Let $f_n:[0,1]\to \mathbb{R}_{\geq 0}$ be a decreasing sequence of smooth functions (i.e. $f_n\leq f_m$ if $n\geq m$) that converges pointwise to $f$. If $m_n = \max f_n$, then we have that the sequence $\{m_n\}$ is decreasing and bounded below by $\sup f$, so it converges and
$$\lim_{n\to\infty} m_n \geq \sup f.$$
Is this an equality? If not, what is a counterexample? If it helps, $f$ is piecewise constant.
Note: @DanielFischer's comment offers a much easier proof.
Let $x_n\in[0,1]$ with $f_n(x_n) = m_n$. As $[0,1]$ is compact, it has a convergent subsequence $x_{n_k}\to x$.
Assume additionally that there exists some $\varepsilon > 0$ such that $f$ is continuous (or in your case constant) on $I = [x - \varepsilon, x + \varepsilon] \cap [0, 1]$. (It depends on your definition of piecewise constant functions.)
As $f_n$ is continuous and decreases monotonically to $f$, the convergence is uniform on $I$ (c.f. Dini's theorem). Thus, we have $m_{n_k} = f_{n_k}(x_{n_k}) \to f(x)$. As $m_n$ decreases monotonically, we also have $m_n\to f(x) \le \sup f$. That is, $m_n \to \sup f$.
