Proving Uniform Convergence of f_n(x) in [a,b]

[estro]

Homework Statement

Suppose:

$$f(x)\ and\ f'(x)\ are\ continuous\ for\ all\ x \in R$$

$$For\ all\ x \in R\ and\ for\ all\ n \in N\ f_n(x)=n[f(x+\frac{1}{n})-f(x)]$$

$$Prove\ that\ when\ a,b\ are\ arbitrary,\ f_n(x)\ is\ uniform\ convergent\ in\ [a,b]$$

The Attempt at a Solution

$$\lim_{n\rightarrow \infty} f_n(x)=\lim_{n\rightarrow \infty} n[f(x+\frac{1}{n})-f(x)] = \lim_{t\rightarrow0} \frac {f(x+t)-f(x)}{t}=f'(x)$$

$$\left \max_{[a,b]}|n[ f(x+\frac{1}{n}) -f(x)]-f'(x)|=|n[ f(x_0+\frac{1}{n}) -f(x_0)]-f'(x_0)|=|\frac {f(x_0+t)-f(x_0)}{t}-f'(x_0)|\rightarrow0 \right$$

I fear that I miss something terribly important.
*I left out all the little technical details to make things shorter.

[Edit] Will appreciate any remarks.

 

[estro]

I have used:
Weierstrass theorem [cont. function has its maximum...], limit comparison test, known lemmas and definitions of uniform convergence.[but left all these explanations from the 'proof' to make it short]

[thread hit second page but I try one last time]

 

[Dick]

One thing you are missing is explicitly saying that fn(x) is continuous. You've said fn(x) converges pointwise to f'(x), that's a start. But the most important thing you haven't done is use that [a,b] is compact. Your theorem wouldn't be true if the interval were unbounded.

 

[estro]

One thing you are missing is explicitly saying that fn(x) is continuous. You've said fn(x) converges pointwise to f'(x), that's a start. But the most important thing you haven't done is use that [a,b] is compact. Your theorem wouldn't be true if the interval were unbounded.

I left this out as a technical detail. [but wrote $$\max_{[a,b]}|f_n(x)-f(x)|$$, if I used rigor it would be too long]
But my reasoning is right? [I fear of misunderstanding uniform convergence, a thus misusing it's lemmas and definitions]Thanks again!

 

[Dick]

I left this out as a technical detail. [but wrote $$\max_{[a,b]}|f_n(x)-f(x)|$$]
But my reasoning is right? [I fear of misunderstanding uniform convergence, a thus misusing it's lemmas and definitions]

Thanks again!

You SAID max [a,b] |fn(x)-f'(x)|->0. But you certainly didn't prove it. Saying it's so doesn't prove it. Maybe leaving out 'technical details' isn't that great an idea.

 

[estro]

You SAID max [a,b] |fn(x)-f'(x)|->0. But you certainly didn't prove it. Saying it's so doesn't prove it. Maybe leaving out 'technical details' isn't that great an idea.

I'll post rigorous proof, and will be very thankful for your opinion

 

[estro]

$$\mbox{f(x) and f'(x) are continuous for all x \in R. (*1)}$$

$$\mbox {Let a,b \in R\ so\ that\ without\ the\ loss\ of\ generality\ a<b.}$$

$$\mbox {Let x \in [a,b]. (*2)}$$

$$\lim_{n\rightarrow \infty} f_n(x)=\lim_{n\rightarrow \infty} n[f(x+\frac {1} {n})-f(x)]=[t_n=1/n]=\lim_{t\rightarrow0} \frac {f(x+t)-f(x)} {t}=f'(x). \ (*3)$$

$$\mbox{(*1) and (*2) and (*3) } \Rightarrow\ g(x)=|f_n(x)-f(x)| \mbox{ is continuous for all x \in R.\ (*4)}$$

$$\mbox{(*2) and (*4)} \Rightarrow\ \mbox {Weierstrass Theorem } \Rightarrow\ \exists\ \max_{[a,b]} g(x)=g(c) \mbox { c \in [a,b].\ (*5)}$$

$$\mbox {(*5) and (*1) } \Rightarrow\ \forall\ x\in[a,b]\ \Rightarrow\ \sup_{[a,b]}|f_n(x)-f(x)|=|\frac {f(c+t)-f(c)} {t} -f'(c)|\rightarrow0$$

$$\Rightarrow \mbox { Basic lemma for uniform convergence } \Rightarrow\ f_n(x) \mbox{ is uniform convergent in [a,b], where a,b are arbitrary.}$$

QED

[#] "Basic lemma for uniform convergence" is sounds little funny but I'm sure you'll understand what I meant.
[#] I'm afraid of having the wrong intuition about uniform convergence, I hope you will be be able to say if I understand the idea from my proof.
[#] If only I could prove g(x) is monotonic [such thing is possible?] then my proof would be a lot shorter, using Dini Theorem.

