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Uniform convergence and derivatives question

  1. Jun 22, 2012 #1
    In Spivak's Calculus, there is a theorem relating the derivative of the limit of the sequence {fn} with the limit of the sequence {fn'}.

    What I don't like about the theorem is the huge amount of assumptions required:

    " Suppose that {fn} is a sequence of functions which are differentiable on [a,b], with integrable derivatives fn', and that {fn} converges (pointwise) to f. Suppose, moreover, that {fn'} converges uniformly on [a,b] to some continuous function g. Then f is differentiable and f'(x) = lim n-->infinity fn'(x). "

    Are really EACH of these assumptions necessary for this to be true? Are there counterexamples for any combination of missing hypotheses? With all these assumptions the proof is quite easy, and I suspect this might be the reason, but in this case, how many of these assumptions can we get rid of?

    I've seen the counterexample of fn = sqrt(x^2+1/n^2), which converges uniformly to |x|, which is not differentiable. And also fn = 1/n*sin(n^2 x) which converges to 0 but the derivatives of fn do not always converge.

    But what about counterexamples involving non-integrable derivatives, non-uniform convergence to a continuous g, or uniform convergence to a function g which is not continuous? And doesn't uniform convergence of the derivatives imply at least pointwise convergence of the functions? etc, etc... I think you get my point (no pun intended)...
     
    Last edited: Jun 22, 2012
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  3. Jun 22, 2012 #2

    micromass

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    That seems a bit stronger than it needs to be. Here is a more general theorem:

    Let [itex](f_n:[a,b]\rightarrow \mathbb{R})[/itex] a sequence of functions such that

    1) [itex]f_n[/itex] is continuous on [a,b].
    2) [itex]f_n[/itex] is differentiable on ]a,b[.
    3) There exists an [itex]x_0\in [a,b][/itex] such that [itex]f_n(x_0)[/itex] converges.
    4) The sequence [itex](f_n\vert_{]a,b[})_n[/itex] converges uniformly.

    Then

    1) [itex](f_n)_n[/itex] is uniformly convergent
    2) The uniform limit f is differentiable on ]a,b[
    3) [itex]f_n^\prime(x)\rightarrow f^\prime(x)[/itex] for all x in ]a,b[

    Note that in complex analysis, if we use complex differentiability, then this statement simplifies even more!!
     
  4. Jun 23, 2012 #3
    Thanks for this reply! I thought about it a lot and I found counterexamples when uniform convergence of {fn'} is not assumed, which means that this condition is in fact essential. Also, I see how we could simplify the theorem to your version, but I still have a problem:

    Basically, if {fn} and {fn'} both converge uniformly on [a,b], I showed that the theorem is true. Now I need to show that uniform convergence of {fn'} together with convergence of {fn(x0)} implies uniform convergence of {fn}. This step seems difficult, and I could only prove it by assuming integrable fn'. The reason is that I can't make any link between fn' and fn if fn' is not integrable, since the FTC does not apply anymore...

    Do you know how this step is proven for non integrable fn' ? (It's already quite hard to think of fn' which are not integrable, and they probably never appear in practice, but I simply want to know, out of curiosity and satisfaction).
     
  5. Jun 23, 2012 #4

    micromass

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    Apply that [itex]f_n(x_0)[/itex] is Cauchy to find

    [tex]\|f_p(x_0)-f_q(x_0)\|\leq \frac{\varepsilon}{2}[/tex]

    Apply uniform convergence of [itex]f^\prime_n[/itex] to find

    [tex]\sup_{y\in ]a,b[}{\|f_p^\prime(y)-f_q^\prime(y)\|}\leq \frac{\varepsilon}{2(b-a)}[/tex]

    Apply the mean-value theorem to get

    [tex]\begin{eqnarray*}
    \|f_p(x)-f_q(x)\| & \leq & \|(f_p-f_q)(x)-(f_p-f_q)(x_0)\|+\|(f_p-f_q)(x_0)\|\\
    & \leq & \sup_{y\in ]a,b[}{\|f_p^\prime(y)-f_q^\prime(y)\|} +\|(f_p-f_q)(x_0)\|\\
    & \leq & \varepsilon
    \end{eqnarray*}
    [/tex]

    So [itex]f_n[/itex] is a uniform Cauchy sequence and thus uniformly convergent by completeness.
     
  6. Jun 25, 2012 #5
    You need uniform convergence of f'n(x) and to make it easier restrict f'n(x) to be continuous.

    As an example f_k(x) = (1\k)sin(x*k) converges uniformly on R to 0 but it's derivative cos(xk) doesn't converge.

    In the reals you don't have any restrictions on the derivative of the function based on the max/min values the function takes.

    In complex analysis you have the cauchy estimate and the cauchy integral formual, which lets you show that if a sequence of holomorphic functions converges to a function g, then g is holomorphic and the derivative of the sequence converges to g'.
     
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