I'm stuck with the following problem
Let $U\subseteq \mathbb{R}$ be a bounded domain, $T>0$ and $u\in C^2(U_T) \cap C(\overline{U_T})$, where $U_T = U \times (0,T]$. If $f=f(x)$ is a continuous function and non positive in $U_T$, prove that any solution of $$\begin{cases} u_t - \Delta u = f(x) & , U_T \\ u=0 & , \partial U_T \end{cases}$$ satisfies $u_t \leq 0$ (Hint: consider $w^\epsilon(x,t) = u(x,t+\epsilon)-u(x,t)$ and apply the maximum principle).
Since $f$ is independent of $t$, I observed that for each $\epsilon>0$ $$(w^\epsilon)_t - \Delta u^\epsilon = (u_t - \Delta u) (x,t+\epsilon) - (u_t-\Delta u)(x,t) = f(x) - f(x) = 0 \qquad , (x,t)\in U_{T-\epsilon}$$ Also, since both $u(x,t), u(x,t+\epsilon)$ are null in $\partial U_T, \partial U_{T-\epsilon}$ respectively, we get $w^\epsilon =0$ in $\partial U_{T-\epsilon}$. If I apply now the maximum principle for $w^\epsilon$, it leads to $w^\epsilon = 0$ in $\overline{U_{T-\epsilon}}$. Taking $\epsilon \to 0^+$, since solutions of the heat equation are conveniently regular $$u_t (x,t) = \lim_{\epsilon \to 0^+} \frac{w^\epsilon (x,t)}{\epsilon} = 0 \qquad , (x,t)\in U_T$$ I don't think this solution is valid for two different reasons: I have not used the hypothesis of $f\leq 0$, and also I'm proving that $u_t = 0$ which seems false to me. I don't know how to work this problem out since the information of $f\leq 0$ is convenient for the PDE itself, maybe I should consider first $v=u-M$, with $M=\sup_{x\in U} f(x) \leq 0$ (¿?)
I will appreciate any suggestion.