We consider the initial and boundary value problem for the heat equation in a bounded interval $[0, \ell]$ with homogenous Dirichlet boundary conditions and $k=1$, and we suppose that the initial value $\phi$ is piecewise continuously differentiable and that $\phi(0)=\phi(\ell)=0$. I want to show that for the solution holds the following inequality.
$$|u(x,t)| \leq C e^{-\left( \frac{\pi}{\ell}\right)^2 t}, 0 \leq x \leq \ell, t \geq 1$$
where $C$ is a constant that depends on the quantity $\int_0^{\ell} [\phi(x)]^2 dx$.
I have thought the following.
Our initial and boundary value problem is the following:
$\left\{\begin{matrix} u_t=u_{xx}, & 0<x<\ell,t>0,\\ u(0,t)=u(\ell,t)=0, & t \geq 0,\\ u(x,0)=\phi(x), & 0 \leq x \leq \ell \end{matrix}\right.$
We know that the solution of the problem is
$$u(x,t)=\sum_{n=1}^{\infty} b_n e^{-\left( \frac{n \pi}{\ell}\right)^2t} \sin{\left( \frac{n \pi x}{\ell}\right)}$$
where $b_n$ are the Fourier coefficients of $\phi$ and $\phi(x)=\sum_{n=1}^{\infty} b_n \sin{\frac{n \pi x}{\ell}}$.
We get that $|u(x,t)| \leq e^{-\left(\frac{\pi}{\ell}\right)^2t} \sum_{n=1}^{\infty} |b_n| e^{-\left( \frac{\pi}{\ell}\right)^2 t (n^2-1)}$.
We have that $\sum_{n=1}^{\infty} |b_n| e^{-\left( \frac{\pi}{\ell}\right)^2 t (n^2-1)} \leq \left( \sum_{n=1}^{\infty} |b_n|^2\right)^{\frac{1}{2}} \left( \sum_{n=1}^{\infty} e^{-2 \left( \frac{\pi}{\ell}\right)^2 t (n^2-1)}\right)^{\frac{1}{2}}$.
From Bessel, we have that $\sum_{n=1}^{\infty} |b_n|^2 \leq \frac{2}{\ell} \int_0^{\ell} \phi^2(x) dx$.
Thus,
$$\sum_{n=1}^{\infty} |b_n| e^{- \left( \frac{\pi}{\ell}\right)^2 t (n^2-1)} \leq \left( \frac{2}{\ell} \int_0^{\ell} \phi^2(x) dx\right)^{\frac{1}{2}} \left( \sum_{n=1}^{\infty} e^{-2 \left( \frac{\pi}{\ell}\right)^2 t(n^2-1)}\right)^{\frac{1}{2}}$$
Finally, for $t \geq 1$, we have
$$\sum_{n=1}^{\infty} e^{-2 \left( \frac{\pi}{\ell}\right)^2 t(n^2-1)} \leq \sum_{n=1}^{\infty} e^{-2 \left( \frac{\pi}{\ell}\right)^2 (n^2-1)}$$
Thus, we get that
$$|u(x,t)| \leq e^{-\left( \frac{\pi}{\ell}\right)^2 t} \sqrt{\frac{2}{\ell}} \left( \int_0^{\ell} [\phi(x)]^2 dx\right)^{\frac{1}{2}} \left( \sum_{n=1}^{\infty} e^{-2\left( \frac{\pi}{\ell}\right)^2 (n^2-1)}\right)^{\frac{1}{2}}$$
But in order to show that $ \sqrt{\frac{2}{\ell}} \left( \int_0^{\ell} [\phi(x)]^2 dx\right)^{\frac{1}{2}} \left( \sum_{n=1}^{\infty} e^{-2\left( \frac{\pi}{\ell}\right)^2 (n^2-1)}\right)^{\frac{1}{2}}$ is a constant, we have to show that $\int_0^{\ell} [\phi(x)]^2 dx<+\infty$ and $\sum_{n=1}^{\infty} e^{-2\left( \frac{\pi}{\ell}\right)^2 (n^2-1)}<+\infty$.
How can we do so?
It looks like $\phi$ is a bounded function on a bounded interval, so its $L^2$ norm has to be finite: $$ \int_0^l |\phi(x)|^2\; dx \leq l (\sup |\phi|)^2. $$ And it also looks like the series converges; it's basically $\sum_n e^{-n^2}$. Lots of standard convergence tests will verify that this is finite.
I should clarify that these "constant" you get in the end of course depend on $\phi$ and $l$.