i need some help to understand point wise and uniformly convergence and solve the following:

Let f be a series of functions defined by $f_n(x) := \dfrac{1}{n}e^{-n²x²}$.

Show that

  1. $f'_n(x)$ converges point wise on $[0, 1]$
  2. but not uniformly.
  3. compute $\lim \limits_{n \to \infty}f'_n(x)$ and $(\lim \limits_{n \to \infty}f_n(x))'$

My results so far:

  1. $$ \ f_n(x) := \dfrac{1}{n}e^{-n²x²} \Rightarrow f'_n(x) := -2nx\dfrac{1}{e^{n²x²}} $$ For x = 0: $$ \lim \limits_{n \to \infty}f'_n(0)=0 $$ For x = 1: $$ \lim \limits_{n \to \infty}f'_n(1)=\dfrac{-2n}{e^{n²}}=0 $$

    How i have to argue now that $f'_n$ converges point wise?

  2. $f'_n$ doesn't converge uniformly: $$ Definition \ of \ uniformly\ convergence \ says: \\ \forall \epsilon >0 \exists N \in \mathbb{N}: | f_n(x) -f(x)| < \epsilon\ ,\ \forall n\geq N\ and \ \forall x\in \mathbb{D} $$ so i have to show that $$ \exists \epsilon >0 \ that \ \forall N \in \mathbb{N} \exists n>N: | f_n(x) -f(x)| > \epsilon\ ,\ n\geq N\ and \ x\in \mathbb{D} $$

    can someone explain or give me some hints how i have to apply this def. to show that $f'_n$ doesnt converge uniformly?

  3. Both limits are 0.


  1. In the case where $x \neq 0, 1$, you can observe that asymptotically, an exponential function majorates any polynomial function, one key fact for taking limits. Proof of it, if you want it: http://www.proofwiki.org/wiki/Exponential_Dominates_Polynomial

  2. Suggestion: Take the second derivative of $f_n$, and use it to find a maxima for $f_n'$, which you will observe does not depend on $n$... Then draw your conclusion.

More thoroughly, $$f_n'' (x) = \frac{2n(2nx^2-1)}{e^{n^2x^2}}$$ has zeroes at $x = \pm 1/\sqrt{2}n$ and $$f_n'(1/\sqrt{2}n) = -\sqrt{2}e^{-1/2}$$ So each $f_n$ is attaining $c := -\sqrt{2}e^{-1/2}$ at some value. So I guess if $\epsilon < c$, there can't be any $N$ such that $\forall n > N$, $|f_n(x)| < \epsilon$ in the unit interval: not uniform convergence.

  • $\begingroup$ I wrote this in a hurry. If you have questions, feel free to ask. $\endgroup$ – user131331 May 1 '14 at 1:57
  • $\begingroup$ thanks dude! helps me a lot! :) $\endgroup$ – Landau May 1 '14 at 21:28

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