Heat equation with periodic initial conditions Let $f:\mathbb{R}^n \to \mathbb{R}, f \in C(\mathbb{R})$ be $[0,1]^n-$periodic.
Consider then the problem:
$$\left\{\begin{matrix}\partial_t u -\Delta_x u = 0\quad t>0\\u(x,0) = f(x)\quad t=0\end{matrix}\right.$$
Show that $$\lim_{t \to \infty}u(x,t) = \int_{[0,1]^n}f(y)dy$$

We can find a solution of the PDE by taking
$$u(x,t) = \int_\mathbb{R^n} \Phi(x-y,t) f(y)dy$$
with $\Phi(x-y,t) = \frac{e^{-\frac{|x-y|^2}{4t}}}{(4\pi t)^{\frac{n}{2}}}$ the heat kernel. I know that $\int_\mathbb{R^n} \Phi(x-y,t)dy=1$ for all $t>0$.
Consider the set of points in $\mathbb{z}^n$ and take the partition of $\mathbb{R}^n = \bigcup_{z \in \mathbb{Z}^n} \left([0,1]^n+z\right)$
We have then that
$$u(x,t) = \sum_{z \in \mathbb{Z}^d}\int_{[0,1]^n+z}\Phi(x-y,t)f(y)dy = \sum_{z \in \mathbb{Z}^d}\int_{[0,1]^n}\Phi(x-y+z,t)f(y+z)dy$$
$$=\sum_{z \in \mathbb{Z}^d}\int_{[0,1]^n}\Phi(x-y+z,t)f(y)dy$$
Now I'd like to find some way to justify something like
$$\int_{[0,1]^n}\Phi(x-y+z,t)f(y)dy \to \left[\int_{[0,1]^n}f(y)dy\right]\left[\int_{[0,1]^n}\Phi(x-y+z,t)dy\right]\mbox{ as }t \to \infty$$
which would solve the question. How would I do that?

Attempt to show it:
$$\left|\int_{[0,1]^n}\Phi(x-y+z,t)f(y)dy - \left[\int_{[0,1]^n}f(w)dw\right]\left[\int_{[0,1]^n}\Phi(x-y+z,t)dy\right]\right| = $$
$$=\left|\int_{[0,1]^n}\Phi(x-y+z,t)\left[f(y) - \int_{[0,1]^n}f(w)dw\right]dy\right|$$
$$\leq \sup_{y \in [0,1]^n}\left[f(y) - \int_{[0,1]^n}f(w)dw\right] \int_{[0,1]^n}\Phi(x-y+z,t)dy$$
We can guarantee that $sup$ is finite because $f$ is periodic on $[0,1]^n$ and continuous, therefore bounded.
Now, as $t \to \infty$ we have that the function $\Phi(x-y+z,t)$ goes to $0$ by the definition. As the integral of $\Phi$ is $1$ for all $t>0$ we must have that, in the bounded square $[0,1]^n$ the integral must go to $0$.
and so we have proved the convergence we needed.
Is this correct? Thanks in advance.

Edit: $\color{red}{\mbox{My attempt is incorrect}}$. As I colleague pointed out to me, my argument would be valid for any value instead of $\int f(y)dy$. I only saw that every term in the summation goes to $0$, which isn't enough. I'd need some form of uniform convergence to pass the limit inside the summation in order to use any results concerning the limit of the terms on summation.
Our professor said to use the fact that periodic functions can be approximated uniformly by trigonometric polynomials, so we'd solve the problem for a trigonometric function and then by the uniform convergence we would have solved for any periodic function. Can anyone help on that? Thanks.
 A: I think it goes more or less as follows or at least I hope it helps,
$$
\begin{split}
u(x,t) 
 & =\lim_{t\to 0}\sum_{z \in \mathbb{Z}^d}\int\limits_{[0,1]^n+z}\Phi(x-y,t)f(y)dy\\
& =\lim_{t\to 0}\sum_{z \in \mathbb{Z}^d}\int\limits_{[0,1]^n+z}\Phi(y-x,t)f(y)dy\\
& =\lim_{t\to 0}\sum_{z \in \mathbb{Z}^d}\int\limits_{[0,1]^n}\Phi(y+z-x,t)f(y+z)dy\\
& =\lim_{t\to 0}\sum_{z \in \mathbb{Z}^d}\int\limits_{[0,1]^n}\Phi\big((z-x)+y,t\big)f(y)dy\\
& =\lim_{t\to 0}\int\limits_{[0,1]^n}f(y)\bigg(\sum_{z \in \mathbb{Z}^d}\Phi\big((z-x)+y,t\big)\bigg)dy\\
& =\lim_{t\to 0}\int\limits_{[0,1]^n}f(y)\bigg(\sum_{z \in \mathbb{Z}^d}\hat{\Phi}(z-x,t)e^{2\pi i zy}\bigg)dy\qquad(\text{Poisson summation formula (PSF))}\\
& =\int\limits_{[0,1]^n}\sum_{z \in \mathbb{Z}^d}\Big(\lim_{t\to 0}\hat{\Phi}(z-x,t)\Big)f(y)e^{2\pi i zy}dy\\
& =\int\limits_{[0,1]^n}\sum_{z \in \mathbb{Z}^d}\Big(\lim_{t\to 0}e^{-t(z-x)^2}\Big)f(y)e^{2\pi i zy}dy\\
& =\int\limits_{[0,1]^n}\sum_{z \in \mathbb{Z}^d}f(y)e^{2\pi i zy}dy\\
& =\int\limits_{[0,1]^n}f(y)\bigg(\sum_{z \in \mathbb{Z}^d}e^{2\pi i zy}\bigg)dy\\
& =\int\limits_{[0,1]^n}f(y)\bigg(\sum_{z \in \mathbb{Z}^d}\delta(z-y)\bigg)dy\qquad\qquad\;\;\qquad(\text{PSF)}\\
&=\int\limits_{[0,1]^n}f(y)dy
\end{split}
$$
