# Everywhere Continuous Non-differentiable Function

Weierstrass introduced the idea that there exist functions that are continuous for every value of $x$ but do not possess a derivative at any value of $x$. We now consider the celebrated **Weierstrass function**, which exhibits this property. In this note, I will demonstrate that such a function exists using Weierstrass’s construction.

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## The Weierstrass Function

We define the function:

$$ f(x)= \sum_{n=0}^{\infty} b^n \cos (a^n \pi x) \tag{1}$$

This can also be written as:

$$ f(x) = \cos(\pi x) + b \cos(a \pi x) + b^2 \cos(a^2 \pi x) + \dots $$

Where:

- $a$ is an odd positive integer,
- $0 < b < 1$,
- $ab > 1 + \frac{3}{2} \pi$.

Under these conditions, the function $f(x)$ is continuous for all $x$ but does not have a finite derivative at any point.

## Continuity of the Weierstrass Function

To show that the function is continuous for all $x$, we first observe that:

$$ |b^n \cos(a^n \pi x)| \leq b^n $$

Since the series $\sum b^n$ is convergent, by Weierstrass's **M-Test for Uniform Convergence**, the series defined by equation (1) is uniformly convergent in every interval. Therefore, the function $f(x)$ is continuous for all $x$.

## Non-Differentiability of the Weierstrass Function

Next, we investigate whether the function is differentiable. The difference quotient for the function is:

$$ \frac{f(x+h) - f(x)}{h} = \sum_{n=0}^{\infty} b^n \frac{\cos[a^n \pi (x+h)] - \cos(a^n \pi x)}{h} \tag{2} $$

Let $S_m$ denote the sum of the first $m$ terms and $R_m$ denote the remainder after $m$ terms. Thus,

$$ \sum_{n=0}^{\infty} b^n \frac{\cos[a^n \pi (x+h)] - \cos(a^n \pi x)}{h} = S_m + R_m $$

## Applying the Mean Value Theorem

Using the **Lagrange Mean Value Theorem**, we have:

$$ \frac{|\cos[a^n \pi (x+h)] - \cos(a^n \pi x)|}{|h|} = |a^n \pi h \sin(a^n \pi(x + \theta h))| \leq a^n \pi |h| $$

Thus,

$$ |S_m| \leq \sum_{n=0}^{m-1} b^n a^n \pi = \pi \frac{a^m b^m - 1}{ab - 1} < \pi \frac{a^m b^m}{ab - 1} $$

## Behavior of the Remainder Term $R_m$

To further analyze the remainder term $R_m$, we make a particular choice of $h$. We write:

$$ a^m x = \alpha_m + \xi_m $$

where $\alpha_m$ is the integer nearest to $a^m x$ and $-1/2 \leq \xi_m < 1/2$. We now choose $h$ such that:

$$ a^m(x+h) = \alpha_m + \xi_m + h a^m $$

We select $h$ such that:

$$ \xi_m + h a^m = 1 \quad \text{which implies} \quad h = \frac{1 - \xi_m}{a^m} \tag{3} $$

As $m \to \infty$, $h \to 0$, and we evaluate the cosine terms as follows:

$$ \cos[a^n \pi (x+h)] = \cos[a^{n-m} \pi (\alpha_m + 1)] = (-1)^{\alpha_{m+1}} $$

Thus, we have:

$$ R_m = \frac{(-1)^{\alpha_m} + 1}{h} \sum_{n=m}^{\infty} b^n [2 + \cos(a^{n-m} \xi_m \pi)] \tag{4} $$

Since each term of the series in (4) is non-negative and the first term is positive, we find:

$$ |R_m| > \frac{b^m}{|h|} > \frac{2a^m b^m}{3} \tag{5} $$

## Concluding: Non-Differentiability

Finally, combining the results for $S_m$ and $R_m$, we find that:

$$ \left| \frac{f(x+h) - f(x)}{h} \right| = |R_m + S_m| \geq |R_m| - |S_m| > \left( \frac{2}{3} - \frac{\pi}{ab-1} \right) a^m b^m $$

Since $ab > 1 + \frac{3}{2}\pi$, we conclude that:

$$ \left( \frac{3}{2} - \frac{\pi}{ab-1} \right) > 0 $$

Thus, as $m \to \infty$ and $h \to 0$, the expression $\frac{f(x+h) - f(x)}{h}$ takes arbitrarily large values. Therefore, the derivative $f'(x)$ does not exist or is at least not finite for any $x$.

### Key Takeaways

**Weierstrass Function**is continuous for all $x$ but non-differentiable at every point.- Uniform convergence guarantees continuity, but the behavior of the difference quotient reveals non-differentiability.
- Hardy extended this result by showing that non-differentiability holds for $ab \geq 1$.