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
-
$f(\cdot,t) : [0,1] \to \mathbb{R}$ is continuous and satisfies $f(0,t)=f(1,t)$ for each fixed $t \in [0,\infty)$.
-
$\int_0^T \lvert f(x,t) \rvert dt <\infty$ 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?
Best Answer
$\newcommand\Th\Theta\newcommand{\Z}{\mathbb Z}$The answer here is no.
E.g., for $(x,t)\in[0,1]\times[0,\infty)$ let \begin{equation} f(x,t):=2x\,1(x<1/2)+(2-2x)\,1(x\ge1/2). \end{equation} Then \begin{equation} (\Th*f)(x,t)=\frac t4-\frac2{\pi^3}\sum_{n=1}^\infty a_n\cos2n\pi x =\frac t4-\frac{c_t}{\pi^3}\,g_t(x), \end{equation} where \begin{equation} a_n:=a_{t,n}:=(1+(-1)^{n-1})\frac{1-e^{-\pi n^2t/2}}{n^4}, \end{equation} \begin{equation} g_t(x):=\sum_{n\in\Z}p_n e^{2\pi ixn}, \end{equation} $p_n:=a_n/c_t$ for $n\ne0$, $p_0:=0$, $c_t:=2\sum_{n=1}^\infty a_n$, so that $p_n\ge0$ for all $n$ and $\sum_{n\in\Z} p_n=1$.
So, $g_t$ is the characteristic function (c.f.) of a random variable $X_t$ such that $P(X_t=2\pi n)=p_n$ for $n\in\Z$. Note that $EX^4=\infty$. So, the fourth derivative of $g_t$ at $0$ does not exist (see e.g. Theorem 2.3.1). So, $(\Th*f)(x,t)$ is not smooth in $x$ at $x=0$ and hence is not jointly smooth in $(x,t)\in[0,1]\times(0,\infty)$. $\quad\Box$