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Theorem - Uniform convergence

  1. Jun 14, 2010 #1
    1. The problem statement, all variables and given/known data

    Suppose:

    [tex]f(x)\ and\ f'(x)\ are\ continuous\ for\ all\ x \in R [/tex]

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

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


    3. The attempt at a solution


    [tex]\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)[/tex]

    [tex]\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[/tex]

    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.
     
    Last edited: Jun 15, 2010
  2. jcsd
  3. Jun 16, 2010 #2
    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]
     
    Last edited: Jun 16, 2010
  4. Jun 16, 2010 #3

    Dick

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    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.
     
  5. Jun 16, 2010 #4
    I left this out as a technical detail. [but wrote [tex]\max_{[a,b]}|f_n(x)-f(x)|[/tex], 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!
     
    Last edited: Jun 16, 2010
  6. Jun 16, 2010 #5

    Dick

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    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.
     
  7. Jun 16, 2010 #6
    I'll post rigorous proof, and will be very thankful for your opinion
     
  8. Jun 16, 2010 #7
    [tex]\mbox{f(x) and f'(x) are continuous for all x \in R. (*1)} [/tex]

    [tex]\mbox {Let a,b \in R\ so\ that\ without\ the\ loss\ of\ generality\ a<b.}[/tex]

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

    [tex]\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)[/tex]

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

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

    [tex]\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[/tex]

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

    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.
     
    Last edited: Jun 17, 2010
  9. Jun 17, 2010 #8

    Dick

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    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.
     
  10. Jun 17, 2010 #9
    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?

    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.
     
    Last edited: Jun 17, 2010
  11. Jun 17, 2010 #10

    Dick

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    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.
     
  12. Jun 17, 2010 #11
    My intuition just suggested me to play with:

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

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

    [Edit] No, its wrong! I go thinking again...
     
    Last edited: Jun 17, 2010
  13. Jun 17, 2010 #12
    Rewrite the expression

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

    as

    [tex]\frac{f(x+\frac{1}{n})-f(x)}{1/n}[/tex].

    Taking the limit as n goes to infinity gives [tex]f'(x)[/tex], 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.
     
    Last edited: Jun 17, 2010
  14. Jun 17, 2010 #13
    Can't you see I already found [tex]f_n(x)[/tex].

    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?
     
  15. Jun 17, 2010 #14
    Yes, look at my edit: can you use the Ascoli theorem?
     
  16. Jun 17, 2010 #15
    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]
     
    Last edited: Jun 17, 2010
  17. Jun 17, 2010 #16
    Did you get it? I wrote it out, here:

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

    [tex] B \le n | f(x + 1/n) - f(x + 1/m) | \le A[/tex]

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

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

    Then, for any [tex]\epsilon > 0[/tex] 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,

    [tex] | f_n(x) - f_m(x) | < \epsilon[/tex] 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!
     
  18. Jun 17, 2010 #17

    Dick

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    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).
     
  19. Jun 17, 2010 #18
    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].
     
  20. Jun 17, 2010 #19
    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, lets 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.
     
    Last edited: Jun 17, 2010
  21. Jun 17, 2010 #20

    Dick

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    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.
     
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