If $Re \lambda \geq 0$ for all $\lambda \in \sigma(A)$ show $Re\langle x,Ax\rangle\geq 0$

Question

Let $H$ be a Hilbert space and A a normal operator with the property $Re\lambda\geq 0$ for all $\lambda \in \sigma(A)$. We want to show that $Re\left<x,Ax\right> \geq 0$ for every $x \in h$.

I have a hint which says to first show $||e^{-tA}||\leq 1$ when $t \geq 0$ which I think I have Post here

The next hint says to differentiate the map $x \mapsto ||e^{-tA}x||^2$

So I suppose $F_x(t)=||e^{-tA}x||^2$ and calculate;

$$\frac{d}{dt}F_x(t)=\frac{d}{dt}||e^{-At}x||^2=\frac{d}{dt}\left<e^{-tA}x,e^{-tA}x\right>$$ $$=?=\left<-Ae^{-tA}x,e^{-tA}x\right>+\left<e^{-tA}x,-Ae^{-tA}x\right>$$

Which is where I'm stuck. I have that this expression is equal to;

$$-2Re\left<e^{-tA}x,Ae^{-tA}x\right>$$ and/or $$-\left<e^{-tA}x,(A^*+A)e^{-tA}x\right>$$

But I can't see how either of these give me any insight about $\left<x,Ax\right>$.

Answer 1

Here is a simple answer using the so called continuous functional calculus, which is applicable since $A$ is normal.

Define the operator $B=A+A^*$. It suffices to show that $\langle Bx,x\rangle\geq0$. Indeed, if we do that then $\langle Ax,x\rangle+\langle A^*x,x\rangle\geq0$, i.e. $\langle Ax,x\rangle+\overline{\langle Ax,x\rangle}\geq0$, i.e. $2\text{Re}\langle Ax,x\rangle\geq0$.

Note that, if $f(z)=z+\bar{z}$, then $B=f(A)$, so $\sigma(B)=f(\sigma(A))=\{z+\bar{z}:z\in\sigma(A)\}=\{2\text{Re}(z):z\in\sigma(A)\}\subset[0,\infty)$, hence $B\geq0$. But positive operators have positive square roots, so $B=C^2$ for some self-adjoint operator $C$ (i.e. use functional calculus again and set $C=g(B)$, where $g(t)=t^{1/2}$), thus $\langle Bx,x\rangle=\langle C^2x,x\rangle=\langle Cx,Cx\rangle=\|Cx\|^2\geq0$.

Answer 2

I guess OP would prefer a different answer; based on folouer of kaklas's comment, indeed the derivative at $0$ is equal to $-2\text{Re}\langle Ax,x\rangle$. But note that $F_x(t)$ is decreasing at 0: if $t>0$, then $$F_x(t)=\|e^{-tA}x\|^2\leq\|e^{-tA}\|^2\|x\|^2\leq\|x\|^2=F_x(0).$$

Thus $\frac{d}{dt}F_x(t)\vert_{t=0}\leq0$, which is obviously equivalent to what you wanted to prove.

Regarding the question mark in your computations, see this post (and don't worry about Banach algebras, just think of the bounded operators instead).