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.

  • $\begingroup$ 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

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 $\sup_{X_n}f_n=1-\frac1n\to1$.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.