1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fourier series coefficients and convergence

  1. Jan 23, 2013 #1
    1. The problem statement, all variables and given/known data

    Third question of the day because this assignment is driving me crazy:

    Suppose that [itex]\left\{ f_{k} \right\} ^{k=1}_{\infty}[/itex] is a sequence of Riemann integrable functions on the interval [0, 1] such that

    [itex]\int ^{0}_{1} |f_{k}(x) - f(x)|dx \rightarrow 0 as k \rightarrow \infty[/itex].

    Show that [itex]\hat{f} _{k} (n) \rightarrow \hat{f} (n) [/itex] uniformly in n as [itex]k \rightarrow \infty[/itex]

    2. Relevant equations



    3. The attempt at a solution

    I can't seem to do this rigorously. I can only approach it intuitively. Since the integral of the absolute value tends to zero, I want to say that [itex]f_{k}(x) \rightarrow f(x)[/itex]. But I'm not sure how to show that. If [itex]f_{k}(x) \rightarrow f(x)[/itex] is indeed true, then is it not trivial that [itex]\hat{f} _{k} (n) \rightarrow \hat{f} (n) [/itex]? Furthermore, how would I show the convergence is uniform? Do I just have to use the epsilon definition? I also want to say for all epsilon greater than zero, there exists fk(x) such that [itex]|f_{k}(x) - f(x)| < \epsilon[/itex] since fk converges to f. But I need to first show that fk converges to f, using the fact that the integral of the absolute value of the difference between the two converges to zero! I am piecing it together but I don't know how to write it down in the form of an answer. Thanks in advance.

    Edit: f hat is the Fourier coefficient, I guess in complex form.
     
  2. jcsd
  3. Jan 23, 2013 #2

    MathematicalPhysicist

    User Avatar
    Gold Member

    Well it's really simple:

    [itex] |\hat{f_k}-\hat{f}|=|\int (f_k(x)-f(x))exp(-i n x) dx |[/itex]

    Now use triangle inequality, and the fact that |exp(-inx)|=1.
     
  4. Jan 23, 2013 #3
    [itex]

    \int ^{1}_{0} |f_{k}(x) - f(x)|dx \geq | \int ^{1}_{0} f_{k}(x) - f(x)dx | = | \int ^{1}_{0} ( f_{k}(x) - \int ^{1}_{0} f(x) ) dx | = | \int ^{1}_{0} ( f_{k}(x) (e^{-inx}) - \int ^{1}_{0} f(x) (e^{-inx}) ) dx | [/itex]

    because [itex] (e^{-inx}) = 1[/itex] for all [itex]n, x \geq 0[/itex]. We know n starts at 1, and the question tells us [itex] x \in [0, 1] [/itex].

    [itex]

    | \int ^{1}_{0} ( f_{k}(x) (e^{-inx}) - \int ^{1}_{0} f(x) (e^{-inx}) ) dx | = | \hat{f} _{k}(n) - \hat{f} (n) |

    [/itex]

    So by the comparison test, [itex] \int ^{0}_{1} |f_{k}(x) - f(x)|dx \rightarrow 0 [/itex] as [itex]k \rightarrow \infty \Rightarrow | \hat{f} _{k}(n) - \hat{f} (n) | \rightarrow 0 \Rightarrow \hat{f} _{k}(n) \rightarrow \hat{f} (n) [/itex]???

    If this is right (I don't think my last statement is correct), do I just use the epsilon definition of uniform convergence by finding N so that [itex]|f_{k}(n) - f(n)| < \epsilon[/itex]?
     
  5. Jan 23, 2013 #4

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No, it is not true that [itex]e^{-inx} = 1[/itex]. It is true if you add absolute value signs: [itex]|e^{-inx}| = 1[/itex]. Try starting like this:
    $$\left|\hat{f}_k - \hat{f}\right| = \left|\int_0^1 f_k(x) e^{-inx} dx - \int_0^1 f(x) e^{-inx} dx\right| = \left|\int_0^1 (f_k(x) - f(x))e^{-inx} dx\right| \leq \int_0^1 \left| (f_k(x) - f(x))e^{-inx} \right| dx = \ldots$$
     
  6. Jan 23, 2013 #5
    My mistake.

    [itex]

    \ldots = \left|\int_0^1 (f_k(x) - f(x))e^{-inx} dx\right| \leq \int_0^1 \left| (f_k(x) - f(x))e^{-inx} \right| dx = \int ^{1}_{0} |f_{k} (x) - f(x)||e^{-inx}| dx = \int ^{1}_{0} |f_{k} (x) - f(x)|(1) dx[/itex] which converges to zero. By the comparison test, [itex] \left|\hat{f}_k - \hat{f}\right| [/itex] also converges to zero.

    Then [itex] | \hat{f} _{k}(n) - \hat{f} (n) | \rightarrow 0 \Rightarrow \hat{f} _{k}(n) \rightarrow \hat{f} (n)[/itex]???
     
  7. Jan 23, 2013 #6

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, that's certainly true. If the absolute value of some quantity is approaching zero, then the quantity itself must also approach zero.
     
  8. Jan 23, 2013 #7
    Alright. Thanks for your help thus far. I have shown that [itex]| \hat{f} _{k}(n) \rightarrow \hat{f} (n)[/itex], but I need to show the convergence is uniform. My question is, will my conclusion help me show uniform convergence? I.e., can I work with what I've gotten to show uniform convergence, or must I start from scratch with the definition?
     
  9. Jan 23, 2013 #8

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Your proof already shows that the convergence is uniform, because your upper bound, [itex]\int ^{1}_{0} |f_{k} (x) - f(x)| dx[/itex], does not depend on [itex]n[/itex].
     
  10. Jan 24, 2013 #9
    Right! Well it wasn't really my proof. You and MathematicalPhysicist basically did it. While I do understand it, I (obviously) have a hard time getting things started. I still have another question I haven't done and the other one you're helping me with!

    Thanks again.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Fourier series coefficients and convergence
Loading...