Define $h(x)=|x|$ on $[-1,1]$ and extend it to $\mathbb R$ by defining $h(2+x) = h(x)$. This is a sawtooth function that is $0$ at even and $1$ at odd integers.

Furthermore define $h_n(x) = (1/2)^n h(2^n x)$ and $$ g(x) = \sum_{n=0}^\infty {1\over 2^n }h(2^n x) = \sum_{n \ge 0}h_n(x)$$

I showed that $g$ is defined everywhere and not differentiable at dyadic rationals. Now I have a question about an exercise. Let $x \in \mathbb R$ be a point that is not a dyadic rational. Define $x_m = p_m/2^m$ and $y_m = (p_m+1)/2^m$ for $p_m \in \mathbb Z$ so that $$ x_m < x < y_m$$ and $$ \lim x_m = x = \lim y_m$$

I am not sure about the following exercise:

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Why in (a) is it necessary to show $|g'_m(x) - g'_{m+1}(x)|=1$ and why is the first $<$ in (b) needed? My proof is as follows:

Because $g_m'(x) = \sum_{0\le n\le m} 2^{-n}h'(2^n x) = h'(x) \sum 1 = (m+1) h'(x)$ because of the second inequality in (b):

$$ g'(x) = \lim_{m \to \infty}{g(x_m) -g(x)\over x_m -x} \ge g'_m(x) \ge (m+1) h'(x) \to \infty$$

therefore $g'(x)$ does not exist.

Please can you tell me what is wrong with my proof?

  • $\begingroup$ possible duplicate of Sum of sawtooth function not differentiable at dyadic rational points $\endgroup$ – Semsem Mar 9 '14 at 14:17
  • $\begingroup$ Assuming your calculations are correct (which I did not check too thoroughly), you don't have to follow the instructions if you want to show that $g$ is not differentiable in the point under consideration, and your reasoning would be fine. You would not have done what the exercise asked you to do, I don't know how important that is for you. $\endgroup$ – Thomas Mar 9 '14 at 14:29
  • $\begingroup$ @Semsem Not a duplicate. The other question is about non-differentiability at dyadic rationals, this is about non-differentiability at other points. $\endgroup$ – user127096 Mar 9 '14 at 19:51
  • $\begingroup$ @ I took my vote back $\endgroup$ – Semsem Mar 9 '14 at 20:10

what is wrong with my proof?

The inequality $g_m'(x)\ge (m+1)h'(x)$. Why do you think this is true? The definition of $g_m$ implies that at points that are not dyadic rationals, $$g_m'(x) = \sum_{n=0}^m h'(2^nx)$$ Each term on the right is either $1$ or $-1$, but different terms may well have different signs.

Why is ... in (a), (b) needed?

The idea suggested to you is to relate (by two-sided inequality (b)) the divided differences of $g$ to the values of $g_m'(x)$. If $g'(x)$ existed, then the divided differences on the two sides of (b) would converge to $g'(x)$ as $m\to\infty$. But then, by the squeezing lemma, the sequence $g_m'(x)$ would have a limit. Part (a) shows that it does not.

By the way, $g$ is known as Takagi function and its graph as Takagi curve or blancmange curve. Dave Richeson made a Geogebra applet illustrating partial sums of the series.

  • $\begingroup$ Thank you for great answer. It is useful to know the name of the function. $\endgroup$ – blue Mar 25 '14 at 11:36
  • $\begingroup$ But where is (a) needed? If (b) holds taking the limit on both sides should suffice $g'(x)$ does not exist by squeezing lemma? $\endgroup$ – blue May 1 '14 at 14:55

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