For which exponents $\gamma$, the function $|x|^{1/2}$ is $\gamma$-Holder continuous?

Question

I have to prove the following.

Let $$u(x):=|x|^{1/2}$$ if $$|x|\le 1$$

For which exponents $\gamma\in (0,1]$, $u\in C^{0,\gamma}([-1,1]).$ The answer should be $\gamma\in (0,\frac{1}{2}]$, but I don't know how to prove it. Any suggestions?

Answer 1

I consider the function $u\upharpoonright[0, 1]$. Let $\gamma\in\mathbb{R}$. Then:

• $u$ is $\gamma$-Hölder-cts iff $\exists{C\in\mathbb{R}^{+}:~}\forall{x,y\in[0,1]:~}|u(y)-u(x)|\leq C\cdot|y-x|^{\gamma}$

Clearly for $\gamma>1/2$ the statement fails, since otherwise for some $C$, setting $y=0$, would hold: $x^{1/2}\leq C\cdot x^{\gamma}$ for $x\in(0;1]$, thus $1\leq C\cdot x^{\gamma-1/2}\longrightarrow C\cdot 0 =0$ for $x\longrightarrow 0^{+}$ $\Longrightarrow$ Contradiction!.

One considers thus only $\gamma\leq 1/2$.

Note that $|y-x|=|(\sqrt{y}-\sqrt{x})(\sqrt{y}+\sqrt{x})|=|u(y)-u(x)|\cdot |u(y)+u(x)|$ for $x,y\geq 0$. Thus:

• $u$ is $\gamma$-Hölder-cts iff $\exists{C\in\mathbb{R}^{+}:~}\forall{x,y\in[0,1]:~}|u(y)-u(x)|^{1-\gamma}\leq C\cdot|u(y)+u(x)|^{\gamma}$

It holds $|u(y)-u(x)|\leq |u(y)+u(x)|$. Setting now $C:=2^{1-2\gamma}$, it holds for $x,y\in[0,1]$ that $|u(y)+u(x)|^{1-2\gamma}\leq |u(1)+u(1)|^{1-2\gamma}=C$ and thus $|u(y)-u(x)|^{\gamma}\leq|u(y)+u(x)|^{1-\gamma}=|u(y)+u(x)|^{1-2\gamma}\cdot|u(y)+u(x)|^{\gamma}\leq C\cdot|u(y)+u(x)|^{\gamma}$.

Hence $u$ ist $\gamma$-Hölder-cts.

Thus Hölder-Continuity holds exactly for $\{\gamma:\gamma\leq 1/2\}$.

The original function $u\upharpoonright[-1,1]$ can be dealt with similarly, only that one first splits the cases for $x,y$ into two: when $x,y$ have the same sign (proceed as above), and when their signs are different (then the expression holds as a trivial consequence of the same-signed case).