Uniform Converges of Continuous Increasing Functions

[e(ho0n3]

Homework Statement
Let f, f₁, f₂, f₃, ... be continuous real-valued functions on the compact metric space E, with f = lim f_n. Prove that if f_i ≤ f_j whenever i ≤ j, then f₁, f₂, ... converges uniformly.

The attempt at a solution
I was trying to reverse engineer the proof, but I'm stuck trying to figure out how the hypothesis, viz. f_i ≤ f_j whenever i ≤ j, comes into play. Any tips?

 

[tim_lou]

I assume you mean f = lim fn in the point-wise limit, and you want to show that
lim sup|f-f_n| = 0

I thought about it a little bit more, and I think a better approach would be to look at
gn(x)=f(x)-fn(x), gn>0.

these maps are continuous, you only need to show that gn converges uniformly to zero.
think about this map intuitively, and suppose it maps from [0,1] to [0,1].

draw a horizontal line at y=ϵ, your job is to show that eventually for some n, all of gn(x) lies below this line.
This proof becomes simple then as long as you think in terms of contradictions. What if for some x, none of gn(x) lies below this line?

 

[tim_lou]

And also, was f required to be continuous? otherwise, I don't think this proposition works because you can have fn(x)=x^2 on [0,1] that is monotonically nonincreasing (as n increases) and pointwise convergent to a discontinuous function. This convergence obviously cannot be uniform.

 

[e(ho0n3]

Yes, f is continuous (see first post).

Now if for some x, none of gn(x) lie below the line y = ϵ, then gn(x) = f(x) - fn(x) > ϵ for all n. But I think this contradicts the fact that fn converges point-wise to f: we can always find an N such that |f(x) - fn(x)| < ϵ for all n > N. We have not used the fact that f1(x) ≤ f2(x) ≤ ... though.

 

[tim_lou]

The fact that fn are increasing means that:

g(x) are always positive
gn(x) are always decreasing

also, think about it, how can you conclude that |gn(x)| < ϵ when n>n0 after you've chosen a n0? Here, you've implicitly used that inequality.

 

[e(ho0n3]

You made a typo by writing g(x) instead of gn(x) and decreasing instead of non-increasing. Anyways, I think I understand your point:

We have that g1(x) >= g2(x) >= ... and since E is compact, each gn reaches its maximum, say Mn, which means M1 >= M2 >= ... Furthermore, for each n, Mn >= 0 so the sequence M1, M2, ... is a bounded monotonic, and hence convergent, sequence. In effect, for any ϵ > 0, we can find an N such that |gn(x)| <= Mn < ϵ for all n > N, for all x.

That should do it. Thanks a bunch.

 

[e(ho0n3]

I made a mistake in my last post: I'm assuming that M1, M2, ... converges to 0. I've been attempting to prove this without any success. Could it be that M1, M2, ... doesn't necessarily converge to 0? If so, then my proof is ruined...

 

[Dick]

I'm not quite sure what the M's are, but there are three main points in the proof. i) Since the approach to the limit is monotonic, if |f_N(x)-f(x)|<e then |f_n(x)-f(x)|<e for all n>=N. ii) since f_n and f are continuous, if |f_N(x)-f(x)|<e then there is a neighborhood of x, B(x,N,e) where, say, |f_N(y)-f(y)|<2e for y in B(x,N,e). iii) E is compact, so it can be covered by a finite number of these neighborhoods. Can you put all of this together?

 

[e(ho0n3]

Mn = max{gn(x)}. So you're saying that E is the union of B(x_i, N_i, e), for i = 1, 2, ..., n, say. I don't know how to tie that together with i) and ii). Sorry.

 

[Dick]

I don't think Mn=max{gn(x):x in E} is very useful. Pick an e>0. For every x in E there is a N(x) such that |f_n(x)-f(x)|<e for all n>=N. x has a neighborhood where |f_n(x)-f(x)|<2e. E is compact. It's covered by a finite number of these neighborhoods. Isn't that enough for you to find an M such that |f_n(x)-f(x)|<2e for n>=M and all x in E?

 

[tim_lou]

didn't you already prove the result?

Yes, f is continuous (see first post).

Now if for some x, none of gn(x) lie below the line y = ϵ, then gn(x) = f(x) - fn(x) > ϵ for all n. But I think this contradicts the fact that fn converges point-wise to f: we can always find an N such that |f(x) - fn(x)| < ϵ for all n > N.

you've just shown sup|gn(x)|->0, didn't you?

 

[e(ho0n3]

I don't think Mn=max{gn(x):x in E} is very useful. Pick an e>0. For every x in E there is a N(x) such that |f_n(x)-f(x)|<e for all n>=N. x has a neighborhood where |f_n(x)-f(x)|<2e. E is compact. It's covered by a finite number of these neighborhoods. Isn't that enough for you to find an M such that |f_n(x)-f(x)|<2e for n>=M and all x in E?

Let me reiterate as much as I understand of your argument. Fix an x in E and an e > 0. By convergence, there is an N(x,e) such that |f_n(x) - f(x)| < e for all n > N(x,e). Let N > N(x,e). By continuity of f_n and f, there is a d(N,e) such that, where D is the metric on E, if D(y,x) < d(N,e), then |f_N(y) - f_N(x)| < e and |f(y) - f(x)| < e, yielding |f_N(y) - f(y)| < 2e. Let B(x, N, e) be the open ball of radius d(N,e). By compactness, E may be covered by a finite number of the B's, say B(x_i, N_i, e), i=1, 2, ..., n. Thus x belongs to B(x_i, N_i, e) for some i, so |f_{N_i}(x) - f(x)| < 2e. From here, how do I get that |f_n(x) - f(x)| < 2e for all n > N(e), where N(e) depends solely on e.

didn't you already prove the result?
you've just shown sup|gn(x)|->0, didn't you?

