Operators such that $\langle Ax,x \rangle=-\langle x,Ax \rangle$

217 Views Asked by At

Let $X$ be a Banach space. We consider the differential equation: $$x'(t)=Ax(t), \ \ \ t\in\mathbb{R}$$ where $A$ is a bounded operator on $X$.

If $X$ is a Hilbert space, and $x(t)$ is a solution of the differential equation, then $$\frac{d}{dt}\|x(t)\|^2=\langle Ax(t),x(t) \rangle+\langle x(t),Ax(t) \rangle$$ If the operator $A$ has the property $\langle Ax,x \rangle=-\langle x,Ax \rangle$, then we will get $$\frac{d}{dt}\|x(t)\|^2=0,$$ which means $|x(t)|$ is constant.

So the condition $\langle Ax,x \rangle=-\langle x,Ax \rangle$ makes $|x(t)|$ constant for every $x(t)$ solution.

Can we find a weaker property on $A$ which will make $|x(t)|$ constant only for the bounded solutions on $\mathbb{R}$? Is there any reference which deals with these problems ?

Note: I am not looking for evident conditions like: The only bounded solution is $0$.