# Convergence of suprema of sequence of functions

Let's say that I have a sequence of continuous, bounded functions $$\lbrace f_n \rbrace$$ which converge uniformly to some continuous, bounded function $$f$$. $$f_n$$, $$f : X \to \mathbb{R}$$, where $$X$$ is a compact subset of $$\mathbb{R}^k$$. Can I say that $$\sup_X f_n$$ converges to $$\sup_X f$$?

Now let's say that there is a compact set $$G \subset \mathbb{R}^k$$ and $$f_n \to f$$ uniformly on $$G$$. There is a sequence of compact sets $$\lbrace X_n \rbrace$$ which converges to $$X$$, where $$X_n$$, $$X \subset G$$. Can I also say that $$\sup_{X_n} f_n$$ converges to $$\sup_X f$$?

• How do you define $X_n$? Feb 27 '19 at 17:40
• $X_n$ is some compact subset of $\mathbb{R}^k$. I say that $X_n \to X$ in the sense that $X = \lim \inf_{n \to \infty} X_n = \lim \sup_{n \to \infty}$. Feb 27 '19 at 17:43
• What does it mean for $f_n\to f$ uniformly if $f_n$ is defined on $X_n$ and $f$ is defined on $X$? Feb 27 '19 at 17:43
• Ah, sorry. The problem isn't well-defined. I should say that $X$, $X_n \subset G$, where $G$ is a compact subset of $\mathbb{R}^k$ and $f_n \to f$ uniformly on $G$. Feb 27 '19 at 17:45

For your first part, the answer is yes. The idea is basically the same as Pebeto's answer, with just slightly more detail needed.

Let $$M_n=\sup_Xf_n$$ and $$M=\sup_Xf$$. By compactness, there exists $$\{x_n\}$$ such that $$f_n(x_n)=M_n$$. If $$M_n\not\to M$$, there would exist $$\varepsilon>0$$ and a subsequence $$\{n_k\}$$ such that $$|M_{n_k}-M|\ge\varepsilon$$ for all $$k$$. But since $$\{x_{n_k}\}$$ is a sequence in the compact set $$X$$, there is a further subsequence $$\{n_{k_j}\}$$ and $$x^*\in X$$ such that $$x_{n_{k_j}}\to x^*$$. By uniform convergence and continuity, one has $$f_{n_{k_j}}(x_{n_{k_j}})\to f(x^*)$$. This implies, if $$y\in X$$,

$$f(y) = \lim_{j\to\infty}f_{n_{k_j}}(y)\le \lim_{j\to\infty}f_{n_{k_j}}(x_{n_{k_j}})=f(x^*),$$

implying $$f(x^*)=M$$. But this implies $$M_{n_{k_j}}\to M$$, a contradiction since by assumption $$|M_{n_{k_j}}-M|\ge\varepsilon$$. Thus, $$M_n\to M$$.

The answer for the second part is no. For a counterexample, take $$G=[0,1]$$ and $$X_n=\{0\}\cup\{1-\frac1n\}$$ for $$n\ge1$$, and let $$f_n(x)=f(x)=x$$ for each $$n\ge1,x\in G$$. The set-theoretic limit of $$\{X_n\}$$ is $$X=\{0\}$$, and $$\sup_Xf=0$$, but $$\sup_{X_n}f_n=1-\frac1n\to1$$.

$$\sup_{X} f_n$$ converges towards $$\sup_{X} f.$$
Let $$x_n$$ be such that $$f_n(x_n) = \sup_{x \in X} f_n(x)$$, then $$f_n(x_n) \geq f_n(x)$$. By compactness of $$X$$, we can take a convergent subsequence $$x_{n'}$$ towards $$x^*$$. Then using uniform continuity, we get $$\lim f_{n'}(x_{n'}) = f(x^*) \geq \lim f_n(x) = f(x).$$