TITLE: Energy estimate of the differential equation $\dot{x}=Ax$
QUESTION [6 upvotes]: Conside the differential equation
$$\dot{x}=Ax,\qquad x(t):{\bf R}\to{\mathcal H}$$ where $\mathcal{H}$ is a Hilbert space and $A$ is a bounded linear operator on $\mathcal{H}$. With the initial condition $x(0)=x_0$, one can have $x(t)=e^{At}x_0$ (Is this legal in the infinite dimension case?). With the spectral method (under the assumption that $x_0$ is an eigenvector of $A$), one has the estimate
$$\|x(t)\|\leq e^{\omega t}\|x_0\|.$$
Here is my question:

With some additional assumption, can one estimate $\|x(t)\|$ without using the spectral method, say, simply taking the inner product?

I have thought that the following sub-questions may be helpful:

What is $\frac{d}{dt}\langle x(t),x(t)\rangle$?
In the finite dimension case, this can be done with the product rule. What about the infinite dimension case?

REPLY [8 votes]: Yes, if $A$ is a bounded linear operator $e^{At}$ is quite OK.  It can be defined by the Taylor series, for example, and [EDIT: for $t \ge 0$ ] we have 
$$\|e^{At}\| \le e^{\|A\| t}.$$
If $x_0$ is an eigenvector for $A$ with eigenvalue $\lambda$, then $x(t) = e^{\lambda t} x_0$.
As for your sub-question, it is still true that 
$$\frac{d}{dt} \langle x(t),\, x(t) \rangle = \langle x'(t), x(t) \rangle + \langle x(t), x'(t) \rangle = 2 \Re \langle x(t), A x(t) \rangle.$$  The proof using difference quotients is basically the same as in single-variable calculus.