Let $\Theta(x,t)$ be the Jacobi-Theta function: \begin{equation} \Theta(x,t):=1+\sum_{n=1}^\infty e^{-\pi n^2 t} \cos(2\pi n x) \end{equation}

Usually, the heat equation with the periodic boundary conditions is solved by the "spatial" convolution with $\Theta(x,t)$. However, I wonder what would happen if we perform convolution with respect to $t$ as well.

More specifically, let $f(x,t) : [0,1] \times [0,\infty) \to \mathbb{R}$ be a everywhere-defined function such that

1. $f(\cdot,t) : [0,1] \to \mathbb{R}$ is continuous and satisfies $f(0,t)=f(1,t)$ for each fixed $t \in [0,\infty)$.

2. $\int_0^T \lvert f(x,t) \rvert dt <\infty$ for any $0<T<\infty$ and $x \in [0,1]$.

Then, define a "spacetime" convolution between $f$ and $\Theta$ as \begin{equation} (\Theta * f)(x,t):=\int_{t/2}^t \int_0^1 \Theta(x-y,t-\tau)f(y,\tau)dy d\tau \end{equation} for $x \in [0,1]$ and $t \in (0,\infty)$.

Now, my question is that: is this $(\Theta * f)$ still jointly smooth on $ [0,1] \times (0, \infty)$ with respect to $(x,t)$? Note that I excluded $0$ from the domain for $t$, to avoid possible complications.

I am aware that $\Theta(x,t) \to \sum_{n \in \mathbb{Z}} \delta(x-n)$ in the sense of distribution as $t \to 0^+$. So, I am still concerned about the time integral in the above convolution.

Could anyone please clarify for me?