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user479223
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Define $$\psi(x)=\begin{cases} 1/|\log x| &\text{if } x\in (0,1/2] \\ 0 &\text{if }x\leq 0\\ 1/\log 2& \text{ if } x>1/2.\end{cases}$$

Note that $\psi$ is increasing and bounded (and even continuous).

Consider an enumeration of the rationals $q_n$ and the function

$$f(x)=\sum_{n=1}^\infty 2^{-n} \psi(x-q_n).$$

$f$ is a sum of increasing positive functions hence is increasing. Hence differentiable ae.

However for any $q_n$, any $x>q_n$ and any $\alpha\in (0,1]$ we have that $$\frac{|f(x)-f(q_n)|}{|x-q_n|^\alpha}\geq 2^{-n} \frac{1}{|x-q_n|^\alpha|\log(x-q_n)|}.$$

As $x\to q_n$ this diverges. Hence nowhere locally $\alpha$ Hölder for any $\alpha$.

Edit, by replacing $\log$ by a continuous increasing function that diverges arbitrarily slowly you can construct a continuous, increasing and bounded function with arbitrarily low local regularity.

Define $$\psi(x)=\begin{cases} 1/|\log x| &\text{if } x\in (0,1/2] \\ 0 &\text{if }x\leq 0\\ 1/\log 2& \text{ if } x>1/2.\end{cases}$$

Note that $\psi$ is increasing and bounded (and even continuous).

Consider an enumeration of the rationals $q_n$ and the function

$$f(x)=\sum_{n=1}^\infty 2^{-n} \psi(x-q_n).$$

$f$ is a sum of increasing positive functions hence is increasing. Hence differentiable ae.

However for any $q_n$, any $x>q_n$ and any $\alpha\in (0,1]$ we have that $$\frac{|f(x)-f(q_n)|}{|x-q_n|^\alpha}\geq 2^{-n} \frac{1}{|x-q_n|^\alpha|\log(x-q_n)|}.$$

As $x\to q_n$ this diverges. Hence nowhere locally $\alpha$ Hölder for any $\alpha$.

Define $$\psi(x)=\begin{cases} 1/|\log x| &\text{if } x\in (0,1/2] \\ 0 &\text{if }x\leq 0\\ 1/\log 2& \text{ if } x>1/2.\end{cases}$$

Note that $\psi$ is increasing and bounded (and even continuous).

Consider an enumeration of the rationals $q_n$ and the function

$$f(x)=\sum_{n=1}^\infty 2^{-n} \psi(x-q_n).$$

$f$ is a sum of increasing positive functions hence is increasing. Hence differentiable ae.

However for any $q_n$, any $x>q_n$ and any $\alpha\in (0,1]$ we have that $$\frac{|f(x)-f(q_n)|}{|x-q_n|^\alpha}\geq 2^{-n} \frac{1}{|x-q_n|^\alpha|\log(x-q_n)|}.$$

As $x\to q_n$ this diverges. Hence nowhere locally $\alpha$ Hölder for any $\alpha$.

Edit, by replacing $\log$ by a continuous increasing function that diverges arbitrarily slowly you can construct a continuous, increasing and bounded function with arbitrarily low local regularity.

Holder -> Hölder
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LSpice
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Define $$\psi(x)=\begin{cases} 1/|\log x| &\text{if } x\in (0,1/2] \\ 0 &\text{if }x\leq 0\\ 1/\log 2& \text{ if } x>1/2\end{cases}.$$$$\psi(x)=\begin{cases} 1/|\log x| &\text{if } x\in (0,1/2] \\ 0 &\text{if }x\leq 0\\ 1/\log 2& \text{ if } x>1/2.\end{cases}$$

Note that $\psi$ is increasing and bounded (and even continuous).

Consider an enumeration of the rationals $q_n$ and the function

$$f(x)=\sum_{n=1}^\infty 2^{-n} \psi(x-q_n)$$$$f(x)=\sum_{n=1}^\infty 2^{-n} \psi(x-q_n).$$

$f$ is a sum of increasing positive functions hence is increasing. Hence differentiable ae.

However for any $q_n$, any $x>q_n$ and any $\alpha\in (0,1]$ we have that $$\frac{|f(x)-f(q_n)|}{|x-q_n|^\alpha}\geq 2^{-n} \frac{1}{|x-q_n|^\alpha|\log(x-q_n)|}$$$$\frac{|f(x)-f(q_n)|}{|x-q_n|^\alpha}\geq 2^{-n} \frac{1}{|x-q_n|^\alpha|\log(x-q_n)|}.$$

As $x\to q_n$ this diverges. Hence nowhere locally $\alpha$ HolderHölder for any $\alpha$.

