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Homework Help: Can anybody explain this to me? (Analysis)

  1. Oct 23, 2012 #1
    For x in [0, infinity), let [itex]f_n(x)= \frac{x}{n}[/itex]...

    Determine whether [itex]f_n[/itex] converges uniformly to f (the limit, which is equal to 0) on [0,1].

    Answer:

    Let [itex]\epsilon > 0[/itex] be given. Let [itex]N= \frac{1}{\epsilon} [/itex]. Then for n>N, [itex] | f_n(x) - 0 | = | \frac{x}{n} | \leq \frac{1/ \epsilon}{n} = \frac{1}{n \epsilon} < \epsilon[/itex] as desired.

    My questions:

    1) Why is [itex]| \frac{x}{n} | \leq \frac{1/ \epsilon}{n} [/itex]? How do we know that x is less than 1 over epsilon?

    2) Why is [itex] \frac{1}{n \epsilon} < \epsilon[/itex]?

    3) Finally, how did they know that N was supposed to be 1/epsilon?

    Thanks in advance
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 23, 2012 #2
    This "proof" is false.This sequence does not converge uniformly in that interval,because the suprimum for x is not final so certainly does not converge to 0.
     
  4. Oct 23, 2012 #3

    But this answer is from here (page 49, exercise 24.2 b)

    http://www.scribd.com/doc/70434268/Ross-Solutions
     
  5. Oct 23, 2012 #4

    HallsofIvy

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    I don't know what you mean by "not final". Since the sequence converges, it MUST converge uniformly on any closed and bounded (i.e. compact) interval.
     
  6. Oct 23, 2012 #5

    jbunniii

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    (I assume you mean page 47.) The proof for part (b) says "Claim: [itex]f_n \rightarrow f[/itex] uniformly on [0,1]." Part (c) shows by contradiction that it is not true on [itex][0,\infty)[/itex].

    I am not sure which part you are trying to solve:

     
  7. Oct 23, 2012 #6
    I was actually talking about part (b)...about why it converges uniformly on [0,1].
     
  8. Oct 23, 2012 #7

    HallsofIvy

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    Oops! I misread the problem.
     
  9. Oct 23, 2012 #8

    jbunniii

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    I read through the proof at your link, and it's a mess. My comments below:

    In general, it's not true. All you know is that [itex]0 \leq x \leq 1[/itex]. If [itex]0 < \epsilon < 1[/itex] (assumption not stated in the proof), then [itex]x \leq 1 < 1/\epsilon[/itex].

    This is certainly not true. Take [itex]\epsilon = 1/10[/itex], [itex]N = 10[/itex], and [itex]n = 11[/itex]. Then
    [tex]\frac{1}{n \epsilon} = \frac{1}{11 /10} = \frac{10}{11}[/tex]
    which is certainly not less than 1/10.

    I suggest ignoring that "solution" altogether. The statement is true, but the proof is bogus.
     
  10. Oct 23, 2012 #9

    jbunniii

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    I suggest starting with
    [tex]|f(x) - 0| = \left|\frac{x}{n}\right| \leq \left|\frac{1}{n}\right| \ldots[/tex]
    where the inequality is true because [itex]|x| \leq 1[/itex].
     
  11. Oct 23, 2012 #10

    So can I prove it like this?

    [itex] | f_n(x) - 0| = | \frac{x}{n} | \leq \frac{1}{n} [/itex]

    So when is 1/n equal to epsilon? It is when n=1/epsilon...so n needs to be greater than 1/epsilon. Do you think my answer is correct?
     
  12. Oct 23, 2012 #11

    jbunniii

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    Yes, if [itex]n > 1/\epsilon[/itex], then [itex]1/n < \epsilon[/itex], which is what you need. The key is that this choice of [itex]n[/itex] works for any [itex]x \in [0,1][/itex].
     
  13. Oct 23, 2012 #12
    Thanks a lot. I also have another question, if you don't mind...

    From the same link that I gave...for exercise 24.6 (b), we are asked if f_n converges uniformly on [0,1]. And the solution (in the link) says that [itex] | \frac{1-2xn}{n^2} | \leq \frac{1}{\sqrt{n}} [/itex]. I wasn't able to understand why that is true...
     
  14. Oct 23, 2012 #13

    jbunniii

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    That's because it is false. Consider [itex]x = 1[/itex], [itex]n = 2[/itex]. Then
    [tex]\left|\frac{1 - 2xn}{n^2}\right| = \left|\frac{1-4}{4}\right| = \frac{3}{4}[/tex]
    which is not less than [itex]1/\sqrt{2} \approx 0.7071[/itex].

    Whoever wrote the solution manual you are using is just plain wrong.

    You can obtain a bound that will work as follows:
    [tex]|1 - 2xn| \leq |1| + |2xn| = 1 + 2n|x| \leq 1 + 2n \leq n + 2n = 3n[/tex]
    Try using this to solve the problem. You'll have to use a different N from what the solution chose, but it should work.
     
  15. Oct 24, 2012 #14


    Thanks a lot :)
     
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