Let $u(x,t)$ solve $$\begin{cases} u_t = u_{xx} \quad (0<x<1, t\gt 0)\\ u(0,t) = u(1,t) = 0 \\ u(x,0) = f(x) \end{cases}\quad \ \ (1)$$ for $f\in C[0,1]$. Show that for all $\tau\ge 0$, $$\int_0^1|u(x,\tau)|^2 \ dx \le\int_0^1|f|^2 \ dx \qquad (2)$$
My attempt: As a consequence of the Cauchy-Schwartz inequality, we see that $u(x,t)\le f(x)$ for $x\in [0,1]$. This implies that $u(x,\tau) \downarrow$ on $[0,1]$. Now, we have to prove this claim. This is where my argument breaks down. Looking at the explicit representation formula for $u(x,t)$ satisfying $(1)$, I don’t think this logic is correct.
Is there an inequality that you can apply to attain $(2)$, or am I on the right track? I’m sure that I am missing something very obvious.
Hint. Seek $u$ in the form $$ u=\sum_{k=1}^\infty c_k(t)\sin( k\pi x). $$ Then $c_k(t)=c_k(0)e^{-k^2t}$ and $$ \int_0^1|u(x,\tau)|^2dx=\frac12\sum_{k=1}^\infty |c_k(\tau)|^2\le \frac12\sum_{k=1}^\infty |c_k(0)|^2=\int_0^1|f(x)|^2dx. $$