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Theorem 3.1. The Bolzano function $\scr B$ is continuous and nowhere differentiable on the interval $[a, b]$.
Proof. First we want to show that the function $\scr B$ is continuous. For fixed $k \in \mathbb{N}$ consider the function ${\scr B}_k$. Let us find the slopes $M_{k}=\left\{M_{k, m}\right\}$ of each of the linear functions on the subintervals. Not to have too many indices we will just write $M_{k}$ instead of $M_{k, m}$. For $k=1$ it is immediate from the definition that $M_{1}=\frac{B-A}{b-a}$ for all of $[a, b]$. Let $k \geq 2$. For each linear part $\left[a_{k}, b_{k}\right]$ of ${\scr B}_k$ we have the following:
1. For $I=\left[t_{1}, t_{2}\right]=\left[a_{k}, a_{k}+\frac{3}{8}\left(b_{k}-a_{k}\right)\right]$,
$$
M_{k+1}^{(1)}=\frac{{\scr B}_k\left(t_{2}\right)-{\scr B}_k\left(t_{1}\right)}{t_{2}-t_{1}}=\frac{\frac{5}{8}\left(B_k-A_{k}\right)}{\frac{3}{8}\left(b_{k}-a_{k}\right)}=\frac{5}{3} \frac{B_k-A_{k}}{b_{k}-a_{k}}=\frac{5}{3} M_{k} ;
$$
2. for $I=\left[t_{2}, t_{3}\right]=\left[a_{k}+\frac{3}{8}\left(b_{k}-a_{k}\right), \frac{1}{2}\left(a_{k}+b_{k}\right)\right]$,
$$
M_{k+1}^{(2)}=\frac{{\scr B}_k\left(t_{3}\right)-{\scr B}_k\left(t_{2}\right)}{t_{3}-t_{2}}=\frac{\left(\frac{1}{2}-\frac{5}{8}\right)\left(B_k-A_{k}\right)}{\left(\frac{1}{2}-\frac{3}{8}\right)\left(b_{k}-a_{k}\right)}=\frac{-\frac{1}{8}}{\frac{1}{8}} \frac{B_k-A_{k}}{b_{k}-a_{k}}=-M_{k} ;
$$
3. for $I=\left[t_{3}, t_{4}\right]=\left[\frac{1}{2}\left(a_{k}+b_{k}\right), a_{k}+\frac{7}{8}\left(b_{k}-a_{k}\right)\right]$,
$$
M_{k+1}^{(3)}=\frac{{\scr B}_k\left(t_{4}\right)-{\scr B}_k\left(t_{3}\right)}{t_{4}-t_{3}}=\frac{\left(1+\frac{1}{8}-\frac{1}{2}\right)\left(B_k-A_{k}\right)}{\left(\frac{7}{8}-\frac{1}{2}\right)\left(b_{k}-a_{k}\right)}=\frac{\frac{5}{8}}{\frac{3}{8}} \frac{B_k-A_{k}}{b_{k}-a_{k}}=\frac{5}{3} M_{k} ;
$$
4. for $I=\left[t_{4}, t_{5}\right]=\left[a_{k}+\frac{7}{8}\left(b_{k}-a_{k}\right), b_{k}\right]$,
$$
M_{k+1}^{(4)}=\frac{{\scr B}_k\left(t_{5}\right)-{\scr B}_k\left(t_{4}\right)}{t_{5}-t_{4}}=\frac{-\frac{1}{8}\left(B_k-A_{k}\right)}{\left(1-\frac{7}{8}\right)\left(b_{k}-a_{k}\right)}=\frac{-\frac{1}{8}}{\frac{1}{8}} \frac{B_k-A_{k}}{b_{k}-a_{k}}=-M_{k} \text {. }
$$
Let $\left\{I_{n, k}\right\}=\left\{\left[I_{n}\left(s_{k}\right), I_{n}\left(t_{k}\right)\right]\right\}$ be the collection of subintervals of $[a, b]$ where ${\scr B}_n$ is linear and define
$$
L_{n}=\sup _{I \in\left\{I_{n+1, k}\right\}}\left(I\left(t_{k}\right)-I\left(s_{k}\right)\right) \quad \text { and } \quad M_{n}=\sup _{\substack{I \in\left\{I_{n+1, k}\right\} \\ i=1,2,3,4}}\left|M_{n}^{(i)}(I)\right| .
