A Lipschitz function of order $1/2$ The following exercise is from an old prelim exam that I was never able to solve when I was a graduate student, and I still think about it every now and then.
Suppose $f:\mathbb{R}\rightarrow\mathbb{R}$ is a $2\pi$-periodic and continuous function.
Define the function
$$\varphi(x)=\int^\pi_{-\pi}\frac{f(x-t)}{\sqrt{|t|}}dt$$
Show that $\varphi$ is a Lipschitz $1/2$-function. More precisely, show that
$$|\varphi(x)-\varphi(y)|\leq 8\|f\|_\infty|x-y|^{1/2}$$
for all $x,y$.

The obvious thing is to extend $\alpha(t)=\sqrt{|t|}$ on $(-\pi,\pi]$  periodically to all of $\mathbb{R}$, This yields a $2\pi$-periodic function, and allows is to write
$$\phi(x)=\int^\pi_{-\pi}\frac{f(t)}{\alpha(x-t)}dt$$
FRom this, we have that
$$|\phi(x+h)-\phi(x)|\leq\|f\|_\infty\int^{\pi}_{-\pi}\Big|\frac{1}{\alpha(x+h-t)}-\frac{1}{\alpha(x-t)}\Big| dt$$
I was not able to go much further. A hint or a full solution to this problem will be appreciated.
 A: It remains to estimate
$$
\int^{\pi}_{-\pi}\left|\frac{1}{\alpha(x+h-t)}-\frac{1}{\alpha(x-t)}\right| \, dt
$$
in terms of $\sqrt{|h|}$. With the substitution $s = x-t $ and using the $2\pi$-periodicity of $\alpha$ this is
$$
I(h) = \int^{\pi}_{-\pi}\left|\frac{1}{\alpha(s+h)}-\frac{1}{\alpha(s)}\right| \, ds \, ,
$$
which can be computed explicitly. We can assume that $|h| \le \pi$, and I'll do the calculation for the case $0 < h \le \pi$.
For $-\pi \le s \le -h$ is
$$
 \alpha(s+h) = \sqrt{-s-h} \le \sqrt{-s} = \alpha(s)
$$
so that the contribution of this interval to $I(h)$ is
$$
 \int_{-\pi}^{-h} \left(\frac{1}{\sqrt{-s-h}} - \frac{1}{\sqrt{-s}}\right) \, ds \\
= \left [ -2 \sqrt{-s-h} +2 \sqrt{-s}\right ]_{s=-\pi}^{s=-h} \\
= 2 \sqrt{h} + 2\sqrt{\pi-h} - 2\sqrt{\pi} \, .
$$
The integrals over the intervals $[-\pi, -h/2]$, $[-h/2, 0]$, $[0, \pi-h]$, $[\pi-h, \pi-h/2]$, and $[\pi -h/2, \pi]$ can be computed similarly, and one gets
$$
 I(h) = 8 \sqrt{h/2} + 8 \sqrt{\pi - h/2} - 8 \sqrt{\pi} \le   8 \sqrt{h/2} \, .
$$
for $0 < h \le \pi$. The case $-\pi \le h < 0$ works similarly.
Combining that with your calculations gives
$$
|\phi(x+h)-\phi(x)|\leq \frac{8}{\sqrt 2} \|f\|_\infty\ |h|^{1/2} \, .
$$
which is the desired result (with a slightly better constant).