 

[Dick]

You are suppressing a dependence. If c is the point where |f_n(x)-f'(x)| is a max then c depends on n. Write it c_n. Now it's true that |f_n(a)-f'(a)|->0 as n->infinity at any FIXED point a. But it's not necessarily true that |f_n(c_n)-f'(c_n)|->0. The point c_n could be jumping around.

 

[estro]

I have problems understanding the dependency between n and x with maximum of a function.
Do I need simply construct sequence of points in [a,b] where for every n, there is a point c_n where g(c_n) is maximum?

...
But it's not necessarily true that |f_n(c_n)-f'(c_n)|->0. The point c_n could be jumping around.

This I don't understand.
I know that f_n(x_0)->f'(x_0) so |f_n(x_0)-f'(x_0)|->0 [Am I right?] so I need somehow find one n who is good for all x in [a,b].
I know that g(x) is continuous in [a,b] so g(x) have a point where it get its maximum...

Little bit confused, but I try to think more about it.

 

[Dick]

It isn't the difference between f_n(x_0)->f'(x_0) and |f_n(x_0)-f'(x_0)|->0, they say the same thing. It's the difference between f_n(x_0)->f'(x_0) and f_n(x_n)->f'(x_n) where the function argument depends on n. A simple example that show what could go wrong is f_n(x)=1/cosh(x-n). f_n(x_0)->0 for every x_0 as n->infinity. But if you define c_n=n, then f_n(c_n)=1.

 

[estro]

My intuition just suggested me to play with:

I choose n from which this is true:
$$| \max_{[a,b]} g(x) - \min_{[a,b]} g(x) |< \epsilon$$

So now I have n who is good for all x.

[Edit] No, its wrong! I go thinking again...

 

[some_dude]

Rewrite the expression

$$n[f(x+\frac{1}{n})-f(x)]$$

as

$$\frac{f(x+\frac{1}{n})-f(x)}{1/n}$$.

Taking the limit as n goes to infinity gives $$f'(x)$$, which exists everywhere by your construction. And then to show uniform convergence on a closed interval [a, b] just apply one of the elementary calculus theorems (cts function on compact set).

EDIT:

Oops, you got that much yourself. And it's not quite so simple I just realized. Do you know Arzela-Ascoli theorem? If so, you can use that. And if not, you need to do a diagonalization type proof I think.

 

[estro]

Rewrite the expression

$$n[f(x+\frac{1}{n})-f(x)]$$

as

$$\frac{f(x+\frac{1}{n})-f(x)}{1/n}$$.

Can't you see I already found $$f_n(x)$$.

Taking the limit as n goes to infinity gives $$f'(x)$$, which exists everywhere by your construction. And then to show uniform convergence on a closed interval [a, b] just apply one of the elementary calculus theorems (cts function on compact set).

The heart of the problem is to find N who is good for all x.
I'm fully aware of "elementary calculus theorems" the problem is not applying them but doing it with sense. Are you sure you know what is uniform convergence?

 

[some_dude]

Can't you see I already found $$f_n(x)$$.
The heart of the problem is to find N who is good for all x.
I'm fully aware of "elementary calculus theorems" the problem is not applying them but doing it with sense. Are you sure you know what is uniform convergence?

Yes, look at my edit: can you use the Ascoli theorem?

 

[estro]

Yes, look at my edit: can you use the Ascoli theorem?

I didn't see such theorem in my book, so I should be able to prove it using more basic tools and more advanced reasoning.
My main goal is to understand the concept of uniform convergence.

Anyway I appreciate you desire to help, thank you.

