Inequality Involving $\int_V \left(\int_{B(x,\epsilon)} \eta_{\epsilon}(x-y) |f(y)|^p dy\right) dx $

Question

I was reading the book "Partial Differential Equation" written by Lawrence C. Evans, coming up with a question.

On page 718, Evans wrote

$$\int_V \left(\int_{B(x,\epsilon)} \eta_{\epsilon}(x-y) |f(y)|^p dy\right) dx \\ \leq \int_W |f(y)|^p \left(\int_{B(y,\epsilon)} \eta_{\epsilon}(x-y) dx\right) dy$$

Where $V,W,U$ are open sets, $V\subset\subset W\subset \subset U$, and $\eta_{\epsilon}(x) := \frac{1}{\epsilon^n} \eta(\frac{x}{\epsilon})$, $\eta$ is the standard mollifier, $f \in L_{loc}^p(U)$

I want to ask why the inequality holds. I was trying to prove it so hard but I have no ideas.

Answer 1

This is an application of Fubini's theorem, followed by the fact that the integrand is positive and $V$ is a subset of $W$.

Answer 2

Since $V\subset \cup_{x\in V} B(x,\epsilon) \subset W$ if $\epsilon$ small enough, we can show: $\{(x,y): x\in V , y\in B(x,\epsilon)\} \subset \{(x,y): y \in W, x\in B(y,\epsilon)\}.$

$\forall (x,y) \in \{(x,y): x\in V , y\in B(x,\epsilon)\}$, we know $x \in V$, then $y \in B(x,\epsilon)\subset \cup_{x\in V} B(x,\epsilon)\subset W$. And $y \in B(x,\epsilon)$ imples $x\in B(y,\epsilon)$. Therefore $(x,y) \in \{(x,y): y \in W, x\in B(y,\epsilon)\}$, which proves "$\subset$".

Then by the above fact and the Fubini Thm,

$\int_V \int_{B(x,\epsilon)} (\cdot)dy\,dx = \int_{ \{(x,y): x\in V , y\in B(x,\epsilon)\}}(\cdot)dy\, dx \le \int_{\{(x,y): y \in W, x\in B(y,\epsilon)\}}(\cdot) dy\,dx = \int_{\{(x,y): y \in W, x\in B(y,\epsilon)\}}(\cdot) dx\,dy = \int_W \int_{B(y,\epsilon)}(\cdot)dx\,dy$

Answer 3

Choise some small $\varepsilon>0$ such that $B(x,\varepsilon)\subset V$, we have $$\int_V\bigg(\int_{B(x,\varepsilon)}\eta_\varepsilon(x-y)|f(y)|^pdy\bigg)dx=\int_V\bigg(\int_V\eta_\varepsilon(x-y)|f(y)|^p~\chi_{B(x,\varepsilon)}(y)~dy\bigg)dx$$

where $\chi_{B(x,\varepsilon)}(y)$ is characteristic function.

Now, using the Fubini's theorem, $$\int_V\bigg(\int_V\eta_\varepsilon(x-y)|f(y)|^p~\chi_{B(x,\varepsilon)}(y)~dy\bigg)dx=\int_V|f(y)|^p\bigg(\int_V\eta_\varepsilon(x-y)~\chi_{B(x,\varepsilon)}(y)~dx\bigg)dy$$

fixed variable $y$, the characteristic function $\chi_{B(x,\varepsilon)}(y)=1$ only at $x\in B(y,\varepsilon)$, thus $$\int_V|f(y)|^p\bigg(\int_V\eta_\varepsilon(x-y)~\chi_{B(x,\varepsilon)}(y)~dx\bigg)dy=\int_V|f(y)|^p\bigg(\int_{B(y,\varepsilon)}\eta_\varepsilon(x-y)dx\bigg)dy.$$

Finally, since the fact that the integrand is positive, then $$\int_V|f(y)|^p\bigg(\int_{B(y,\varepsilon)}\eta_\varepsilon(x-y)dx\bigg)dy\le \int_W|f(y)|^p\bigg(\int_{B(y,\varepsilon)}\eta_\varepsilon(x-y)dx\bigg)dy.$$