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I've been trying this problem from Stein, but with no luck.

Consider the function $$f_{1}(x)=\sum_{n=0}^{\infty}{2^{-n} e^{2\pi i 2^{n} x} }.$$ a) Prove that $f_{1}$ satisfies $|f_{1}(x)-f_{1}(y)| \leq A_{\alpha}|x-y|^{\alpha}$ for each $\alpha \in (0,1)$.

b) $f_{1}$ is nowhere differentiable hence not of bounded variation.

It sounds beautiful and I was wondering if there's any nice proof. A friend tells me there's a more general theory about some so-called Hilbert functions which justify this, but I'm interested in something easier!

Thanks!

André Caldas
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Anna
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  • What is $\alpha$? – Martin Argerami Mar 28 '12 at 05:03
  • Do you mean that for each $\alpha$ there exists $A_\alpha$ such that $|f_1(x) - f_1(y)|\leq A_\alpha|x - y|^\alpha$? Or do you mean that there exists an $\alpha\in(0,1)$ and $A_\alpha > 0$ such that the above inequality holds? (I think it's the latter, since the former would imply that $f_1$ is Lipschitz). If it is the latter, then you can simply say that $f_1$ is $\alpha$-Hölder continuous; see here: http://en.wikipedia.org/wiki/H%C3%B6lder_condition. –  Mar 28 '12 at 05:04
  • The former actually! Edited – Anna Mar 28 '12 at 05:11
  • @WNY: Why would the former imply that $f_1$ is Lipschitz? – Did Mar 28 '12 at 05:20
  • @WNY: The former would only imply that $f$ is Lipschitz if the $A_\alpha$ are bounded, I think. – martini Mar 28 '12 at 05:35
  • @Didier, martini: The way I stated it, for $\alpha = 1$ there exists $A_1 > 0$ such that for all $x,y$, $|f_1(x) - f_1(y)| \leq A_1|x - y|$. Notice that I didn't make the assumption in the "former" that $\alpha\in(0,1)$. –  Mar 28 '12 at 05:54
  • @Anna: Ok, fine, so $f_1$ is $\alpha$-Hölder continuous for each $\alpha\in(0,1)$. It's clear now, thanks. –  Mar 28 '12 at 05:57
  • @WNY Your first comment leaves the impression that if a function is $\alpha$-Hölder for every $\alpha$ in $(0,1)$, then it is Lipschitz. (Note that the OP introduces $\alpha$ in $(0,1)$, not in $(0,1]$.) Which is not true. Your second comment makes things explicit. – Did Mar 28 '12 at 06:08
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    @Didier: The OP didn't introduce $\alpha$ and $A_\alpha$ in the original post, and I was making guesses; it was only after I posted my comment that the OP edited the original question. Sorry for the confusion. –  Mar 28 '12 at 06:11
  • @WNY That's true, and I had missed that. Sorry as well... :-) – Did Mar 28 '12 at 06:33
  • Regarding (b), this function is treated in Problem 8 in Chapter 5 in Stein and Shakarchi's Fourier Analysis. There you can find step-by-step hints. – Petra Axolotl May 20 '22 at 17:11

2 Answers2

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Triangle Inequality and Mean Value Theorem yield $$ \begin{align} |f_1(x)-f_1(y)| &\le\sum_{n=0}^\infty2^{-n}\left|e^{2\pi i2^nx}-e^{2\pi i2^ny}\right|\\ &=\sum_{n=0}^\infty2^{-n}\left|e^{2\pi i2^nx}-e^{2\pi i2^ny}\right|^{1-\alpha}\left|e^{2\pi i2^nx}-e^{2\pi i2^ny}\right|^\alpha\\ &\le\sum_{n=0}^\infty2^{-n}2^{1-\alpha}(2\pi2^n)^\alpha|x-y|^\alpha\\ &=2\pi^\alpha\frac{1}{1-2^{\alpha-1}}|x-y|^\alpha\\ &=A_\alpha|x-y|^\alpha \end{align} $$ Note that as $\alpha\to1^-$, $A_\alpha\to\infty$.

Hardy proves in Theorem $1.31$ of Weierstrass's Non-Differentiable Function that $f_1$ is nowhere differentiable.

robjohn
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Lets divide the sum into two: $$ \sum_{n=0}^{N}{2^{-n} e^{2\pi i 2^{n} x} } +\sum_{n=N+1}^{\infty}{2^{-n} e^{2\pi i 2^{n} x} }=S_1(x)+S_2(x). $$ The difference of the first sum can be estimated by the mean value theorem: $$ |\Delta S_1(x)|\le \sum_{n=0}^{N}{2^{-n} (2\pi 2^{n} |\Delta x|)} =2\pi (N+1)|\Delta x|, $$ and the second are marjorized by the sum of an infinite geometric progression: $$ |\Delta S_2(x)|\le \sum_{n=N+1}^{\infty}{2^{-n}}=2^{-N}. $$ Now for given $\Delta x$ one can choose $N$ s.t. both summands satisfy the required estimate.

Andrew
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  • It might be useful to compute the value of $A_\alpha$ that this method gives. Also note that $$ |\Delta S_2(x)|\le 2\cdot\sum_{n=N+1}^{\infty}{2^{-n}}=2^{1-N} $$ – robjohn Mar 28 '12 at 20:46