I am attempting to come up with a uniformly convergent sequence of differentiable functions $g_{n}:(0,1)\to \mathbb R$ such that the sequence $\{g_{n}'\}$ does not converge.

I was thinking that $g_{n}(x)= \frac{\sin(nx + 3)} { \sqrt(n+1)}$ would work, however my friend does not think so. Why wouldn't this work? I am having trouble thinking of an example where this would be true.

  • $\begingroup$ "the sequence g′1, g′2, g′3, ... does not converge." Does not converge for every $x\in (0,1)$? At least one $x$? Does not converge uniformly? $\endgroup$ – zhw. Sep 1 '15 at 2:34
  • $\begingroup$ Does not converge pointwise in (0,1) $\endgroup$ – FillermaoGalms Sep 1 '15 at 20:12
  • $\begingroup$ So we just need divergence of $g_n'(x)$ for one $x\in (0,1)?$ $\endgroup$ – zhw. Sep 1 '15 at 20:32
  • $\begingroup$ a function $g_{n}$(x) such that $g_{n}$(x) converges uniformly in (0,1) but the sequence composed of its derivatives does not converge at all in (0,1) $\endgroup$ – FillermaoGalms Sep 2 '15 at 0:55

I think your example works. I'd like to simplify it though. Let's let $g_n(x) = \sin (nx)/\sqrt n, n = 1,2,\dots $ Clearly $|g_n(x)|\le 1/\sqrt n$ for all $x \in \mathbb {R},$ so $g_n \to 0$ uniformly on $\mathbb {R}.$

On the other hand, $g_n'(x) = \sqrt n \cos (nx).$ Claim: For every $x\in (0,1),$ $g_n(x)\ge \sqrt {n/2}$ for infinitely many $n,$ and $g_n(x)\le -\sqrt {n/2}$ for infinitely many $n.$ Thus the sequence $g_n(x)$ oscillates unboundedly as $n\to \infty,$ hence certainly doesn't converge.

The claim follows from this lemma: If $A$ is an arc on the unit circle of arc length greater than $1,$ and $x\in (0,1),$ then $ e^{inx} \in A$ for infinitely many $n.$ Proof: The sequence $e^{i nx}$ marches around the circle infinitely many times in steps of fixed arc-length $x<1.$ Now remember what your parents taught you: You can't step over a puddle that's larger than your stride. Apply this to our sequence: Because the steps are smaller than the length of $A,$ we have to land in $A$ at least once every orbit. That's the idea, and you can make it perfectly rigorous.

To get the claim from the lemma, consider the arcs $A= \{e^{it}: -\pi/4 < t < \pi/4\}, B = \{e^{it}: 3\pi/4 < t < 5\pi/4\}.$ The lengths of these arcs each equal $\pi/2>1.$ Apply the lemma and the claim follows.


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