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Let $S(x)=\sum f_n(x)$ be pointwise convergent and differentiable on some $A\subseteq\Bbb R$, with all the $f_n$ differentiable on $A$. In the comments to this answer a user asked for an $S$ such that $\sum f_n'$ is pointwise convergent but not identical to $S'$. Out of curiosity I gave it a try. For such an $S$, I thought it's convenient to start from $T=\sum f_n'$, assuming the $f_n'$ are integrable and $T$ does not converge uniformly, in order to integrate the $f_n'$ on a convenient interval and take the sum of these resulting functions as $S$.

Soon enough I came up with $$T(x)=\sum_{n=1}^\infty f_n'(x)=\sum_{n=1}^\infty-x(1-x)^n=\begin{cases}x-1 &x\in(0,1] \\ 0&x=0.\end{cases} $$which suggests$$S(x)=\sum_{n=1}^\infty\frac{(1-x)^{n+1}(nx+x+1)}{(n+1)(n+2)}.$$ On $[0,1]$ one has $S(x)=\frac12(x-1)^2$ and thus $S'(x)=x-1$. In particular $S'(0)=-1\ne T(0),$ so $S'\ne T$.

I briefly tried to find a much more pathological case, but it wasn't apparent to me.

Question: Defining $S$ and $T$ as in the first paragraph, what is a specific $S$ such that $S'\ne T$ on the whole $A$, except maybe for countably many points?

Vincenzo Oliva
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  • maybe $e^{-1/x^2}=1-1/x^2+1/(2x^4)-...$ – janmarqz Jun 17 '18 at 02:08
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    @janmarqz: No. That is $f_n(x)=\frac{(-1/x^2)^n}{n!}$, $f_n'(x)=\frac{2nx(-1/x^2)^{n+1}}{n!}$ so $\sum f_n'(x)=\frac{2e^{-1/x^2}}{x^3}=\frac{d}{dx}\left(e^{-1/x^2}\right).$ – Vincenzo Oliva Jun 17 '18 at 02:31
  • for the derivatives at zero use http://www.wolframalpha.com/input/?i=D%5Bexp(-1%2Fx%5E2),x%5D – janmarqz Jun 17 '18 at 02:34
  • check this http://www.wolframalpha.com/input/?i=lim_%7Bx-%3E0%7D+D%5BD%5Bexp(-1%2Fx%5E2),x%5D,x%5D – janmarqz Jun 17 '18 at 02:37
  • and http://www.wolframalpha.com/input/?i=lim_%7Bx-%3E0%7D+D%5BD%5BD%5Bexp(-1%2Fx%5E2),x%5D,x%5D,x%5D to see the pattern at zero – janmarqz Jun 17 '18 at 02:38
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    @janmarqz The OP looks for $S'≠T$ holding globally, not at a single point. – Ѕᴀᴀᴅ Jun 17 '18 at 02:42
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    Udayan B. Darji obtained several results on this topic as part of his 1991 Ph.D. dissertation, with these results being published in 1996 as Limits of differentiable functions. I mentioned some of his results in this 18 January 2001 sci.math post, and later in this 29 December 2006 sci.math post I posted a more leisurely overview of his results than is given in his published paper. – Dave L. Renfro Jun 17 '18 at 18:08
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    A special case of one of Darji's results is the following. There exists a sequence ${f_n}$ of differentiable functions from $[0,1]$ to $\mathbb R$ that converges uniformly to a differentiable function $f$ such that ${f'_n}$ converges pointwise to some function $g$ and ${x: f'(x) \neq g(x)} = [0,1].$ Incidentally, one can easily translate these sequence results to series results by considering the sequence $f_1,$ $f_2 - f_1,$ $f_3 - f_2,$ $\ldots$ and then considering the sum (= limit of partial sums) of this new sequence. – Dave L. Renfro Jun 17 '18 at 18:15
  • @DaveL.Renfro: Very interesting, thank you. – Vincenzo Oliva Jun 18 '18 at 10:26

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First thing to do I think is to switch from series of functions to sequneces of functions by $$ g_n := \sum_{k=0}^n f_k, $$ since it's equiavlent, but easier to deal with.

I define my $g_n$ on $[0,1]$ by saying that I want $g_n$ to be $$ g_n(x) = \left\{ \begin{array}{lr} 0 : &x \in [0,1]\setminus(\bigcup_{k=1}^{n-1} [\frac{k}{n}-\frac{1}{n^{3/2}},\frac{k}{n}+\frac{2}{n^{3/2}}])\\ 1 : &x=\frac{k}{n}, k=1,\ldots,n-1 \\ -1 : &x=\frac{k}{n}+\frac{1}{n^{3/2}}, k=1,\ldots,n-1 \end{array} \right. $$ and we fill the gaps by linear functions. So these are teeth, that stay the same height, but are thinner and thinner. Now if we integrate those $g_n$s, we get some small "bubbles" on the intervals $[\frac{k}{n}-\frac{1}{2n^{3/2}},\frac{k}{n}+\frac{2}{2n^{3/2}}]$ that converge uniformly to 0, but $g_n$'s obviously don't converge to zero.

The last point is to show that it actually fails to converge in almost every point. Take an irrational $x \in (0,1)$, by Dirichlet's approximation theorem we have that for infinitely many irreducible rationals $$ \left|x-\frac{k}{n}\right| < \frac{1}{n^2}. $$ Take such a rational, say $k/n$. Now we have $$ g_n(x) = g_n\left(\frac{k}{n} + \left(x-\frac{k}{n}\right)\right) \approx 1 + n^{3/2}\left(x-\frac{k}{n}\right) < 1 + \frac{1}{n^{1/2}}, $$ so for infinitely many $n$'s $g_n(x)$ is arbitrarly close to 1.

Nikodem Dyzma
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