Consider the problem \begin{equation} \left\{ \begin{aligned} &u_{t} - u_{xx} = 0, \ \ \ \ x \in (0,L), \ \ t > 0,\\ & u(0,t) = u(L,t) = 0,\ \ \ \ t > 0 \\ & u(x,0) = f(x), \ \ \ \ x \in (0,L) \end{aligned} \right. \label{modelo} \end{equation} Suposing that $u \in C^{2,1}((0,L) \times (0, T)) \ \cap \ C^{0}([0,L] \times [0, T)) $ and \begin{equation*} |u(x,t)| \leq Ce^{-t^2}, \ \ \forall \ t \geq 0, \end{equation*} Show that $f \equiv 0$.
Question: If initial temperature distribution is identically null, then $u \equiv 0$? Else, how to use the hypothesis $|u(x,t)| \leq Ce^{-t^2}$ to prove this result?
If you assume that $f$ is piecewise smooth (both $f$ and $f'$ are piecewise continuous) on $[0,L]$, then your solution can be written as a series $$ u(x,t)=\sum_0^{\infty}A_n\sin\frac{n\pi x}{L}e^{-(n\pi L)^2t} $$ where the $A_n$ are the coefficients of the expansion of $f$ as a series of sines. From this formula follows that, unless $f\equiv 0$, some coefficient $A_{n}$ will be non-zero, let $A_{n_0}$ be the first non-zero coefficient. Then $u$ decays exponentially in time $$ \sup_x u(x,t)\sim A_{n_0}e^{-(n_0\pi L)^2t}, $$ as $t\to\infty$, thus violating the bound $$ |u(x,t)|\le Ce^{-t^2} $$ which entails a much faster decay to zero.