Suppose $X$ is a metric space. Let $S$ be a subset of $C(X)$ which is the set of continuous real-valued functions on $X$. If $S$ is equicontinuous and bounded, define $g:X \rightarrow \mathbb{R}$ such that for every $x\in X$, $g(x)=\sup \{f(x)| f\in S\}$

Show that $g\in C(X)$

I am not sure how to do this problem.

I think since $S$ is equicontinuous you have $\forall \epsilon>0$ $\exists \delta>0$ such that $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta$ for all $f\in S$

  • 2
    $\begingroup$ Try writing out the proof when $S$ has just two elements. $\endgroup$ – Ian Apr 21 '16 at 19:16
  • 1
    $\begingroup$ Actually, is $X$ given to be separable? I can give a straightforward proof in this case: take an enumeration $y_m$ of the countable dense subset, take a sequence $f_{n,m}$ in $S$ with $f_{n,m}(y_m) \to g(y_m)$. Then diagonalize. Use equicontinuity to conclude that the limit of the diagonalized sequence is $g$ and then use closedness to conclude that $g \in S$. $\endgroup$ – Ian Apr 22 '16 at 1:22
  • $\begingroup$ Is there a way to prove this using Arzela Ascoli? $\endgroup$ – MathUser123 Oct 18 '17 at 1:59

We can prove lower and upper semicontinuity separately.

We know that supremum of any family of continuous functions is lower semicontinuous. In fact supremum of any family of lower semicontinuous functions is lower semicontinuous. See, for example, these posts: To show that the supremum of any collection of lower semicontinuous functions is lower semicontinuous or Show that the supremum of a collection of lower semicontinuous function is lower semicontinuous.

So it remains to show that $g(x)=\sup\limits_{f\in S} f(x)$ is upper semicontinuous. I.e., we want to show that for any $M$ the set $$g^{-1}(-\infty,M)=\{x\in X; g(x)<M\}$$ is open.

So let $x_0$ be a point such that $g(x_0)<M$. Let us choose $\varepsilon=\frac{M-g(x_0)}2$. From equicontinuity we get that there is a neighborhood $U$ of $x_0$ such that for $x\in U$ and for any $f\in S$ we have $|f(x)-f(x_0)|<\varepsilon$.

Then for every $f\in S$ and $x\in U$ we get $$f(x)=f(x_0)+(f(x)-f(x_0)) \le f(x_0) + |f(x)-f(x_0)| \le g(x_0)+\varepsilon = \frac{M+g(x_0)}2$$ which implies $$g(x) = \sup_{s\in S} f(x) \le \frac{M+g(x_0)}2 < M.$$

We have shown that any point $x_0\in M$ has a neighborhood $U$ such that $x_0\in U\subseteq g^{-1}(-\infty,M)$, which means that the set $g^{-1}(-\infty,M)$ is open.


Using Ian's idea of seperating of assuming only S two element. I did S has $f_1,f_2$

and $f_1<f_2$ for all $x\in X$ and $f_1<f_2$ for all $y\in X$. Also we know that S is equicontinous then $\forall \epsilon>0$ $\exists \delta>0$ such that $|f(x)-f(y)|<\epsilon$ when $|x-y|<\delta$

Because $g(x)=\sup f(x)$ then $g(x)=f_2(x)$ and by a similar logic $g(y)=f_2(y)$

and because $f\in S$ then $g\in S$ so then

$\forall \epsilon >0$ $\exists \delta>0$ such that $|g(x)-g(y)|<\epsilon$ when $|x-y|<\epsilon$

  • $\begingroup$ You can't assume that $f_1<f_2$. Part of the point is that there might be some points where $f_1-f_2$ changes sign, at which point you switch between $f_1$ and $f_2$. You're trying to argue that at these points, $g$ is still continuous. $\endgroup$ – Ian Apr 24 '16 at 2:20

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.