\|\|x\| \\ \epsilon \|x\| \le \|(T-\lambda I+\epsilon I)x\|. $$ It follows that $T-\lambda I+\epsilon I$ is injective and has a bounded inverse on $\mathcal{R}(T-\lambda I+\epsilon I)$ which must be dense because $$ \mathcal{R}(T-\lambda I+\epsilon I)^{\perp}=\mathcal{N}(T-\lambda I+\epsilon I)=\{0\}. $$ In fact, the range is closed because $T-\lambda I+\epsilon$ and its inverse are bounded. Therefore $(-\infty,\lambda)\subset\rho(T)$. Similarly $(\Lambda,\infty)\subset\rho(T)$. Both $\lambda$ and $\Lambda$ are in the spectrum of $T$. $\;\;\blacksquare$

Answer 3

The other answers are better in my opinion, but you can indeed use the spectral theorem to show this.

First, if $\sigma(T)\subset [0,\infty)$, then $$ \langle Tx,x\rangle=\int_{[0,\infty)}\lambda\,d\langle E_\lambda x,x\rangle\geq 0. $$ Conversely, if $\sigma(T)\cap (-\infty,0)\neq \emptyset$, then $E((-\infty,0))\neq 0$ (the support of the spectral measure is $\sigma(T)$). Thus, if $x\in\operatorname{ran}E((-\infty,0))$ is non-zero, then $$ \langle Tx,x\rangle=\int_{(-\infty,0)}\lambda \,d\langle E_\lambda x,x\rangle<0. $$