If partial sums of derivatives diverge then derivative does not exist?

Question

Let $f_n :[a,b]\to \mathbb R$ be differentiable and such that $\sum_n f_n'(x)$ diverges for all $x \in [a,b]$. Let $\sum_n f_n $ converge uniformly on $[a,b]$ to $f$.

Does it follow that $f'$ does not exist for any $x \in [a,b]$? Or can somebody give a counterexample?

Answer 1

Let me give you a counterexample in the case of $\Bbb{C}$ instead of $\Bbb{R}$.

Let $n \in \mathbb{N}$. Differentiate two cases:

1. $n=2k$ is even. Define $f_n := \frac{1}{k^2} \cdot e^{k^2 \cdot i \cdot x}$.
2. $n=2k-1$ is odd. Define $f_n := -\frac{1}{k^2} \cdot e^{k^2 \cdot i \cdot x}$.

You can check that $\sum_n f_n$ converges uniformly (to zero), but the series of derivatives is $\sum_n (-1)^n i e^{(k(n))^2 \cdot i \cdot x}$ for suitable $k(n) \in \Bbb{N}$, which diverges.

But of course, the limit function (zero) is differentiable.

I hope that one can produce a counterexample for the real numbers using essentially the same idea.

EDIT: The problem I encountered when I wanted to produce a proof for the real case was that I took $\sin(k^2 x)$ instead of $e^{i k^2 x}$. But in that case it was not clear to me that $\sum_n \cos((k(n))^2 x) $ diverges for every(!) $x$.

EDIT 2: I have now found an example for the general case (with values in $\Bbb{R}$ instead of $\Bbb{C}$).

Let us first show that $(\cos(2^k \cdot x))_k$ does not converge to zero for any $x \in \Bbb{R}$. For this suppose that it does (for a contradiction).

Then $(\sin(2^k x))^2 \rightarrow 1$ (because of $\sin^2 + \cos^2 = 1$). Using the double angle formula, we derive $$0 = \lim_k \cos(2^{k+1} x) = \lim_k [(\cos(2^k x))^2 - (\sin(2^k x))^2] = -1,$$ a contradiction.

Now again differentiate two cases for $n \in \Bbb{N}$:

1. $n = 2k$ is even. Define $f_n := \frac{1}{2^k} \cdot \sin(2^k x)$.
2. $n=2k-1$ is odd. Define $f_n := -\frac{1}{2^k} \cdot \sin(2^k x)$.

As above, you can easily show that $\sum_n f_n$ converges uniformly (to zero), but the series of derivatives is $\sum_n (-1)^n \cos(2^{k(n)} x)$ and as shown above, the sequence $((-1)^n \cos(2^{k(n)} x))_n$ does not converge to zero [to be completely formal, note that $k(2n) = n$ so that the subsequence $((-1)^{2n} \cos(2^{k(2n)} x))_n = (\cos(2^{n} x))_n$ does not converge to zero, so that the whole sequence can not converge to zero.]