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Here's the problem: Let

$ \varphi_0(x) = \begin{cases} x & \quad \text{if } 0≤x≤1/2 \\ 1-x & \quad \text{if } 1/2≤x≤1 \\ \end{cases}$

Extend $\varphi_0$ by periodicity with period 1 to the whole line R. Then let

$\varphi_n (x) = \frac{\varphi_0(4^n(x))}{4^n}$ and $f(x)= \sum_{n=0}^\infty \varphi_n(x)$

Show that

a) f(x) is continuous b) f(x) is nowhere differentiable

We're given a hint: Consider $\frac{f(x_0±4^{-n})-f(x_0)}{4^{-n}}$

I'm really having trouble figuring out what's going on here. I extrended the function as $\varphi_0(x+n) = \varphi(x) $, for $n= ±1,±2,...$

I guess I want to show that for all $\epsilon>0$ there exists a $\delta >0$ such that $|x-y| <\delta$ implies $|f(x)-f(y)| < \epsilon$.

So I have $|f(x)-f(y)| = \sum_{n=0}^\infty (\frac{\varphi_0(4^n(x))}{4^n} - \frac{\varphi_0(4^n(y))}{4^n})$ but I don't really know where to go from here.

I don't really see how the hint is helpful either.

Thanks

MathNoob
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  • Do you know anything about the uniform limit of continuous functions? Also, have you actually computed $\frac{f(x_0±4^{-n})-f(x_0)}{4^{-n}}$? If so, try using that computation in the limit definition of the derivative $f'(x_0)$. – Greg Martin Apr 30 '17 at 03:44

1 Answers1

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You just need to apply the Weierstrass M test and the fact that if a series of continuous function converges uniformly to a function $f$, then $f$ is also continuous

onyenna
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