$$
That is, $L_n$ is the maximal length of an interval where ${\scr B}_{n+1}$ is linear and $M_{n}$ is the maximum slope (to the absolute value) of ${\scr B}_{n+1}$. Clearly
$$
L_{n} \leq\left(\frac{3}{8}\right)^{n+1}|b-a|\quad \text { and }\quad M_n \leq\left(\frac{5}{3}\right)^{n+1}\left|\frac{B-A}{b-a}\right|
$$
which gives that the maximum increase/decrease of the function from step $n$ to $n+1$ is bounded by $M_{n} L_{n} \leq\left(\frac{5}{8}\right)^{n+1}|B-A|$. Hence, for $k \in \mathbb{N}$,
$$\sup _{x \in[a, b]}\left|{\scr B}_{k+1}(x)-{\scr B}_k(x)\right| \leq\left(\frac{5}{8}\right)^{k+1}|B-A|$$
Let $m, n \in \mathbb{N}$ with $m>n$. We have
\begin{aligned}
\sup _{x \in[a, b]}\left|{\scr B}_m(x)-{\scr B}_n(x)\right| & \leq \sup _{x \in[a, b]}\left(\sum_{k=n+1}^{m}\left|{\scr B}_k(x)-{\scr B}_{k-1}(x)\right|\right) \\
& \leq \sum_{k=n+1}^{m} \sup _{x \in[a, b]}\left|{\scr B}_k(x)-{\scr B}_{k-1}(x)\right| \\
& \leq \sum_{k=n+1}^{m}\left(\frac{5}{8}\right)^{k}|B-A| \\
&=|B-A|\left(\sum_{k=1}^{m}\left(\frac{5}{8}\right)^{k}-\sum_{k=1}^{n}\left(\frac{5}{8}\right)^{k}\right) \\
& \rightarrow|B-A|\left(\frac{5}{3}-\frac{5}{3}\right)=0 \quad \text { as } m, n \rightarrow \infty .
\end{aligned}
Thus $\left\{{\scr B}_k\right\}$ is a uniformly Cauchy sequence on the interval $[a, b]$ and since each ${\scr B}_k$ is continuous it follows from Theorems 2.1 and 2.3 that Bolzano's function is continuous on $[a, b]$.
Secondly, we show that $\scr B$ is not differentiable at any $x \in[a, b]$. Again, let $\left\{I_{n, k}\right\}=\left\{\left[I_{n}\left(s_{k}\right), I_{n}\left(t_{k}\right)\right]\right\}$ be the collection of subintervals of $[a, b]$ where ${\scr B}_{n}$ is linear and define $M$ as the set of all endpoints in $\left\{I_{n, k}\right\}$, i.e.
$$
M=\left\{s, t \mid[s, t] \in\left\{I_{n, k}\right\}\right\} .
$$
We show that $M$ is dense in $[a, b]$. That is, for any $x_{0} \in[a, b], \exists x_{n} \in M$ such that $x_{n} \rightarrow x_{0}$. Let $x_{0} \in[a, b]$ be arbitrary but fixed. If $x_{0}=b$ we are done since $b \in M$. Assume that $x_{0} \neq b$, we proceed as follows.
(i) Step 1: let $L=b-a$ and define
\begin{aligned}
J_{0}^{(0)} &=\left[a, a+\frac{3}{8} L\right), &&& J_{0}^{(1)} &=\left[a+\frac{3}{8} L, a+\frac{1}{2} L\right) \\
J_{0}^{(2)} &=\left[a+\frac{1}{2} L, a+\frac{7}{8} L\right)& \text { and }&& J_{0}^{(3)} &=\left[a+\frac{7}{8} L, b\right)
\end{aligned}
Clearly there exists $i_{0} \in\{0,1,2,3\}$ such that $x_{0} \in J_{0}^{\left(i_{0}\right)}$. We take $J_{0}=J_{0}^{\left(i_{0}\right)}$.
(ii) Step $n$ : we have $x_{0} \in I_{n-1}=\left[a_{n}, b_{n}\right]$. Let $L_{n}=b_{n}-a_{n}$ and define $J_{n}^{(0)}=\left[a_{n}, a_{n}+\frac{3}{8} L_{n}\right), \quad J_{n}^{(1)}=\left[a_{n}+\frac{3}{8} L_{n}, a_{n}+\frac{1}{2} L_{n}\right),\quad J_{n}^{(2)}=\left[a_{n}+\frac{1}{2} L_{n}, a_{n}+\frac{7}{8} L_{n}\right)$ and $J_{n}^{(3)}=\left[a_{n}+\frac{7}{8} L_{n}, b_{n}\right)$.
