Continuous Nowhere Differentiable Function

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There exists a real-valued function $f : \left[{0 \,.\,.\, 1}\right] \to \left[{0 \,.\,.\, 1}\right]$, such that:

$(1): \quad f$ is continuous
$(2): \quad f$ is nowhere differentiable.


Let $C \left[{0 \,.\,.\, 1}\right]$ denote the set of all continuous functions $f: \left[{0 \,.\,.\, 1}\right] \to \R$.

By Continuous Functions on Closed Interval are Complete, $C \left[{0 \,.\,.\, 1}\right]$ is a complete metric space under the supremum norm $\left\|{\cdot}\right\|_\infty$.

Let $X$ consist of the $f \in C\left[{0 \,.\,.\, 1}\right]$ such that:

$f\left({0}\right) = 0$
$f\left({1}\right) = 1$
for any $0 \le x \le 1$, we have $0 \le f\left({x}\right) \le 1$.

Then we have the following lemma:

Lemma 1

$X$, defined as above, is a complete metric space (under $\left\|{\cdot}\right\|_\infty$).

Proof of Lemma 1

For every $n \in \N$, let $f_n \in X$. Furthermore, suppose that in $C\left[{0 \,.\,.\, 1}\right]$ we have:

\((1):\)      \(\displaystyle \lim_{n \to \infty} \left\Vert{f_n - f}\right\Vert_\infty\) \(=\) \(\displaystyle 0\)                    

If we can prove that $f \in X$, we know $X$ contains all its limit points, and hence is closed by Closed Set iff Contains all its Limit Points.

Subsequently, Topological Completeness is Weakly Hereditary assures us that $X$ will then be complete.

Let us now prove that $f \in X$.

If $f\left({0}\right) \ne 0$, we have $\left\|{f_n - f}\right\|_\infty \ge|f_n(0)-f(0)|=|f(0)|>0$ for all $n \in \N$.

This would contradict equation $(1)$, and hence $f \left({0}\right) = 0$.

Similarly, we find that necessarily $f \left({1}\right) = 1$.

Also, for all $n \in \N$ and $x \in \left[{0 \,.\,.\, 1}\right]$, we have that $0 \le f_n\left({x}\right) \le 1$.

Assume that there is an $x \in \left[{0 \,.\,.\, 1}\right]$ such that either $f\left({x}\right) < 0$ or $f\left({x}\right) > 1$.

We see that it must be that $\left\|{f_n - f}\right\|_\infty \ge |f_n(x)-f(x)|>0$ for any $n \in \N$, contradicting $(1)$.

Therefore, $f \in X$, and hence $X$ is complete.


For every $f \in X$, define $\hat f : \left[{0 \,.\,.\, 1}\right] \to \R$ as follows:

$\hat f\left({x}\right) = \begin{cases} \frac 3 4 f \left({3x}\right) & \text{if } 0 \le x \le \frac 1 3 \\ \frac 1 4 + \frac 1 2 f\left({2-3x}\right) & \text{if } \frac 1 3 \le x \le \frac 2 3 \\ \frac 1 4 + \frac 3 4 f\left({3x-2}\right) & \text{if } \frac 2 3 \le x \le 1 \end{cases}$

We have the following lemma:

Lemma 2

$\hat \cdot: X \to X$ is a contraction mapping.

Furthermore, we have the following inequality:

$\forall f,g \in X: \left\|{\hat f - \hat g}\right\|_\infty \le \frac 3 4 \left\|{f - g}\right\|_\infty$

Proof of Lemma 2


The Contraction Mapping Theorem assures existence of a unique $h \in X$ with $\hat h = h$.

This $h$ is a continuous function (as $h \in X \subset C\left[{0 \,.\,.\, 1}\right]$), and we will show that it is nowhere differentiable.

To do this, we establish the following lemma:

Lemma 3

For every $n \in \N$ and $k \in \left\{{1, 2, 3, 4, \ldots, 3^n}\right\}$, we have the following inequality:

$\left|{h \left({ \dfrac {k-1} {3^n} }\right) - h \left({ \dfrac k {3^n} }\right)}\right| \ge 2^{-n}$

Proof of Lemma 3

For any $n \in \N$ and $k \in \left\{{1, 2, 3, \ldots, 3^n}\right\}$:

$1 \le k \le 3^n \implies 0 \le \dfrac{k-1} {3^{n+1}} < \dfrac k {3^{n+1}} \le \dfrac 1 3$
$3^n < k \le 2 \cdot 3^n \implies \dfrac 1 3 \le \dfrac {k-1} {3^{n+1}} < \dfrac k {3^{n+1}} \le \dfrac 2 3$
$2 \cdot 3^n < k \le 3^{n+1} \implies \dfrac 2 3 \le \dfrac {k-1} {3^{n+1}} < \dfrac k {3^{n+1}} \le 1$


Next, fix $a \in \left[{0 \,.\,.\, 1}\right]$.

We will show that $h$ is not differentiable at $a$.

We will construct a sequence $\left\langle{t_n}\right\rangle$ with elements in $ \left[{0 \,.\,.\, 1}\right]$, such that we have the following limit:

$\displaystyle \lim_{n \to \infty} t_n = a$

To this end, let $n \in \N$ be arbitrary.

Now let $k$ be the unique largest element of $\left\{{ 1, 2, 3, 4, \ldots, 3^n}\right\}$ such that we have:

$(k-1) 3^{-n} \le a \le k 3^{-n}$

By the triangle inequality, we have the following:

$\left|{ h\left({ \dfrac {k-1} {3^n} }\right) - h\left({a}\right) }\right| + \left|{ h\left({a}\right) - h\left({ \dfrac {k} {3^n} }\right) }\right| \ge \left|{ h\left({ \dfrac {k-1} {3^n} }\right) - h\left({ \dfrac {k} {3^n} }\right) }\right| \ge 2^{-n}$

Next, let $t_n$ be either $\dfrac {k-1} {3^n}$ or $\dfrac {k} {3^n}$, such that the following equation is satisfied:

$\left|{ h\left({t_n}\right) - h\left({a}\right) }\right| = \max\left\{{ \left|{ h\left({ \dfrac {k-1} {3^n} }\right) - h\left({a}\right) }\right|, \left|{ h\left({a}\right) - h\left({ \dfrac {k} {3^n} }\right) }\right| }\right\}$

This implies that for any $n \in \N$, $t_n \ne a$.


$2\left|{ h\left({t_n}\right) - h\left({a}\right) }\right| \ge 2^{-n}$


$\left|{ t_n - a }\right| \le 3^{-n}$

Hence, for any $n$:

$t_n \in \left[{0 \,.\,.\, 1}\right]$

and also

$\displaystyle \lim_{n \to \infty} t_n = a$

The above inequalities imply that we have:

$\dfrac {\left|{ h\left({t_n}\right) - h\left({a}\right) }\right|} {\left|{t_n - a}\right|} \ge \dfrac 1 2 \left({ \dfrac 3 2}\right)^n$

Therefore, $\displaystyle \lim_{n \to \infty} \dfrac {h \left({t_n}\right) - h \left({a}\right)} {t_n - a}$ cannot exist, as the absolute value of this expression diverges when $n$ tends to $\infty$.

From the definition of differentiability at a point, we conclude that $h$ is not differentiable at $a$.



As the Contraction Mapping Theorem is not constructive, the given proof is not either.


The construction of such functions: continuous, but nowhere differentiable, was first demonstrated by Karl Theodor Wilhelm Weierstrass.

Also known as

This result is sometimes known as Weierstrass's theorem.