Define operator: $$(T(t)f)(\theta)=\frac{1}{2\pi}\int_{-\pi}^{\pi}G(\theta-\xi,t)f(\xi)d\xi, t>0,$$ Where $G(\theta,t)=1+2\sum_{n=1}^{\infty}e^{-n^2t}\cos n\theta$, and $T(0)f=f$.
I have a problem proving $T(s+t)=T(s)T(t)$. I try to use Fourier transform, since $(T(t)f)(\theta)=G(\theta,t)*f(\theta)$, it is sufficient to show that $\mathscr{F}(T(t)T(s)f)(\theta)=\mathscr{F}(T(t+s)f)(\theta)$, where $\mathscr{F}$ is the notation of Fourier transformation. But how about $$ \mathscr{F}(G(\theta, t)*G(\theta, s)*f(\theta))=\mathscr{F}(G(\theta, t))\mathscr{F}(G(\theta, s))\mathscr{F}(f(\theta))=?$$
Let $c_0 = \frac{1}{\sqrt{2\pi}}$ and $c_n=\frac{1}{\sqrt{\pi}}\cos(n\theta)$ for $n=1,2,\cdots$. This is an orthonormal subset of $L^2[-\pi,\pi]$, though it is not complete. In terms of this basis, and the inner product $\langle\cdot,\cdot\rangle$ on $L^2[-\pi,\pi]$, $$ T(t)f = \sum_{n=0}^{\infty}e^{-n^2 t}\langle f,c_n\rangle c_n. $$ Because of this $\langle T(t)f,c_n\rangle = e^{-n^2t}\langle f,c_n\rangle$ for $ > 0$. Therefore, for $s,t > 0$, \begin{align} T(s)T(t)&f=\sum_{n=0}^{\infty}e^{-n^2 s}\langle T(t)f,c_n\rangle c_n \\ &=\sum_{n=0}^{\infty}e^{-n^2s}e^{-n^2t}\langle f,c_n\rangle c_n \\ &=\sum_{n=0}^{\infty}e^{-n^2(s+t)}\langle f,c_n\rangle c_n \\ &= T(s+t)f. \end{align} The cases $T(t)T(0)f=T(0)T(t)f=T(t)f$ are obvious.