Consider the vector space $$\mathcal V=\left\{f:[0,1]\to\mathbb R\ \left|\ \sup_{x\neq y}\frac{|f(x)-f(y)|}{\sqrt{|x-y|}}\lt \infty \ \text{ and }\ f\left(\frac 1 2\right)=0 \right. \right\}$$ Show that the function $f(x)=\sqrt{\left|x-\frac 12\right|}$ belongs to $\mathcal V$.
When $x\neq y$, $\ \sup_{x\neq y}\frac{f(x)-f(y)}{\sqrt{x-y}}=\frac{\mathrm d f}{\mathrm d x}\cdot\sqrt{x-y},\ $ and since $f$ is differentiable for $x\in[0,\frac 12)\cup(\frac 12,1]$, the only point to investigate is $x=\frac 12$. I don't see why the property would be satisfied at $x=\frac 12$ because I think the derivative of $f$ tends to $\pm$ infinity and doesn't exist. I thought that solving $$\lim_{x\to\frac 12}\lim_{y\to \frac 12}\frac{\left|\sqrt{\left|x-\frac 12\right|}-\sqrt{\left|y-\frac 12\right|}\right|}{\sqrt{|x-y|}}$$ or even $$\lim_{x\to\frac 12^+}\lim_{y\to \frac 12^-}\frac{\left|\sqrt{x-\frac 12}-\sqrt{\frac 12-y}\right|}{\sqrt{x-y}}$$ would help to prove that the supremum is real.
It's called the Hölder condition of exponent $1/2$. Such function need not be differentiable everywhere, but is, of course, continuous. To prove the original statement, note that \begin{align} &\frac{\big|\sqrt{|x-1/2|}-\sqrt{|y-1/2|}\big|}{\sqrt{|x-y|}}\\ \le&\frac{\big|\sqrt{|x-1/2|}-\sqrt{|y-1/2|}\big|}{\sqrt{\big||x-1/2|-|y-1/2|\big|}} \end{align} by $|u-v|\ge\big||u|-|v|\big|$. Now we turn to the following inequality for $v>u\ge0$: \begin{align} \frac{\sqrt{v}-\sqrt{u}}{\sqrt{v-u}}&=\frac{\sqrt{v-u}}{\sqrt{v}+\sqrt{u}}\\ &=\frac{\sqrt{1-u/v}}{1+\sqrt{u/v}}\le1 \end{align} And conclude the proof.