Comparison of semi-norms Let $p\in (1,\infty)$ and define $$|f|_1=\left(\int_0^1\int_0^1\frac{|f(x)-f(y)|^p}{|x-y|^p}\,dx\,dy\right)^{1/p}$$
$$|f|_2=\left(\int_0^1\int_0^{1-h}\frac{|f(x+h)-f(x)|^p}{h^p}\,dx\,dh\right)^{1/p}$$
Suppose that $f$ vary in the set where both numbers $|f|_1,|f|_2$ are finite. I need to prove that there exist constants $c_1,c_2>0$, not depending on $f$, such that $$c_1|f|_1\leq |f_2|\leq c_2|f_1|$$
I could prove the second inequality by noting that $$|f|_1^p=|f|_2^p+\int_{-1}^0\int_{-h}^1\frac{|f(x+h)-f(x)|^p}{h^p}\,dx\,dh$$
However the first inequality seems to be a little  trickier to me.
Thanks
 A: Let us denote the integrand
$$\left\lvert \frac{f(y)-f(x)}{y-x}\right\rvert^p$$
by $Q(y,x)$. $Q$ is symmetric, so
$$\begin{align}
\lvert f\rvert_1^p &= \int_0^1\int_0^1Q(y,x)\,dx\,dy\\
&= \int_0^1\int_0^y Q(y,x)\,dx\,dy + \int_0^1\int_y^1Q(y,x)\,dx\,dy\\
&= \int_{0 \leqslant x\leqslant y\leqslant 1} Q(y,x)\,d\lambda + \int_{0 \leqslant y \leqslant x\leqslant 1} Q(y,x)\,d\lambda\\
&= 2 \int_{0 \leqslant x\leqslant y\leqslant 1} Q(y,x)\,d\lambda
\end{align}$$
where $d\lambda$ is the two-dimensional Lebesgue measure.
For $\lvert f\rvert_2$, write $y = x+h$, and note that the domain of integration is
$$\begin{align}
T &= \left\lbrace (x,y) : 0 \leqslant y-x \leqslant 1,\, 0 \leqslant x \leqslant 1 - (y-x) \right\rbrace\\
&= \{ (x,y) : 0 \leqslant x,\, x \leqslant y,\, y \leqslant 1+x,\, y \leqslant 1\}\\
&= \{ (x,y) : 0 \leqslant x \leqslant y \leqslant 1\},
\end{align}$$
thus
$$\lvert f\rvert_2^p = \frac12 \lvert f\rvert_1^p.$$
Or, without the exponents, $\lvert f\rvert_2 = 2^{-1/p}\lvert f\rvert_1$.
