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The problem is follow, I want to know weather there is a function $u\in \text{C}^{0}\left((-1,1)\right)$ but not $\text{C}^1((-1,1))$, such that $\lim\limits_{h\rightarrow 0^+}\frac{|u(x+h)-u(x)|}{\left|h\right|^{1/2}}=\sqrt{|x|},\forall x\in(-1,1)$.

I get this problem while constructing a counterexample for the problem:

If $\alpha+\beta=1,u\in\text{C}^{0,\alpha}(\Omega),$and $v=\frac{u(x+he)-u(x)}{|h|^\alpha}\in\text{C}^{0,\beta}(\Omega)$,where $e$ is a unit vector and $h$ is a small real number, then $u \in\text{C}^1(\Omega)$.

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  • $\begingroup$ Your second statement is rather strange. If $u$ is an arbitrary function from $\text{C}^{0,1/2}(\Omega)$ then $v\in \text{C}^{0,1/2}(\Omega_h)$, $\Omega_h=\{x\in\Omega\,|\,\text{dist}(x,\partial \Omega)>h\}$. But, generally speaking, $v$ wouldn't belong to $\text{C}^{1}(\Omega_h)$. $\endgroup$
    – Andrew
    Jul 5, 2013 at 16:21
  • $\begingroup$ You are right.I just want to construct such a function. And the function $u$ satisfies the equation I described is one. $\endgroup$
    – Thomas
    Jul 6, 2013 at 5:04
  • $\begingroup$ I think I made a mistake. There should be a exponent $\alpha$. $\endgroup$
    – Thomas
    Jul 6, 2013 at 5:08
  • $\begingroup$ For any such function the right derivative would be equal $+\infty$ for $x\ne0$: $$ \lim\limits_{h\rightarrow 0^+}\frac{u(x+h)-u(x)}{h}= \lim\limits_{h\rightarrow 0^+}\sqrt{|x|}{h^{-1/2}}=+\infty. $$ So it is not of $C^1$. $\endgroup$
    – Andrew
    Jul 6, 2013 at 8:46
  • $\begingroup$ You are right. But what I need is the existence of such a function. $\endgroup$
    – Thomas
    Jul 7, 2013 at 9:46

1 Answer 1

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There is no such function. Any function with the property postulated would be monotone increasing on (0,1). But then it would have to be differentiable almost everywhere, which contradicts the assumption.

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  • $\begingroup$ Thanks for your answer. It is helpful. But what if I don't need $u$ to be monotone increasing on $(0,1)$. Just like the question I change,$\lim\limits_{h\rightarrow 0^+}\frac{|u(x+h)-u(x)|}{\left|h\right|^{1/2}}=\sqrt{|x|},\forall x\in(-1,1)$. Then $u$ may not be monotone increasing, in this case. $\endgroup$
    – Thomas
    Jul 7, 2013 at 9:52

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