I'm struggling with this question a little bit and just don't know what to do. I'm trying to figure out under what condition on $u_0(x)$, function :
$u(x,t) = \sum_{k=0}^\infty c_k e^{-k^2 \pi^2 t}cos(k\pi x)$
where $c_k = \int_0^1u_0(x)cos(k\pi x)dx$
is smooth ($u(x,t) \in C^\infty([0,1] \times (0,\infty$)).
Because obviously i can not just go with term by term differentiation...
First, check that $$\sum_{k=0}^\infty \frac{\partial ^{\ell+m}}{\partial x^\ell\partial t^m}c_ke^{-k^2\pi^2t}\cos(k\pi x),$$ converges for all $m,\ell\in\mathbb N$ and all $t>0$ and all $x\in [0,1]$ (which is rather straightforward). Now, take $0<\delta <t$. The strategy is to show that $$\frac{\partial ^{\ell+m}}{\partial x^\ell\partial t^m}\sum_{k=0}^nc_ke^{-k^2\pi^2t}\cos(k\pi x)$$ converges uniformly in $t$ and $x$ to $u$ on $[0,1]\times [\delta ,\infty )$ for all $\delta >0$ and all $\ell,m\in\mathbb N$, what will prove the claim. We have that
\begin{align*} &\left|\frac{\partial ^{\ell+m}}{\partial x^\ell\partial t^m}\sum_{k=0}^nc_ke^{-k^2\pi^2t}\cos(k\pi x)-\sum_{k=0}^\infty \frac{\partial ^{\ell+m}}{\partial x^\ell\partial t^m}c_ke^{-k^2\pi^2t}\cos(k\pi x)\right|\\ &\leq \pi^{2m+\ell}\sum_{k=n+1}^\infty k^{2m+\ell}c_ke^{-k^2\pi^2 t}\\ &\leq \pi^{2m+\ell}\sum_{k=n+1}^\infty k^{2m+\ell}c_ke^{-k^2\pi^2 \delta }. \end{align*} Therefore $$\frac{\partial ^{\ell+m}}{\partial x^\ell\partial t^m}\sum_{k=0}^nc_ke^{-k^2\pi^2t}\cos(k\pi x)\underset{n\to \infty }{\longrightarrow}\sum_{k=0}^\infty \frac{\partial ^{\ell+m}}{\partial x^\ell\partial t^m}c_ke^{-k^2\pi^2t}\cos(k\pi x),$$ uniformly in $t$ and $x$ on $[0,1]\times [\delta ,\infty )$. You can therefore conclude that $u$ is smooth on $[0,1]\times [\delta ,\infty )$ for all $\delta >0$, and thus, $u$ is smooth on $[0,1]\times (0,\infty )$ as wished.