We call a function that assigns a starting value of a time-dependent differentialfunction to a solution of a later timevalue as the evolution operator $E(t)$.
Look at the thermal equation
$$ u_t=u_{xx},~~~u_0(x)=u(0,x) $$
where $u\colon [0,T]\times\mathbb{R}\to\mathbb{R}$ with the evolution operator $E(T)u_0=u(T,0)$.
Show that $u_0\star G_t$ with
$$ G_t(x)=\frac{1}{\sqrt{2\pi t}}\exp\left(-\frac{x^2}{t}\right) $$ is a solution of the above equation.
Advice: You therefore have to show $E(T)u_0=u_0\star G_T$.
I am a little bit helpless.
Did I understand the evolution operator right, when thinking that
$E(T)(x)=u(T,0)$ for every $x$?
Then I have to show $(u_0\star G_T)(x)=u(T,0)$?
If yes: How can I do so? Does the Fourier transformation play any role? I thought of that because of the convolution but it was only a quick thought...
I would be very thankful if anyone could help me to solve this question.
Yes, your suspicions are correct. Try the Fourier transform: if $U$ is the Fourier transform of $u$, then Fourier transforming your PDE yields: $$ \frac{\partial U(k,t)}{\partial t} = -k^2 U(k,t) $$ See if you can work with this new PDE.
Note: I am use the convention that the Fourier transform is: $$ U(k,t) = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} e^{-ikx} u(x,t) \, dx$$