But I didn't use the fact that E is compact or that the f1 <= f2 <= ..., so I figured the proof must be incorrect.

 

[Dick]

Let me reiterate as much as I understand of your argument. Fix an x in E and an e > 0. By convergence, there is an N(x,e) such that |f_n(x) - f(x)| < e for all n > N(x,e). Let N > N(x,e). By continuity of f_n and f, there is a d(N,e) such that, where D is the metric on E, if D(y,x) < d(N,e), then |f_N(y) - f_N(x)| < e and |f(y) - f(x)| < e, yielding |f_N(y) - f(y)| < 2e. Let B(x, N, e) be the open ball of radius d(N,e). By compactness, E may be covered by a finite number of the B's, say B(x_i, N_i, e), i=1, 2, ..., n. Thus x belongs to B(x_i, N_i, e) for some i, so |f_{N_i}(x) - f(x)| < 2e. From here, how do I get that |f_n(x) - f(x)| < 2e for all n > N(e), where N(e) depends solely on e.But I didn't use the fact that E is compact or that the f1 <= f2 <= ..., so I figured the proof must be incorrect.

You have a finite number of neighborhoods covering E where |f(y)-f_n(x_i)(y)|<2e for y in each neighborhood and n>N(x_i). So yes. The N(x_i) depends on x and e. Now take N to be the max{N(x_i)}. It's the max of a finite number of integers.

 

[tim_lou]

You made a typo by writing g(x) instead of gn(x) and decreasing instead of non-increasing. Anyways, I think I understand your point:

We have that g1(x) >= g2(x) >= ... and since E is compact, each gn reaches its maximum, say Mn, which means M1 >= M2 >= ... Furthermore, for each n, Mn >= 0 so the sequence M1, M2, ... is a bounded monotonic, and hence convergent, sequence. In effect, for any ϵ > 0, we can find an N such that |gn(x)| <= Mn < ϵ for all n > N, for all x.

That should do it. Thanks a bunch.

Sorry for quoting the wrong post before. And indeed, you missed proving Mn->0. Dick's approach certainly works. However, this above also works, and I think it would help a lot to see this problem from a different angle.

Suppose Mn ≥ ϵ, Mn is always achieved by compactness, say g(xn)=Mn.

consider the set An={x|gn(x)≥ ϵ}

∩An =∅, why?
but gn(xn)=Mn≥ ϵ for all n means ∩An is not empty, why? (consider limit point compactness, say some subsequence xnj->x. Consider gn(x). gn is decreasing is crucial here)
contradiction.

 

[e(ho0n3]

I don't understand how using max{N(x_i)} helps because in the neighborhood B(x_i, N_i, e) where x lives, the only n for which |f_n(x) - f(x)| < 2e is true is N_i.

 

[Dick]

I don't understand how using max{N(x_i)} helps because in the neighborhood B(x_i, N_i, e) where x lives, the only n for which |f_n(x) - f(x)| < 2e is true is N_i.

If the convergence is monotone, then if |f_N(x)-f(x)|<2e then |f_n(x)-f(x)|<2e for any n>N. The f_n are increasing toward an upper bound of f. The f_n can only get closer to f as n increases.

 

[Dick]

Sorry for quoting the wrong post before. And indeed, you missed proving Mn->0. Dick's approach certainly works. However, this above also works, and I think it would help a lot to see this problem from a different angle.

Suppose Mn ≥ ϵ, Mn is always achieved by compactness, say g(xn)=Mn.

consider the set An={x|gn(x)≥ ϵ}

∩An =∅, why?
but gn(xn)=Mn≥ ϵ for all n means ∩An is not empty, why? (consider limit point compactness, say some subsequence xnj->x. Consider gn(x). gn is decreasing is crucial here)
contradiction.

Ah. I see what you are doing. If Mn>=e for all n, then you've got an intersection of nested nonempty closed sets on a compact space. Sure. That works too.

 

[e(ho0n3]

If the convergence is monotone, then if |f_N(x)-f(x)|<2e then |f_n(x)-f(x)|<2e for any n>N. The f_n are increasing toward an upper bound of f. The f_n can only get closer to f as n increases.

Got it. Thanks.

 

[e(ho0n3]

∩An =∅, why?

No clue.

but gn(xn)=Mn≥ ϵ for all n means ∩An is not empty, why? (consider limit point compactness, say some subsequence xnj->x. Consider gn(x). gn is decreasing is crucial here)

If ∩An is not empty, there is an x such that gn(x) = Mn for all n. I don't see how this is possible though.

 

[e(ho0n3]

∩An =∅, why?

I finally figured this out. If there was an x in the intersection, then gn(x) >= e for all n, but this implies that e <= lim gn(x), which is impossible.

but gn(xn)=Mn≥ ϵ for all n means ∩An is not empty, why?

If x is in An, then it is in A(n+1) because g(n+1)(x) >= gn(x) >= e. It follows that A1 contains A2 contains A3 etc. Also, each An is a closed subset by the continuity of gn. By compactness, as Dick noted, there is at least one element contained in each An so the intersection is not-empty.

Thanks a lot tim_lou.