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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!

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    @WNY That's true, and I had missed that. Sorry as well... :-)2012-03-28

2 Answers 2

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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.

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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.

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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} $2012-03-28