We say that $u: [0,T) \times [-\pi, \pi] \rightarrow \mathbb{C}$ solves the heat equation with initial condition $u_0: [-\pi, \pi] \rightarrow \mathbb{C} $ if:
$$
\begin{cases}
\partial_t u(t,x) = \partial^2_x u (t,x), \quad (t,x) \in (0,T) \times [-\pi, \pi], \\
u(0,x)=u_0(x) \quad x \in [-\pi, \pi]
\end{cases}
$$
(and if periodicity of $u$ and $\partial_x u$ holds).
The proposed solution is:
$$
u(t,x):=\sum_{k \in \mathbb{Z}} c_k(u_0)e^{-k^2t}e^{ikx}
$$
Im trying to understand how to show that $u \in C^{\infty}((0,\infty) \times \mathbb{R}) $. In a first step I’ll assume that the fourier coefficients of the initial condition $c_k(u_0)$ are summable, i.e.:
$$
\sum_{k \in \mathbb{Z}} |c_k(u_0)| < \infty
$$
One way to show $u \in C^{\infty}((0,\infty) \times \mathbb{R}) $ is by demonstrating that $\forall l,h \in \mathbb{N}$, the series:
$$
\sum_{k \in \mathbb{Z}} \partial_t^l \partial_x^h (c_k(u_0)e^{-k^2t}e^{ikx})
$$
converges absolutely (w.r.t. the $L^{\infty}$ norm) on $(\delta,\infty) \times \mathbb{R} $ for all $\delta > 0$.
Then using the theorem:
Let $H$ be a Hilbert space, $(u_k)_{k \in \mathbb{N}} \subset H$ a sequence then: $\sum_{k=0}^{\infty} \|u_k\| < \infty \Rightarrow (\sum_{k=0}^{\infty} u_k$ converges in $H)$ w.r.t. $\|\cdot\|$.
we deduce uniform convergence of all partial derivatives (on $(\delta,\infty) \times \mathbb{R} $ for all $\delta > 0$), which in turn implies using:
$\forall n: f_n$ differentiable and $\exists x: \lim_{N \rightarrow \infty} \sum_{n=1}^N f_n(x)=f(x)$ and $\sum_{n=1}^N f_n' \rightarrow f'$ uniformly then $\sum_{n=1}^N f_n \rightarrow f$ uniformly and $\sum_{n=1}^{\infty} f_n' = f'$
differentiability of $u$ on $(0,\infty) \times \mathbb{R} = \bigcup_{\delta > 0} (\delta,\infty) \times \mathbb{R}$.
Is this really necessary? Couldnt we just show that:
$$
\sum_{k \in \mathbb{Z}} \|c_k(u_0)e^{-k^2t}e^{ikx}\|_{L^{\infty}} < \infty
$$
which would (again using the first theorem) imply convergence of $\sum_{k \in \mathbb{Z}} c_k(u_0)e^{-k^2t}e^{ikx}$ in the Hilbert space $C^{\infty}((0,\infty) \times \mathbb{R})$, which means $ u \in C^{\infty}((0,\infty) \times \mathbb{R})$.
The only issue I see with the second option is that we havent shown:
$$
\partial_t u = \partial_t \sum_{k \in \mathbb{Z}} c_k(u_0)e^{-k^2t}e^{ikx} = \sum_{k \in \mathbb{Z}} \partial_t (c_k(u_0)e^{-k^2t}e^{ikx})
$$
and the analogous result for $\partial_x$ which we would need to show that $u$ satisfies the heat equation. So is it necessary to show absolute convergence of all partial derivatives as in the first approach? Im in part asking the question because the second approach can more easily be used to show the result for $u_0 \in L^2$ (i.e. if I drop the assumption that the $c_k(u_0)$ are summable).
Best Answer
First remark: the theorem you quoted holds more generally for Banach spaces.
Second remark: $C^{\infty}$ is not a Hilbert space. It is not even a Banach space. It is only the $C^k$ spaces which are Banach space. With the topology of uniform convergence on compact sets of partial derivatives of all orders, $C^{\infty}$ forms something more general, called a Frechet space. Fortunately, since smoothness is defined as being in each of these $C^k$ spaces, we can still invoke the theorem you quoted, without having to resort to more abstract mumbo jumbo, to prove the desired smoothness of $u$.
Now, if you apply $\partial_t^l\partial_x^h$ to the summands, then you get $(-k^2)^l(ik)^h\cdot c_ke^{-k^2t}e^{ikx}$, which has absolute value $k^{2l+h}|c_k|e^{-k^2t}$. Note that if you take the supremum over all $(t,x)\in(0,\infty)\times\Bbb{R}$, then the result is $k^{2l+h}|c_k|$, so the series won’t converge unless you put unnecessarily and unreasonably strong decay hypotheses on the coefficients $c_k$. If you merely assume the coefficients $c_k$ are in $\ell^1$ (i.e form an absolutely convergent series) then you will only be able to deal with the case $l=h=0$, i.e prove that $u$ is continuous.
To get an improved result, we proceed as follows: suppose the $c_k$’s grow at-most exponentially, in the sense that there is a $B>0$ such that for all $k$, we have $|c_k|\leq e^{B|k|}$. In this case, for each $\delta>0$, \begin{align} \sum_{k\in\Bbb{Z}}\left\|\partial_t^l\partial_x^h\left(c_k e^{-k^2t}e^{ikh}\right)\right\|_{L^{\infty}((\delta,\infty)\times\Bbb{R})}&=\sum_{k\in\Bbb{Z}}|k|^{2l+h}|c_k|e^{-k^2\delta}\leq \sum_{k\in\Bbb{Z}}|k|^{2l+h}e^{B|k|-k^2\delta}<\infty. \end{align} This series is finite since for large $|k|$, the quadratic term will dominate, so we have a decaying exponential, and this is enough to make the whole series converge (even with the polynomial growth $|k|^{2l+h}$). Since this is true for all $l,h$, and all $\delta>0$, it follows that $u$ is $C^{\infty}$ on $(0,\infty)\times\Bbb{R}$, and that in this region, term-by-term differentiation is valid in both space and time. In other words, things are as nice as you would want them.
Note that this at-most exponential growth condition is very general. If you assume the coefficients $c_k$ lie in some $\ell^p$ space with $p\in [1,\infty]$, then they are necessarily bounded (i.e in $\ell^{\infty}$), so the exponential bound clearly holds.