[Edit] I am working on new proof using Cauchy Criterion for Uniform Convergence. [Basic and powerful theorem]

 

[some_dude]

Did you get it? I wrote it out, here:

For natural numbers n,m with m <= n, you can manipulate $$| f_n(x) - f_m(x) |$$ bu using the definitions to get the inequality:

$$B \le n | f(x + 1/n) - f(x + 1/m) | \le A$$

The real numbers B and A >= 0 exist since f' is cts on compact [a, b], and canceling the n gives:

$$B/n \le | f(x + 1/n) - f(x + 1/m) | \le A/n$$.

Then, for any $$\epsilon > 0$$ just choose N sufficiently large that the difference between the leftside and the right side is less then epsilon, then you know the same is true for the expression in the middle. And so,

$$| f_n(x) - f_m(x) | < \epsilon$$ for n, m greater than N. And therefore the sequence is uniformly Cauchy. But we also know (as you showed in the first post), the sequence is (pointwise) convergent, and therefore Uniformly Cauchy implies it is uniformly convergent.

...now damnit i need to get back to work!

 

[Dick]

You could also use that f(x) has a bounded derivative on [a,b]. That tells you a lot about the possible behavior of f(x).

 

[snipez90]

You certainly don't need Arzela-Ascoli (ok so to be honest I have forgotten what Arzela-Ascoli actually states). Use the fact that f'(x) is continuous on the compact interval [a,b+1] and hence uniformly continuous on that interval. Also apply the mean value theorem to f over the interval [x, x + 1/n].

 

[estro]

Check out my thinking:

I know that |f_n(x)-f'(x)|=g(x) is continuous on [a,b], so it got its maximum somewhere, let's say at c.
Now I'll chose n so that |f_n(c)-f'(c)|<e.
Because the difference between f_n(x) and f'(x) is the biggest in c, the n I choose should be good for all x in [a,b].

What's wrong with my reasoning?
This is what I tried to do in my proof.

 

[Dick]

Check out my thinking:

I know that |f_n(x)-f'(x)|=g(x) is continuous on [a,b], so it got its maximum somewhere, let's say at c.
Now I'll chose n so that |f_n(c)-f'(c)|<e.
Because the difference between f_n(x) and f'(x) is the biggest in c, the n I choose should be good for all x in [a,b].

What's wrong with my reasoning?
This is what I tried to do in my proof.

You picked a c for a single n. Now you want to go back and change n? Then you have to change c! You have GOT to see that there is something wrong there. Here's a general outline of how you actually want to proceed. For any given x you can choose an N(x) such that for all n>N(x) |f_n(x)-f'(x)|<e. Then you want to show that the properties of f_n and f' let you extend that so you can actually say than |f_n(y)-f'(y)|<e' for y in an open neighborhood of x (where e' is some function of e) for all n>N(x). Now you have an open cover of [a,b]. Use compactness.

 

[estro]

You picked a c for a single n. Now you want to go back and change n?

No, I wanted to use the same n.

So you say that if $$\max_{[a,b]} g(x)=g(c)$$ it is still possible that for a bigger n, $$\max_{[a,b]} g(x)> g(c)$$ right? If yes my idea is wrong.

Sorry if I'm too slow here, but I think this is very important point I don't fully understand.

 

[Dick]

No, I wanted to use the same n.

So you say that if $$\max_{[a,b]} g(x)=g(c)$$ it is still possible that for a bigger n, $$\max_{[a,b]} g(x)> g(c)$$ right? If yes my idea is wrong.

Sorry if I'm too slow here, but I think this is very important point I don't fully understand.

Well, sure. There's nothing about the problem that would say the convergence is monotone, is there?

 

[estro]

Now, I understand why I'm wrong.
I'll keep thinking.

 

[estro]

$$f_n(x) \rightarrow f'(x) \Rightarrow\ \forall\ n>N\ |f_n(c)-f'(c)|<\epsilon$$

$$t_n=c-1/n\ \Rightarrow\ |f_n(t_n)-f'(t_n)| \rightarrow 0$$

Now I only need to use uniform continuity of g(x) in [a,b].

