when limit of a maximum of a function equals maximum of limit

Question

Let $f:\mathbb{R}\times[0,1]\rightarrow[0,1]$ be a continuous function. When can I conclude that $$\lim_{m\rightarrow\infty}\max_{p\in[0,1]}f(m,p)=\max_{p\in[0,1]}\lim_{m\rightarrow\infty}f(m,p)$$

The context is the following. I am trying to prove that equation $f(m,p)=c>0$, in $p$, given $m$, has no solution 'for large $m$'. I have shown that for any $p\in[0,1]$, $\lim_{m\rightarrow\infty}f(m,p)=0$. From here I thought my desired result follows. What follows is that the right hand side of the equality above is zero. However, in order to prove my desired result, I need to argue that the left hand side is zero.

Do I need additional properties of $f$ or is continuity enough?

After reading answer by @Gono, let me include what I think is a countrexample to continuity being enough. Let $f:\mathbb{R}\times[0,1]\rightarrow[0,1]$ be given by $$f(m,p)=\begin{cases}\exp{(-1)}p&\text{ if }m<1\\\exp{(-pm)}&\text{ if }m\geq1\text{ and }p\geq\frac{1}{m}\\\exp{(-1)}pm&\text{ if }m\geq1\text{ and }p<\frac{1}{m}\\\end{cases}$$ Plot of $f$ is below. It is clearly continuous, $\lim_{m\rightarrow\infty}f(m,p)=0$ for any $p\in[0,1]$ and $\max_{p\in[0,1]}f(m,p)=\exp{(-1)}$ for any $m\in\mathbb{R}$. Hence $$\lim_{m\rightarrow\infty}\max_{p\in[0,1]}f(m,p)=\exp{(-1)}\neq0=\max_{p\in[0,1]}\lim_{m\rightarrow\infty}f(m,p)$$

Answer 1

First I thought it work this way:

With $f$ continuous also $$h(m) = \max_{p \in [0,1]} f(m,p)$$ is continuous hence $$\begin{align*}\lim_{m\rightarrow\infty}\max_{p\in[0,1]}f(m,p)&= \lim_{m\to \infty} h(m) \\ &= h(\lim_{m\to\infty} m) \\&= \max_{p\in[0,1]}f( \lim_{m\rightarrow\infty} m,p) \\&= \max_{p\in[0,1]}\lim_{m\rightarrow\infty} f( m,p)\end{align*}$$ and all we needed was the continuity of $h$ and $f$ to move limit into (or outside) function(s).

But obviously it doesn't but if we write $f_m(p) = f(m,p)$ and consider $f_m$ being a sequence of functions with limit $$f = \lim_{m\to\infty} f_m$$ your equation holds iff $$\lim_{m\to\infty} \max_{p \in [0,1]} f_m(p) = \max_{p\in [0,1]} f(p)$$

And for that you need $f_m\to f$ uniformly on $[0,1]$.

Your counterexample converges to $f(p) = \exp(-1)$ as $m\to\infty$ but not uniformly.