Is $\lim (\text{max}\{ f \}) = \text{max}\{(\lim f)\}$?

Question

I'm trying to prove some statement from my differential equations class, and for some part of the proof I used that, being each $f_n(x,y)$ a continuous function defined over a compact set $\Omega$,

$$ \lim_{n \rightarrow \infty} \text{max}\{|f_n(x,y)| : (x,y)\in\Omega\} = \text{max}\{|\lim_{n\rightarrow\infty} f_n(x,y)|: (x,y)\in\Omega\ \}.$$ I thought this is true, but I don't know how to justify it, or if it's only true when $f$ is continuous or anything. Is this true in general? If not, under what conditions is it true? Any help will be appreciated, thanks in advance.

Answer 1

Based on the useful suggestion of Zwim, I propose here a proof for your identity under the uniform convergence assumption. I.e.

Theorem. Assume that

• $\Omega$ is compact
• Continuous functions $f_n$ uniformly converges to continuous function $f$, i.e. $$\sup_{x\in \Omega} |f_n(x)-f(x)| \xrightarrow{n\rightarrow +\infty} 0.$$

Let $x^*_n$ and $x^*$ be the maximizer of $f_n$ and $f$, respectively (they are well defined since $f_n$ and $f$ are continuous and $\Omega$ is compact). Then $$\lim_n f_n(x^*_n)= f(x^*)$$

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Proof.

First, notice that $x^*_n$ is a miximizer of $f_n$ and $f_n$ converges to $f$ uniformly, we obtain \begin{align} \lim_n f_n(x^*_n) \geq \lim_n f_n(x^*)= f(x^*). \end{align}

Second, as $x^*$ is a minimizer of $f$ we obtain \begin{align} f(x^*) & \geq f(x^*_n) \Rightarrow f(x^*) \geq \lim_n f(x^*_n)= \lim_n f_n(x^*_n). \end{align} Here the last equality is a consequence of the uniform convergence \begin{align} |f(x^*_n)-f_n(x^*_n)|\leq \sup_{x\in \Omega} |f(x)-f_n(x)| \xrightarrow{n\rightarrow +\infty} 0 \end{align}

The proof is completed.

Answer 2

Take $f_n(x)=n\,x^n(1-x)$ on $[0,1]$ then $f_n$ converge simply to $0$ while the other limit is $\frac 1e$

(i.e. max is reached in $f_n(\frac n{n+1}$)) : https://www.desmos.com/calculator/ckapsmstdn

Adding another variable $y$ does not change the deal.

Since max is involved or equivalently $||\cdot||_\infty$ norm, then uniform convergence for $f_n$ is most probably required.