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

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 $$\int_0^1 v^{-1/p} \|f'\|_p dv = \frac{p}{p-1}\|f'\|_p.$$