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.