# The solvability of a Hölder ODE [closed]

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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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

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

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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