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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 $\sup_X f$?

Thanks for your help! Answers or directions to texts/resources are both helpful.

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  • $\begingroup$ How do you define $X_n$? $\endgroup$
    – Pebeto
    Feb 27 '19 at 17:40
  • $\begingroup$ $X_n$ is some compact subset of $\mathbb{R}^k$. I say that $X_n \to X$ in the sense that $X = \lim \inf_{n \to \infty} X_n = \lim \sup_{n \to \infty}$. $\endgroup$ Feb 27 '19 at 17:43
  • $\begingroup$ What does it mean for $f_n\to f$ uniformly if $f_n$ is defined on $X_n$ and $f$ is defined on $X$? $\endgroup$
    – Jason
    Feb 27 '19 at 17:43
  • $\begingroup$ 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 $G$. $\endgroup$ Feb 27 '19 at 17:45
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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 $\sup_{X_n}f_n=1-\frac1n\to1$.

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For your first part, surely

$\sup_{X} f_n$ 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 $\lim f_{n'}(x_{n'}) = f(x^*) \geq \lim f_n(x) = f(x).$

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