Uniform bound for integral in terms of $\left\lVert f' \right\rVert_4^4$

Question

Show that there is a constant $C>0$ such that for any compactly supported $C^1$ function $f: \mathbb{R} \to \mathbb{R}$, we have $$\int_{\mathbb{R}} \left(\frac{f(x)-f(y)}{x-y}\right)^4dy \le C \left\lVert f' \right\rVert_4^4\qquad\text{for all }x \in \mathbb{R}.$$

This is an old quals problem that I don't know how to do. One hint is that I may use integration by parts, but I don't know how to apply the hint either. Any approach would be much appreciated.

Answer 1

This is more or less a copy of the argument here, which seems to be a lot more harder than just integration by part: Using the fundamental theorem of calculus and Cauchy Schwarz's inequality (twice),

\begin{align} (f(y) - f(x))^4 &= \left( \int_x^y f'(t) dt\right)^4\\ &\le \left( \int_x^y |f'(t)|^2 dt\right)^2 \left( \int_x^y 1^2 dt\right)^2 \\ \Rightarrow \left( \frac{f(y)-f(x)}{y-x} \right)^4 &\le \left( \frac{1}{y-x} \int_x^y |f'(t)|^2 dt\right)^2 \\ &= \frac{1}{(y-x)^2} \left(\int_x^y (t-x)^{-1/4} (t-x)^{1/4} |f'(t)|^2 dt\right)^2 \\ &\le \frac{1}{(y-x)^2} \left| \int_x^y (|t-x|^{-1/2} dt \right|\cdot \int_x^y |t-x|^{1/2} |f'(t)|^4 dt \\ &= \frac{2}{|y-x|^{3/2}}\int_x^y |t-x|^{1/2} |f'(t)|^4 dt. \end{align} Integrating with respect to $y$ and use Fubini,
\begin{align} \int_{x}^\infty \left( \frac{f(y)-f(x)}{y-x} \right)^4dy &\le \int_x^\infty \frac{2}{|y-x|^{3/2}}\int_x^y |t-x|^{1/2} |f'(t)|^4 dt dy \\ &= 2 \int_x^\infty \left( \int_t^{\infty} \frac{1}{|y-x|^{3/2}} dy\right) |t-x|^{1/2} |f'(t)|^4 dt \\ &= 4 \int_x^\infty |t-x|^{-1/2} |t-x|^{1/2} |f'(t)|^4 dt = 4\int_x^\infty|f'(t)|^4 dt. \end{align}

and similarly $$ \int_{-\infty}^x \left( \frac{f(y)-f(x)}{y-x} \right)^4 dy\le 4 \int_{-\infty}^x |f'(t)|^4 dt.$$

Thus we have $$ \int_{\mathbb R} \left( \frac{f(y)-f(x)}{y-x} \right)^4 dy \le 4\| f'\|_4^4.$$

Answer 2

Here's another proof, brought to you by the Wikipedia article on Hardy's inequality.

Write $\frac{f(y)-f(x)}{y-x} = \frac1{y-x}\int_x^yf'(u)du = \int_0^1 f'(x+(y-x)v)dv. $

Now let $p>1$. By Minkowski's inequality you have that $$\bigg[\int_{\Bbb R} \bigg| \int_0^1 f'(x+(y-x)v)dv\bigg|^p dy \bigg]^{1/p}$$

$$\leq \int_0^1 \bigg[\int_{\Bbb R} |f'(x+(y-x)v)|^pdy\bigg]^{1/p}dv.$$

Now for the inner integral, make a change of variable $z:=vy+(x-vx)$ and you see that $dy = v^{-1}dz$. Thus the last expression equals $$\int_0^1 v^{-1/p} \|f'\|_p dv = \frac{p}{p-1}\|f'\|_p.$$