Just a rough idea, what you think? You gave me obvious hint, only now I get it. [I think]

 

[Dick]

$$f_n(x) \rightarrow f'(x) \Rightarrow\ \forall\ n>N\ |f_n(c)-f'(c)|<\epsilon$$

$$t_n=c-1/n\ \Rightarrow\ |f_n(t_n)-f'(t_n)| \rightarrow 0$$

Just a rough idea, what you think? You gave me obvious hint, only now I get it. [I think]

That's WAY too rough. I'm not even sure what it means. Look, you know for any x you can make f'(x)-f_n(x) as small as you want for n>N(x). Now what you want to show is that you can make f'(y)-f_n(y) as small as you want for y in some open interval around x. You can make f'(y)-f'(x) small just because f' is continuous. But you can't control the radius because that's ALL you know about f'. You can make f_n(y)-f_n(x) small to match that radius because f_n has a bounded derivative. Have you shown this yet? You have more control on f_n than you have on f'. This isn't a terribly easy problem if you aren't used to thinking about stuff like this. Don't feel stupid. But do stop using really formal language to describe a proof until you can get a gut feeling for what's going on.

 

[Dick]

PS. As some-dude pointed out earlier, this is a special case of Arzela-Ascoli. It probably wouldn't hurt to look that up and convince yourself the the f_n are equicontinuous. Then once you know how to prove it using that theorem, go back and try to prove it from scratch.

 

[estro]

I'm not sure what N(x) means.
Got so much hints and guidance but I struggle to get intuition about how to deal with $$\max_{[a,b]} |f_n(x)-f'(x)|$$ and get rid of dependency in x.
This problem makes me ponder about my mathematical ability...
I'm really frustrated.

 

[Dick]

I'm not sure what N(x) means.
Got so much hints and guidance but I struggle to get intuition about how to deal with $$\max_{[a,b]} |f_n(x)-f'(x)|$$ and get rid of dependency in x.
This problem makes me ponder about my mathematical ability...
I'm really frustrated.

N(x) means that when you pick N for an x using pointwise convergence, that N is going to depend on x. Hence it's best to write N(x) so you don't think it applies to all x. You might want to pick a final N to be the max of all N(x), but that won't work because there are an infinite number of x. The idea is to show you can make f'-f_n small on an open interval around x. Then use compactness to pick a finite subcover. Now you only have a finite number of N's so you can use the max. I don't know what else to say.

 

[Dick]

Would you try, please, to prove you can make f' and f_n close on an open interval around x. This isn't just a substitute into formula problem. Think. Ok?

 

[estro]

Would you try, please, to prove you can make f' and f_n close on an open interval around x. This isn't just a substitute into formula problem. Think. Ok?

$$\mbox {Let } \epsilon<0,\ x_0 \in [a,b]$$

$$\lim_{n\rightarrow\infty} f_n(x)=f'(x)\ \Rightarrow\ \forall\ n>N\ |f_n(x_0)-f'(x_0)|< \frac {\epsilon}{3} \mbox { (*1)}$$

$$f_n(x) \mbox { is continuous in R } \Rightarrow\ \forall\ |x-x_0|< \delta_n,\ |f_n(x)-f_n(x_0)|< \frac {\epsilon}{3} \mbox { (*2)}$$

$$f'(x) \mbox { is continuous in R } \Rightarrow\ \forall\ |x-x_0|< \delta,\ |f'(x_0)-f'(x)|< \frac {\epsilon}{3} \mbox { (*3)}$$

$$t_n=\min \{\delta_n, \delta \}$$ $$\Rightarrow\ \mbox {(*1) and (*2) and (*3) } \Rightarrow\ \forall\ |x-x_0|< t_n$$

$$|f_n(x)-f'(x)|=|f_n(x)-f_n(x_0)+f_n(x_0)-f'(x_0)+f'(x_0)-f'(x)| \leq |f_n(x)-f_n(x_0)|+|f_n(x_0)-f'(x_0)|+|f'(x_0)-f'(x)|< \frac {\epsilon}{3}+\frac {\epsilon}{3}+\frac {\epsilon}{3}<\epsilon$$

I have hard time to get sound intuition about this because the neighborhood around x is not fixed [changes for every n], and this makes me uncomfortable.

...
The idea is to show you can make f'-f_n small on an open interval around x. Then use compactness to pick a finite subcover. Now you only have a finite number of N's so you can use the max. I don't know what else to say.

