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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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closed as off-topic by Michael Renardy, Andrey Rekalo, Daniel Moskovich, Willie Wong, Andres Caicedo Jul 12 '13 at 5:54

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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 $\text{C}^{1}(\Omega_h)$. –  Andrew Jul 5 '13 at 16:21
    
You are right.I just want to construct such a function. And the function $u$ satisfies the equation I described is one. –  Thomas Jul 6 '13 at 5:04
    
I think I made a mistake. There should be a exponent $\alpha$. –  Thomas Jul 6 '13 at 5:08
    
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 $C^1$. –  Andrew Jul 6 '13 at 8:46
    
You are right. But what I need is the existence of such a function. –  Thomas Jul 7 '13 at 9:46
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1 Answer 1

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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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. –  Thomas Jul 7 '13 at 9:52
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