Define $$\psi(x)=\begin{cases} 1/|\log x| &\text{if } x\in (0,1/2] \\ 0 &\text{if }x\leq 0\\ 1/\log 2& \text{ if } x>1/2\end{cases}.$$

Note that $\psi$ is increasing and bounded (and even continuous).

Consider an enumeration of the rationals $q_n$ and the function

$$f(x)=\sum_{n=1}^\infty 2^{-n} \psi(x-q_n)$$

$f$ is a sum of increasing positive functions hence is increasing. Hence differentiable ae.

However for any $q_n$, any $x>q_n$ and any $\alpha\in (0,1]$ we have that $$\frac{|f(x)-f(q_n)|}{|x-q_n|^\alpha}\geq 2^{-n} \frac{1}{|x-q_n|^\alpha|\log(x-q_n)|}$$

As $x\to q_n$ this diverges. Hence nowhere locally $\alpha$ Holder for any $\alpha$.

Define $$\psi(x)=\begin{cases} 1/|\log x| &\text{if } x\in (0,1/2] \\ 0 &\text{if }x\leq 0\\ 1/\log 2& \text{ if } x>1/2.\end{cases}$$

Note that $\psi$ is increasing and bounded (and even continuous).

Consider an enumeration of the rationals $q_n$ and the function

$$f(x)=\sum_{n=1}^\infty 2^{-n} \psi(x-q_n).$$

$f$ is a sum of increasing positive functions hence is increasing. Hence differentiable ae.

However for any $q_n$, any $x>q_n$ and any $\alpha\in (0,1]$ we have that $$\frac{|f(x)-f(q_n)|}{|x-q_n|^\alpha}\geq 2^{-n} \frac{1}{|x-q_n|^\alpha|\log(x-q_n)|}.$$

As $x\to q_n$ this diverges. Hence nowhere locally $\alpha$ Hölder for any $\alpha$.

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user479223
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Define $$\psi(x)=\begin{cases} 1/|\log x| \quad &\text{if } x\in (0,1/2] \\ 0 &\text{if }x\leq 0\\ =1/\log 2& \text{ if } x>1/2\end{cases}.$$$$\psi(x)=\begin{cases} 1/|\log x| &\text{if } x\in (0,1/2] \\ 0 &\text{if }x\leq 0\\ 1/\log 2& \text{ if } x>1/2\end{cases}.$$

Note that $\psi$ is increasing and bounded (and even continuous).

Consider an enumeration of the rationals $q_n$ and the function

$$f(x)=\sum_{n=1}^\infty 2^{-n} \psi(x-q_n)$$

$f$ is a sum of increasing positive functions hence is increasing. Hence differentiable ae.

However for any $q_n$, any $x>q_n$ and any $\alpha\in (0,1]$ we have that $$\frac{|f(x)-f(q_n)|}{|x-q_n|^\alpha}\geq 2^{-n} \frac{1}{|x-q_n|^\alpha|\log(x-q_n)|}$$

As $x\to q_n$ this diverges. Hence nowhere locally $\alpha$ Holder for any $\alpha$.

Define $$\psi(x)=\begin{cases} 1/|\log x| \quad &\text{if } x\in (0,1/2] \\ 0 &\text{if }x\leq 0\\ =1/\log 2& \text{ if } x>1/2\end{cases}.$$

Note that $\psi$ is increasing and bounded.

Consider an enumeration of the rationals $q_n$ and the function

$$f(x)=\sum_{n=1}^\infty 2^{-n} \psi(x-q_n)$$

$f$ is a sum of increasing positive functions hence is increasing. Hence differentiable ae.

However for any $q_n$, any $x>q_n$ and any $\alpha\in (0,1]$ we have that $$\frac{|f(x)-f(q_n)|}{|x-q_n|^\alpha}\geq 2^{-n} \frac{1}{|x-q_n|^\alpha|\log(x-q_n)|}$$

As $x\to q_n$ this diverges. Hence nowhere locally $\alpha$ Holder for any $\alpha$.

Define $$\psi(x)=\begin{cases} 1/|\log x| &\text{if } x\in (0,1/2] \\ 0 &\text{if }x\leq 0\\ 1/\log 2& \text{ if } x>1/2\end{cases}.$$

Note that $\psi$ is increasing and bounded (and even continuous).

Consider an enumeration of the rationals $q_n$ and the function

$$f(x)=\sum_{n=1}^\infty 2^{-n} \psi(x-q_n)$$

$f$ is a sum of increasing positive functions hence is increasing. Hence differentiable ae.

However for any $q_n$, any $x>q_n$ and any $\alpha\in (0,1]$ we have that $$\frac{|f(x)-f(q_n)|}{|x-q_n|^\alpha}\geq 2^{-n} \frac{1}{|x-q_n|^\alpha|\log(x-q_n)|}$$

As $x\to q_n$ this diverges. Hence nowhere locally $\alpha$ Holder for any $\alpha$.

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user479223
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