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- Apr 14, 2013

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I am looking at the following example of a continuous function $f:\mathbb{R}\rightarrow \mathbb{R}$ that is not differentiable at any $x\in \mathbb{R}$.

For $x\in [-1,1]$ we define $\phi (x)=|x|$ and then we extend $\phi$ to the whole $\mathbb{R}$ such that $\phi (x+2)=\phi (x)$.

Then the desired $f:\mathbb{R}\rightarrow \mathbb{R}$ is $$f(x)=\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx\right )$$

Then $f$ is continuous and bounded on $\mathbb{R}$.

For $x_0\in \mathbb{R}$ and $k\in \mathbb{N}^{\star}$ take $\delta_k=\pm \frac{1}{2}4^k$, where the sign is chosen such that there is no integer between $4^kx_0$ and $4^k(x_0+\delta_k)$ (it is $4^k\mid \delta_k$).

Define $\gamma_n=\frac{1}{\delta_k}\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]$.

For $n>k$ it is $\gamma_n=0$ and for $0\leq n\leq k$ it is $|\gamma_n|\leq 4^n$.

Since $|\gamma_k|=4^k$ we have that $$\left |\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\right |=\left |\sum_{n=0}^k\left (\frac{3}{4}\right )^n\gamma_n\right |\geq 3^k-\sum_{n=0}^{k-1}3^n=\frac{1}{2}\left (3^k+1\right )$$ so for $k\rightarrow +\infty$, $\delta_k\rightarrow 0$ and $3^k\rightarrow +\infty$, it follows that....

This is not very clear for me.

Let's start from the beginning.

Why do we define in that they the function $\phi$ ?

Then $f$ is continuous and bounded. Could you give me a hint how we can see that?