As before, there exists $i_{n} \in\{0,1,2,3\}$ such that $x_{0} \in J_{n}^{\left(i_{n}\right)}$. We take $J_{n}=J_{n}^{\left(i_{n}\right)}$.
Hence $M$ is dense in $[a, b]$ since
$$
\cssId{eq1}{\left|x_{0}-a_{n+1}\right| \leq\left(\frac{3}{8}\right)^{n+1}|b-a| \rightarrow 0 \text { as } n \rightarrow \infty^{\href{javascript:void(location.hash="#fn1")}{1}}}
$$
Now we show that $\scr B$ is non-differentiable for every $x_{0} \in M$. Let $x_{0} \in M$ be arbitrary but fixed, we consider two cases that exhaust all possibilities.
For $x_{0}=a$ : Let $x_{n}=a+\left(\frac{3}{8}\right)^{n}|b-a|$. Then $x_{n} \rightarrow a$ as $n \rightarrow \infty$ and $x_{n} \in M$ for every $n \in \mathbb{N}$. By the construction of the function $\scr B$ it is clear that ${\scr B}\left(x_{n}\right)={\scr B}_{n+1}\left(x_{n}\right)$ for every $n \in \mathbb{N}$. Also, ${\scr B}(a)=A$ and ${\scr B}_{n+1}\left(x_{n}\right)=A+\left(\frac{5}{3}\right)^{n}\left(\frac{3}{8}\right)^{n}|b-a|$. Hence
$$
\frac{{\scr B}\left(x_n\right)-B(a)}{x_{n}-a}=\frac{A+\left(\frac{5}{3}\right)^{n}\left(\frac{3}{8}\right)^{n}|b-a|-A}{\left(\frac{3}{8}\right)^{n}|b-a|}=\left(\frac{5}{3}\right)^{n} \rightarrow \infty \text { as } n \rightarrow \infty
$$
and therefore ${\scr B}'\left(x_{0}\right)$ does not exist.
For $x_{0} \in M \backslash\{a\}$ : let $x_{n}=x_{0}-\left(\frac{1}{8}\right)^{n+q}|b-a|, q \in \mathbb{N}$. Since $x_{0} \in M$, there exists $r \in \mathbb{N}$ such that ${\scr B}\left(x_{0}\right)={\scr B}_{p}\left(x_{0}\right)$ for all $p \geq r$. We can choose $q>r$ so that $x_{n} \in(a, b]$ for every $n \in \mathbb{N}$. From the construction of $\scr B$ we see that ${\scr B}\left(x_{n}\right)={\scr B}_{n+1}\left(x_{n}\right)={\scr B}_{n}\left(x_{0}\right)+(-1)^{n} K\left(\frac{1}{8}\right)^{n+q}$ where $K \in \mathbb{R}$ with $K \geq|b-a| /|B-A| \neq 0$. Moreover, since $q>r,{\scr B}\left(x_{0}\right)={\scr B}_{n}\left(x_{0}\right)$ for every $n \in \mathbb{N}$. This implies that
$$
\left({\scr B}\left(x_{0}\right)-{\scr B}_{n}\left(x_{0}\right)\right) 8^{n+q}=\left({\scr B}_{n}\left(x_{0}\right)-{\scr B}_{n}\left(x_{0}\right)\right) 8^{n+q}=0 .
$$
So for $n \in \mathbb{N}$,
$$
\begin{aligned}
\frac{{\scr B}\left(x_{0}\right)-{\scr B}\left(x_{n}\right)}{x_{0}-x_{n}} &=8^{n+q}\left({\scr B}\left(x_{0}\right)-{\scr B}_{n}\left(x_{0}\right)-(-1)^{n} K\left(\frac{1}{8}\right)^{n+q}\right) \\
&=\left({\scr B}\left(x_{0}\right)-{\scr B}_{n}\left(x_{0}\right)\right) 8^{n+q}-(-1)^{n} K=(-1)^{n+1} K .
\end{aligned}
$$
But $(-1)^{n+1} K$ does not converge as $n \rightarrow \infty$ and thus $B^{\prime}\left(x_{0}\right)$ does not exist. In no way is it clear from this that $\scr B$ is nowhere differentiable, only that it is non-differentiable on a dense subset of $[a, b]$ (which theoretically means that it might still be possible that $\scr B$ is differentiable almost everywhere). We will not complete the proof here but merely give a reference: the complete proof can be found in Jarník [34].
1 And also $\left|x_{0}-b_{n+1}\right| \rightarrow 0$ as $n \rightarrow \infty$.
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