properties of solution of heat-equation I am currently self studying the heat equation
$$\begin{cases}
u_t(x,t)-u_{xx}(x,t)=0&x\in \mathbb T, t>0\\
u(x,0)=f(x)
\end{cases}
$$with $\mathbb T=\mathbb R/\omega\mathbb Z$, $\omega>0$,
and an exercise in my book asked me to do study the following: 


*

*determine a solution $u$ as a Fourier series

*show $u\in C^\infty(\mathbb T)$

*under which conditions for $f$ does the function $t\mapsto u(t,x)$ be $C^1$-regular from $[0,\infty)$ to $L^2(\mathbb T)$?

*under which conditions for $f$ does the function $t\mapsto u(t,x)$ be
$C^1$-regular from $[0,\infty)$ to $C(\mathbb T)$?


My attempt: a


*

*Let $u(x,t)=a(x)b(t)$. Then $\frac{a''(x)}{a(x)}=\frac{b'(x)}{b(x)}=-c^2$ for some $c\in\mathbb R$. We look for non-trivial solutions. 
If $c=0$, then $a''(x)=0$, hence $a(x)=\alpha x+\beta$. Since we are considering $\mathbb T$, we have $a(0)=a(\omega)$, hence $a(x)=\beta$. This gives $u=const$.
If $-c^2<0$, consider $p(z)=z^2+c^2$. We have $p(z)=0\iff z=\pm ic $. Thus we get the real solution $a(x)=\alpha\cos(c x)+\beta\sin(c x)$. We have $a(-\omega)=a(0)=a(\omega)$ and also, since we are considering $\mathbb T$, $a'(-\omega)=a(0)=a(\omega)$. Comparing the left and right sides yields $c=c_n=\frac{2\pi n}\omega$. Hence we get 
$$
u_n(x,t)=\exp(-c_n^2 t)[\alpha_n\cos(c_nx)+\beta_n\sin(c_nx)]
$$By superposition we get the general solution
$$
u(x,t)=A+\sum_{n\in\mathbb Z}\exp(-c_n^2 t)[\alpha_n\cos(c_nx)+\beta_n\sin(c_nx)]\qquad (1)
$$
Now just the case $p(z)=0\iff z=\pm c$ is left. In this case we have $a(x)=\alpha\exp(cx)+\beta\exp(-cx)$ which also leads to a trivial solution, right?
So we get the solution (1) where $\alpha_n,\beta_n$ are choosen such that $u(x,0)=f(x)$.

*First I've tried to calculate $\alpha_n,\beta_n$. We have $u(x,0)=f(x)=A+\sum_{n\in\mathbb Z}u_n(x,0)$. Consider $(f,u)_{L^2}$. Then $A=\frac1\omega\int_0^\omega f$ and $\alpha_n=\frac2\omega\int_0^\omega f(x)\cos(c_n x)dx$ and $\beta_n=\frac2\omega\int_0^\omega f(x)\sin(c_n x)dx$. Why do I know now $u\in C^\infty(\mathbb T)$? We have $\cos(c_nx),\sin(c_nx)\in C^\infty(\mathbb T)$. The terms in the sum of the right side of (1) are analytic, can we conclude $u\in C^\infty(\mathbb T)$?

*I have to check if there is a continous map $t\mapsto L(t)$ such that $\|\frac1h(u(\cdot,t+h)-u(\cdot,t))-L(t)\|_2\rightarrow 0$ for $h\rightarrow 0$. So first I wanted to consider $t>0$. When I put (1) into this definition, I get really ugly terms and can't handle the limit $h\rightarrow0$. Any hints?

*See 3.
I am really thankful about any hint or sketch of a proof.
 A: I'll assume the usual $2\pi$ interval for $\mathbb{T}$. To simplify notation, let
$$
     \langle f,g\rangle=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x')\overline{g(x')}x'
$$
be the inner product on $L^2(\mathbb{T})$, normalized so that $\|1\|=1$. Then $\{ e^{inx}\}_{n=-\infty}^{\infty}$ is a complete orthonormal basis of $L^2(\mathbb{T})$, which allows you to expand $u(x,t)$
$$
          u(x,t) =\sum_{n=-\infty}^{\infty}c_n(t)e^{inx},\;\;\;\; c_n(t)=\langle u(x',t),e^{inx'}\rangle .
$$
To satisfy the PDE,
$$
               u_t = u_{xx},\;\;\; u(x,0)=f(x),\\
              \sum_n c_n'(t)e^{inx} = \sum_n (-n^2)c_n'(t)e^{inx} \\
               c_n'(t)=-n^2 c_n(t),\;\; c_n(0)=\langle f,e^{inx'}\rangle, \\
                   c_n(t) = \langle f,e^{inx'}\rangle e^{-n^2t} \\
                u(x,t)=\sum_{n=-\infty}^{\infty}e^{-n^2t}\langle f,e^{inx'}\rangle e^{inx}
$$
All of the operations are justified for $t > 0$ because of the presence of the terms $e^{-n^2t}$. The sum on the right converges in $L^2(\mathbb{T})$, but it also converges absolutely and uniformly in $x$ for $t \in [t_0,\infty)$ where $t_0 > 0$, which follows from a crude estimate
$$
                   |\langle f,e^{inx}\rangle| \le \|f\|_{L^2}\|e^{inx}\|_{L^2}=\|f\|_{L^2},\;\;\; n \in \mathbb{Z}.
$$
The same is true of all partial derivatives with respect to $x$ and or $t$, which is then enough to prove that $u(x,t)$ is jointly infinitely differentiable for $t \ge t_0 > 0$ and $x\in\mathbb{T}$. This answers questions about vector differentiability in $L^2(\mathbb{T})$ and in $C(\mathbb{T})$, but only for $t > 0$.
To investigate differentiability at $t=0$, assume that $f$ is periodic and absolutely continuous with $f' \in L^2(\mathbb{T})$. This is equivalent to the condition that
$$
         \|f'\|_{L^2}^2=  \sum_{n=-\infty}^{\infty}n^2|\langle f,e^{inx}\rangle|^2 < \infty.
$$
In that case,
\begin{align}
    &\|u_{x}(x,t)-f'\|_{L^2}^2 \\
  &=\|\sum_{n=-\infty}^{\infty}(-in)e^{-n^2t}\langle f,e^{inx}\rangle e^{inx}-\sum_{n=-\infty}^{\infty}(-in)\langle f,e^{inx}\rangle e^{inx}\|_{L^2}^2 \\
  &=\|\sum_{n=-\infty}^{\infty}in(e^{-n^2t}-1)\langle f,e^{inx}\rangle e^{inx}\|^2 \\
  &=\sum_{n=-\infty}^{\infty}n^2(e^{-n^2t}-1)^2|\langle f,e^{inx}\rangle|
\end{align}
By the Lebesgue dominated convergence theorem applied to the discrete sum,
$$
        \lim_{t\downarrow 0}\|u_{x}(x,t)-f'\|_{L^2(\mathbb{R})}=0.
$$
That's the gist of how to handle such arguments and to keep the notation clean.
