1. The problem statement, all variables and given/known data Let X be a compact metric space, F a family of uniformly bounded, equicontinuous real-valued functions on X. Is the function g(x) := supf ∈ F f(x) necessarily continuous? 3. The attempt at a solution The hypotheses about the function family seem to point to some use of Ascoli's theorem. If we have that g(x) = sup f(x) for f in the family then we have an increasing sequence fn(x) -> g(x) for fixed x. I tried starting with proving that g is continuous: by Ascoli's theorem this corresponding sequence fn has a (uniformly) convergent subsequence; in particular, g(x) is attained at x by some function f in the closure of F. Perhaps there's a continuity argument to be made there? Any function in the closure of F would have to be continuous because X is compact and the set of bounded continuous functions with compact domain is closed. But the connection to sup f is not clear since sup f need not be the uniform limit of any sequence in F. I also tried a triangulation argument like this: Let xn → x. There's a sequence of functions in F such that fk(x) approaches g(x) pointwise. Choose ε > 0, then [fk(x) - g(x)| < ε/3 for sufficiently large k. Since the fk are equicontinuous there exists δ > 0 such that |xn - x| < δ implies [fk(xn) - fk(x)| < ε / 3 for any k. If I could only guarantee [fk(xn) - g(xn)| < ε/3 for sufficiently large k and n, I will have arrived. But I just can't bridge the gap. Again, it looks like I need some sort of continuity-based argument here, but I don't know how I can make one without assuming that g is continuous. Then I try to think of counterexamples, but the rather dense hypotheses make counterexamples a little hard to visualize for me.