Corollary of Holder inequality

Question

I don't understand how applicate H-inequality to prove this thing: $|(\rho_n\star f)(x)-f(x)|\leq\int|f(x-y)-f(x)|\rho_n(y)dy\leq\biggl(\int|f(x-y)-f(y)|^p\rho(y)dy\biggr)^{1/p},$

where $\rho_n$ is a Friedrichs mollifier (with unitary integral and compact support $\overline B(0,n)$),

$f\in L^p$ and with the Lebesgue measure.

(my problem is the last inequality only)

CFR. Brezis, proof of Theorem 4.26

Answer 1

• The inequality $$\left|\left(\rho_{n} \star f\right)(x) -f(x)\right| \leq \int \left|f(x -y) -f(x)\right| \rho_{n}(y) \ dy$$ is a direct consequence of $\int \rho_{n} \ d\lambda = 1$ and $\left| \int . \right| \leq \int \left| . \right|$ - where $\lambda$ denotes the Lebesgue measure on $\mathbb{R}^{N}$.
• Now for the second inequality:
  Let $\mu_{n}$ be the Borel probability measure defined by $$\mu_{n}(A) = \int_{A} \rho_{n} \ d\lambda \ \text{ for all } A \in \mathcal{B}\left(\mathbb{R}^{N}\right)$$
  Then - by Hölder's ineaquality - we have $$\int \left|f(x -y) -f(x)\right| \ d\mu_{n}(y) \leq \left(\int \left|f(x -y) -f(x)\right|^{p} \ d\mu_{n}(y)\right)^{\frac{1}{p}}$$ i.e $$\int \left|f(x -y) -f(x)\right| \rho_{n}(y) \ dy \leq \left(\int \left|f(x -y) -f(x)\right|^{p} \rho_{n}(y) \ dy\right)^{\frac{1}{p}}$$