Question about self adjoint operator to be a positive operator

Question

Let me first define the self-adjoint operator in a Hilbert space
$A \in B(H)$ be a self-adjoint operator on a Hilbert space $H$ if $A=A^*$ where $A^*$ is the adjoint of $A$.
and $A \in B(H)$ be a self-adjoint operator the $A$ is a positive operator if $<Ax,x>\geq0$
My question let we have $A$ as a self-adjoint operator then I need to prove that $A$ is a positive operator if and only if every spectral value of $A$ is a non-negative real number.
Let me define the spectral value, let $H$ be a Hilbert space over a field $K$ and $A\in B(H)$ the set $\sigma(A)$={$\lambda\in K$ such that ($A-\lambda I)$is not invertible in $B(H)$} is called the spectrum of $A$ and the elements of $\sigma(A)$ are known as the spectral value of $A$

Answer 1

Theorem: Suppose that $A$ is a bounded self-adjoint linear operator on a complex Hilbert space $X$ such that $\langle Ax,x \rangle \ge 0$ for all $x\in X$. Then the spectrum of $A$ is a subset of the non-negative real axis.

Proof: By standard results, the spectrum $\sigma(A)$ is a subset of $\mathbb{R}$. Because of the positivity of $A$, it follows that, for any real $\lambda > 0$ and $x\in X$, one has $$ \lambda\|x\|^2=\lambda\langle x,x\rangle \le \langle (A+\lambda I)x,x\rangle \le \|(A+\lambda I)x\|\|x\| \\ \lambda \|x\|\le \|(A+\lambda I)x\|. \tag{*} $$ From this it follows that $\mathcal{N}(A+\lambda I)=\{0\}$, and $A+\lambda I$ has a bounded inverse on its range. The range of $A+\lambda I$ is dense for all $\lambda > 0$ because $$ \mathcal{R}(A+\lambda I)^{\perp}=\mathcal{N}(A+\lambda I)=\{0\}. $$ To see that the range $\mathcal{R}(A+\lambda I)$ is closed for $\lambda > 0$, suppose that $\{ x_n \}$ is a sequence such that $(A+\lambda I)x_n$ converges to some $y$ as $n\rightarrow\infty$. Then $\{ (A+\lambda I)x_n \}$ is a Cauchy sequence, which also forces $\{ x_n \}$ to be a Cauchy sequence by inequality $(*)$. So $\lim_n x_n = x$ exists and $$y=\lim_n (A+\lambda I)x_n = (A+\lambda I)x,$$ which proves that the range of $A+\lambda I$ is closed. Therefore, $A+\lambda I$ has a bounded inverse, which proves that $-\lambda\in\rho(A)$ for all $\lambda > 0$. Therefore, $\sigma(A)\subseteq [0,\infty)$. $\;\;\blacksquare$

Answer 2

I'll prove one direction. First I need to define the notion of a bounded below operator. A continuous linear operator $T: H \rightarrow H$ is called bounded below if there exists a positive number $\delta>0$ such that for any $x\in H$ we have $||Tx||\geq||x|| \delta $. You can easily see that a bounded below operator is one-one and has a closed range.

Now assume that for any $h \in H, <Ah,h> \geq 0$. Let $\alpha<0$, then we show that $A-\alpha I$ is invertible therefore it is not the spectrum of $A$. For any $h \in H$,

\begin{equation} \begin{split} ||(A-\alpha I)h||^2=<(A-\alpha I)h,(A-\alpha I)h> \\ &=||Ah||^2+|\alpha|^2||h||^2-2\alpha<Ah,h> \\ &\geq |\alpha|^2||h||^2. \end{split} \end{equation} Therefore $||(A-\alpha I)h||\geq|\alpha|||h|| $, which means $A-\alpha I$ is bounded below, hence it is one-one ($ker(A-\alpha I)=0$) and has a closed range. To prove invertibility it remains to show that the range is $H$.

To prove the last part, I use the fact that $T(H)=ker(T^*)^\perp$ For any bounded operator $T$ with a closed range.

So, we have $(A-\alpha I(H))=ker((A-\alpha I)^*)^\perp=ker(A^*-\bar\alpha I)^\perp=ker(A-\alpha I)^\perp=0^\perp=H$, hence $A-\alpha I$ is on-to.

To prove the other direction, the only proof I know requires knowledge about the existence of a positive square root of a positive operator (Here by positive I mean an operator with nonnegative spectrum) which comes from the functional calculus theorem. So I will not discuss it.