Now I'm trying to think about this. I'm not familiar with compactness theorem. [But I think Weierstrass Theorem can also help]

I have an idea now to play with $$t_n= \{x\ |\ \max_{[a,b]} |f_n(x)-f'(x)|\}$$, like I did with $$\mbox {x_0}$$ in the above proof.

 

[Dick]

$$\mbox {Let } \epsilon<0,\ x_0 \in [a,b]$$

$$\lim_{n\rightarrow\infty} f_n(x)=f'(x)\ \Rightarrow\ \forall\ n>N\ |f_n(x_0)-f'(x_0)|< \frac {\epsilon}{3} \mbox { (*1)}$$

$$f_n(x) \mbox { is continuous in R } \Rightarrow\ \forall\ |x-x_0|< \delta_n,\ |f_n(x)-f_n(x_0)|< \frac {\epsilon}{3} \mbox { (*2)}$$

$$f'(x) \mbox { is continuous in R } \Rightarrow\ \forall\ |x-x_0|< \delta,\ |f'(x_0)-f'(x)|< \frac {\epsilon}{3} \mbox { (*3)}$$

$$t_n=\min \{\delta_n, \delta \}$$ $$\Rightarrow\ \mbox {(*1) and (*2) and (*3) } \Rightarrow\ \forall\ |x-x_0|< t_n$$

$$|f_n(x)-f'(x)|=|f_n(x)-f_n(x_0)+f_n(x_0)-f'(x_0)+f'(x_0)-f'(x)| \leq |f_n(x)-f_n(x_0)|+|f_n(x_0)-f'(x_0)|+|f'(x_0)-f'(x)|< \frac {\epsilon}{3}+\frac {\epsilon}{3}+\frac {\epsilon}{3}<\epsilon$$

I have hard time to get sound intuition about this because the neighborhood around x is not fixed [changes for every n], and this makes me uncomfortable.




Now I'm trying to think about this. I'm not familiar with compactness theorem. [But I think Weierstrass Theorem can also help]

I have an idea now to play with $$t_n= \{x\ |\ \max_{[a,b]} |f_n(x)-f'(x)|\}$$, like I did with $$\mbox {x_0}$$ in the above proof.

You are absolutely right to be worried about the n dependence. Suppose you could show |f_n(x)-f_n(y)|<=M*|x-y| for some constant M, independent of n. Would that help?

 

[estro]

$$\mbox {Let } \epsilon<0,\ \ x_1,x_2 \in [a,b]$$

$$\lim_{n\rightarrow\infty} f_n(x)=f'(x)\ \ \ (*1)$$

$$(*1)\ \Rightarrow\ \forall\ n>N_1\ \ |f_n(x_1)-f'(x_1)|< \frac {\epsilon}{3}\ \ \ (*2)$$

$$(*1)\ \Rightarrow\ \forall\ n>N_2\ \ |f'(x_2)-f_n(x_2)|< \frac {\epsilon}{3}\ \ \ (*3)$$$$f'(x) \mbox{ is continuous in R } \Rightarrow\ \forall\ \ |x_1-x_2|<\delta\ \ |f'(x_1)-f'(x_2)|< \frac {\epsilon}{3}\ \ \ (*4)$$$$\mbox{(*2) and (*3) and (*4) } \Rightarrow\ \forall\ \ |x_1-x_2|<\delta\ and\ \forall\ n > N= \max\{ N_1,N_2 \}$$

$$|f_n(x_1)-f_n(x_1)|=|f_n(x_1)-f'(x_1)+f'(x_1)-f'(x_2)+f'(x_2)-f_n(x_2)| \leq |f_n(x_1)-f'(x_1)|+|f'(x_1)-f'(x_2)|+|f'(x_2)-f_n(x_2)|<\epsilon$$I'm almost sure this proof is right, but I only proved that f_n(x) is uniformly convergent on every open interval around some x in [a,b].

I'm thinking now to do what I did in the above proof with: x_1 = where f'(x) gets its maximum and x_2=where f'(x) gets its minimum.

 

[Dick]

Sorry, I'm not going to buy that. I don't think you've proved anything. What you have proved is that i) you really don't understand the issue and ii) that you are pretty good at ignoring good advice. |f_n(x)-f_n(y)|<=M*|x-y| for some constant M, independent of n. That is what you really need. Wouldn't that be helpful? Would you know how to prove it if you had to?