Prove/disprove that g(x) = sup f(x), f in F, continuous

1. Aug 7, 2015

QIsReluctant

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.

2. Aug 8, 2015

Zondrina

I'm sorry, but could you explain exactly what this line means:

$$g(x) := \text{sup}_{f \in F} f(x)$$

You're choosing one of the functions from $F$, and treating $f(x)$ like a set?

Did you mean:

$$g(x) := \text{sup}\{f \in F \}$$

As in $g(x)$ is chosen to be the biggest function in the set $F$?

I really think this should read $g(x) := \text{max}\{f \in F \}$ anyway.

3. Aug 8, 2015

QIsReluctant

No -- I meant that, at each individual x, we choose the supremum of the set of values at x.
For example, if F = {the functions x → ax: a ∈ (0, 1)}, then g(0.5) = 0.5 and g(-0.5) = 0.

4. Aug 9, 2015

Zondrina

So $g(x)$ takes on the value of one of the equicontinuous functions $f_i \in F, \space 1 \leq i \leq n$ for each $x$. The function $f_i$ chosen for the value of $x$ must maximize $g(x) = f_i(x)$.

We could write this as $g(x) := \displaystyle \sup_{1 \leq i \leq n} f_i(x) = \sup \{ f_1(x), f_2(x), \cdots, f_n(x) \}$.

So we choose a value of $x$, then we choose a value of $i$ such that $f_i(x)$ is maximized.

How do you usually go about showing a function is continuous? How does this apply to $g(x)$? I think you want to show:

$$\displaystyle \lim_{x \to a} g(x) = \displaystyle \lim_{x \to a} \displaystyle \sup_{1 \leq i \leq n} f_i(x) = \sup \{ f_1(a), f_2(a), \cdots, f_n(a) \}$$

More formally:

$$\forall \epsilon > 0, \exists \delta > 0 \space | \space 0 < | x - a | < \delta \Rightarrow |g(x) - \sup \{ f_1(a), f_2(a), \cdots, f_n(a) \}| < \varepsilon$$

You know each $f_i$ is equicontinuous, and you also know $|\sup \{ f_1(a), f_2(a), \cdots, f_n(a) \}| \leq M$ because the functions in $F$ are uniformly bounded.

Last edited: Aug 9, 2015
5. Aug 9, 2015

pasmith

There is nothing in the question to suggest that $F$ is finite or even countable.

6. Aug 9, 2015

Zondrina

Yes, I suppose this would change the notation in my last post a bit. I was intending to mean there are infinitely many of these functions, but I did not express it as such. It's hard to think when you first wake up with no coffee, my apologies.

The previous post I wrote should read something like:

7. Aug 9, 2015

WWGD

Never mind, sorry, I don't see the delete option.

8. Aug 10, 2015

Zondrina

After sleeping on this a bit, I think the notation used in the original post is okay. I can't understand why, but I looked at it differently yesterday than I do today. The limit in my last post should have read:

$$\displaystyle \lim_{x \to a} g(x) = \displaystyle \lim_{x \to a} \displaystyle \sup_{f \in F} f(x) = \sup \{ f_1(a), f_2(a), f_3(a), ... \}$$

Still, the point is to show $|g(x) - \sup \{f_1(a), f_2(a), f_3(a), ... \}| < \varepsilon$.

A graphical argument might help to see what is really happening.

Last edited: Aug 10, 2015
9. Aug 11, 2015

haruspex

Throughout this thread I see references to equicontinuous functions, but as far as I am aware there's no such beast. It's the family that's equicontinuous, and from that fact, it seems to me the desired result follows fairly swiftly.