I am having trouble with the following problem.
Let
$$F(x,t)=\sum_{k \in \mathbb{Z}}d_k e^{-4 \pi^2 k^2 t} e^{2 \pi i kx}$$ $$(..., d_{-1}, d_0, d_1, ...) \in \ell^2(\mathbb{Z})$$
Where $d_k, x \in \mathbb{R}, t >0$, and assume that, for some positive real numbers $a,b$ it is true that $|d_k| \leq ae^{-bk^2}$
I want to prove that $F$ is continuous and infinitely differentiable with respect to each of its two independent variables $x$ and $t$.
For continuity, if I look at $|f(x,t) - f(x_0, t_0)|$ I see that
$$\left| \sum_{k \in \mathbb{Z}} d_k \left( e^{-4 \pi^2 k^2 t} e^{2 \pi i kx} - e^{-4 \pi^2 k^2 t_0} e^{2 \pi i kx_0} \right) \right| \leq \sum_{k \in \mathbb{Z}} \left| d_k e^{-4 \pi^2 k^2 t} e^{2 \pi i kx} - e^{-4 \pi^2 k^2 t_0} e^{2 \pi i k x_0}\right| \to 0 $$ as $(x,t)\to(x_0,t_0)$. But is this enough for continuity?
Similarly for $1$-differentiability, I just have that
$$\frac{\mathrm{d} f(x,t)}{\mathrm{d} x} = \lim_{h \to 0} \frac{f(x+h,t) - f(x,t)}{h} = \lim_{h \to 0} \frac{\sum_k d_k e^{-4 \pi^2 k^2 t} e^{2 \pi i kx} (e^{2\pi i k h} - 1)}{h} = \lim_{h \to 0} \int_\mathbb{Z} \frac{ d_k e^{-4 \pi^2 k^2 t} e^{2 \pi i kx} (e^{2\pi i k h} - 1)}{h} \mathrm{d} \delta(k)$$
where $\delta$ is the counting measure.
Now, we can find a doinating function
$$\left| \frac{d_k e^{-4 \pi^2 k^2 t} e^{2 \pi i kx} (e^{2\pi i k h} - 1)}{h} \right| \leq |d_k e^{-4 \pi^2 k^2 t} e^{2 \pi i kx}| $$
From the inequality $|e^{ix} - 1| \leq |x| : x \in \mathbb{R}$
and this is less than or equal to $$2|d_k|$$ since $e^{-4\pi^2k^2 t} < 1$ for $t >0$ and $e^{2\pi i kx} \leq 2$ for $x \in \mathbb{R}$
Thus, we can use the dominated convergence theorem to interchange the integral and the limit and find our limit by using the limit of the exponential function.
Is this all that it's needed? I feel that I am missing something, especially in the continuity part. Thank you!