Let $C \closedint 0 1$ denote the set of all real functions $f: \closedint 0 1 \to \R$ which are continuous on $\closedint 0 1$. By Continuous Function on Closed Interval is Complete, $C \closedint 0 1$ is a complete metric space under the supremum norm $\norm {\,\cdot \,}_\infty$. Let $X$ consist of the $f \in C \closedint 0 1$ such that: :$\map f 0 = 0$ :$\map f 1 = 1$ :$\forall x \in \closedint 0 1: 0 \le \map f x \le 1$ Then we have the following lemma: === Lemma 1 === $X$, defined as above, is a complete metric space under $\norm {\,\cdot \,}_\infty$. <section title="Proof of Lemma 1" header="3" foldable="true"> For every $n \in \N$, let $f_n \in X$. Furthermore, suppose that in $C \closedint 0 1$: {{begin-eqn}} {{eqn | n = 1       | l = \lim_{n \mathop \to \infty} \norm {f_n - f}_\infty       | r = 0 }} {{end-eqn}} If we can prove that $f \in X$, we know $X$ contains all its limit points. Hence by Closed Set iff Contains all its Limit Points, $X$ is closed. {{explain|Provide a link to the appropriate page which defines "closed".}} From Topological Completeness is Weakly Hereditary, $X$ is complete. {{explain|Provide a link to the appropriate page which defines "complete" -- topologically complete, or complete in the sense that a complete metric space is complete? Utmost clarity needed here.}} It is now to be proved that $f \in X$. Suppose $\map f 0 \ne 0$. Then: :$\forall n \in \N: \norm {f_n - f}_\infty \ge \size {\map {f_n} 0 - \map f 0} = \size {\map f 0} > 0$ This would contradict equation $(1)$. Hence $\map f 0 = 0$. Similarly, it is necessary that $\map f 1 = 1$. Also, for all $n \in \N$ and $x \in \closedint 0 1$, we have that: :$0 \le \map {f_n} x \le 1$ Suppose there is an $x \in \closedint 0 1$ such that either: :$\map f x < 0$ or: :$\map f x > 1$ We see that it must be that: :$\forall n \in \N: \norm {f_n - f}_\infty \ge \norm {\map {f_n} x - \map f x} > 0$ which contradicts $(1)$. {{Handwaving|"we see that it must be that"}} Therefore, $f \in X$, and hence $X$ is complete. </section>{{qed|lemma}} For every $f \in X$, define $\hat f: \closedint 0 1 \to \R$ as follows: :$\map {\hat f} x = \begin{cases} \dfrac 3 4 \map f {3 x} & : 0 \le x \le \dfrac 1 3 \\ \dfrac 1 4 + \dfrac 1 2 \map f {2 - 3 x} & : \dfrac 1 3 \le x \le \dfrac 2 3 \\ \dfrac 1 4 + \dfrac 3 4 \map f {3 x - 2} & : \dfrac 2 3 \le x \le 1 \end{cases}$ We have the following lemma: === Lemma 2 === $\hat \cdot: X \to X$ is a contraction mapping. Furthermore, we have the following inequality: :$\forall f, g \in X: \norm {\hat f - \hat g}_\infty \le \dfrac 3 4 \norm {f - g}_\infty$ <section title="Proof of Lemma 2" foldable="true" header="3"> {{stub}} </section>{{qed|lemma}} The Contraction Mapping Theorem assures existence of a unique $h \in X$ with $\hat h = h$. We have that $h \in X \subset C \closedint 0 1$. Thus, by definition, $h$ is a continuous real function. It remains to be shown that $h$ is nowhere differentiable. To do this, we establish the following lemma: === Lemma 3 === For every $n \in \N$ and $k \in \set {1, 2, 3, 4, \ldots, 3^n}$, the following inequality holds: :$\size {\map h {\dfrac {k - 1} {3^n} } - \map h {\dfrac k {3^n} } } \ge 2^{-n}$ <section title="Proof of Lemma 3" foldable="true" header="3"> For all $n \in \N$ and $k \in \set {1, 2, 3, \ldots, 3^n}$: :$1 \le k \le 3^n \implies 0 \le \dfrac{k - 1} {3^{n + 1} } < \dfrac k {3^{n + 1} } \le \dfrac 1 3$ :$3^n < k \le 2 \cdot 3^n \implies \dfrac 1 3 \le \dfrac {k - 1} {3^{n + 1} } < \dfrac k {3^{n + 1}} \le \dfrac 2 3$ :$2 \cdot 3^n < k \le 3^{n + 1} \implies \dfrac 2 3 \le \dfrac {k - 1} {3^{n + 1} } < \dfrac k {3^{n + 1} } \le 1$ {{stub|induction}} </section>{{qed|lemma}} Let $a \in \closedint 0 1$ be arbitrarily selected. It is to be shown that $h$ is not differentiable at $a$. This is to be achieved by constructing a sequence $\sequence {t_n}$ with elements in $\closedint 0 1$, which has the following limit: :$\displaystyle \lim_{n \mathop \to \infty} t_n = a$ To this end, let $n \in \N$ be arbitrary. Let $k$ be the unique largest element of $\set {1, 2, 3, 4, \ldots, 3^n}$ such that: :$\paren {k - 1} 3^{-n} \le a \le k 3^{-n}$ By the triangle inequality: {{begin-eqn}} {{eqn | o =        | r = \size {\map h {\frac {k - 1} {3^n} } - \map h a} + \size {\map h a - \map h {\frac k {3^n} } }       | c =  }} {{eqn | o = \ge       | r = \size {\map h {\frac {k - 1} {3^n} } - \map h {\frac k {3^n} } }       | c =  }} {{eqn | o = \ge       | r = 2^{-n}       | c =  }} {{end-eqn}} Next, let $t_n$ be either $\dfrac {k - 1} {3^n}$ or $\dfrac k {3^n}$, such that the following equation is satisfied: :$\size {\map h {t_n} - \map h a} = \max \set {\size {\map h {\dfrac {k - 1} {3^n} } - \map h a}, \size {\map h a - \map h {\dfrac k {3^n} } } }$ This implies: :$\forall n \in \N: t_n \ne a$ Furthermore: :$2 \size {\map h {t_n} - \map h a} \ge 2^{-n}$ and: :$\size {t_n - a} \le 3^{-n}$ Hence, for any $n$: :$t_n \in \closedint 0 1$ and also: :$\displaystyle \lim_{n \mathop \to \infty} t_n = a$ The above inequalities imply that: :$\dfrac {\size {\map h {t_n} - \map h a} } {\size {t_n - a} } \ge \dfrac 1 2 \paren {\dfrac 3 2}^n$ But the absolute value of this expression diverges when $n$ tends to $\infty$. Therefore $\displaystyle \lim_{n \mathop \to \infty} \dfrac {\map h {t_n} - \map h a} {t_n - a}$ cannot exist. From the definition of differentiability at a point, we conclude that $h$ is not differentiable at $a$. {{qed}}