# Construction of a continuous real function nowhere differentiable

Rudin PMA p.154

Define $\phi$ as:

$$\phi(x)=|x| ,\;\;\forall x\in [-1,1] \;\; \text{and}\;\; \phi(x+2)=\phi(x)$$

Let $\delta_m=\frac{1}{2} 4^{-m}, \forall m\in\mathbb{Z}^+$ and fix $x\in\mathbb{R}$. Then, Rudin states that:

$$|\sum_{n=0}^m (\frac{3}{4})^n \frac{\phi(4^n (x+\delta_m)) - \phi(4^n x)}{\delta_m}|≧ 3^m - \sum_{n=0}^{m-1} 3^n$$

I don't understand why this inequality holds.

Please help me understand this..

Thank you in advance

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## 1 Answer

Just to be clear: Rudin defines $$\gamma_n = \frac{\varphi(4^n(x+\delta_m))-\varphi(4^nx)}{\delta_m}.$$ You want to show that $$\left\vert \sum_{n=0}^m \left(\frac{3}{4}\right)^n \gamma_n \right\vert \ge 3^m - \sum_{n=0}^{m-1} 3^n.$$

Did you read the sentence immediately before? It says:

When $0 \le n \le m$, (36) implies that $|\gamma_n| \le 4^n$. Since $|\gamma_m|=4^m$, we conclude that ...

So $$\left\vert \sum_{n=0}^m \left(\frac{3}{4}\right)^n \gamma_n \right\vert = \left\vert \pm 3^m + \sum_{n=0}^{m-1} \left(\frac{3}{4}\right)^n \gamma_n \right\vert$$ depending on whether $\gamma_m$ is positive or negative. Now you can proceed...

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I don't get the part $|\gamma_m|=4^m$.. Why is that? – Katlus Dec 15 '12 at 10:16
@Katlus: Take a look at (39) and the text that proceeds it. Read it carefully. – wj32 Dec 15 '12 at 10:30