I have a question on the proof of Evan's PDE Appendix C.5.Theorem 7(Properties of mollifiers), showing $f^{\epsilon} \to f$ in $L_{loc}^p(U).$ The proof starts by choosing open sets compactly contained in each so that $V \subset\subset W \subset\subset U$ and showing $\|f^{\epsilon}\|_{L^p(V)} \leq \|f\|_{L^p(W)}.$
After showing $|f^{\epsilon}(x)|^p \leq \int_{B(x,\epsilon)} \eta_{\epsilon}(x-y)|f(y)|^p dy$ on $V$ which I attached in the link lllll , Evan's proceeds as
$$ \begin{align*} \int_{V} |f^{\epsilon}(x)|^p dx &\leq \int_V \left(\int_{B(x,\epsilon)} \eta_{\epsilon} (x-y)|f(y)|^p dy\right) dx\\
& \leq \int_W |f(y)|^p \left(\int_{B(y,\epsilon)} \eta_{\epsilon} (x-y) dx\right) dy = \int_W |f(y)|^p dy,
\end{align*}$$
provided $\epsilon > 0 $ is sufficiently small.
The first inequality comes from the part $|f^{\epsilon}(x)|^p \leq \int_{B(x,\epsilon)} \eta_{\epsilon}(x-y)|f(y)|^p dy$ on $V,$ and $\int_{B(y,\epsilon)} \eta_{\epsilon} (x-y) dx = 1,$ so the last inequality follows, but
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Firstly, is my understanding correct about the second inequality? Here $f$ is locally integrable (i.e. $f \in L^p_{loc}(U)$), so we are simply interchanging the order of integrals by Fubini's theorem, and then the inequality comes in from expanding the space $V$ to $W,$ right? Then why would we want to work on $W$ in this proof? The later proof uses continuous approximation of $f$ on $W$ to obtain $\lim \sup_{\epsilon \to 0} \|f^{\epsilon} – f\|_{L^p(V)}$ is arbitrarily small, which concludes the proof, but doesn't seem to need $W$ if it weren't for using the inequality $\|f^{\epsilon}\|_{L^p(V)} \leq \|f\|_{L^p(W)}.$
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Further, where do we need $\epsilon$ to be sufficiently small?
Best Answer
In the first place, you are totally right about the second inequality; if we consider the integral $$ \int_{V} \left( \int_{B(x, \epsilon)} \eta_{\epsilon}(x-y)|f(y)|^{p} dy\right) dx,$$ we can identify that the integrand is a non-negative function and since we are dealing with $\sigma$-finite measures (Lebesgue measure on $\mathbb{R}^{n}$) all the hypotheses of Fubini's theorem are fulfilled and then we can change the order of integration. Moreover, since $V \subset W$ we also have $$ \int_{V} \left( \int_{B(x, \epsilon)} \eta_{\epsilon}(x-y)|f(y)|^{p} dy\right) dx \leq \int_{W} \left( \int_{B(x, \epsilon)} \eta_{\epsilon}(x-y)|f(y)|^{p} dy\right) dx,$$ so, combining this with Fubini's theorem the second inequality is obtained.
Now, I think that Evans introduce the open set $V \subset \subset W$ just because it gives us an edge in the last part of the proof. In the book Evans presents the following estimate: He first fixes $\delta >0$ and then chooses $g \in C(V)$ such that $\|f-g \|_{L^{p}(V)} < \delta$ (which is possible since $C(V)$ is dense in $L^{p}(V)$) then he states that $$ \|f^{\epsilon} - f\|_{L^{p}(V)} \leq \color{blue}{{\|f^{\epsilon} - g^{\epsilon}\|_{L^{p}(V)}}} + \|g^{\epsilon} - g\|_{L^{p}(V)}+\|g - f\|_{L^{p}(V)}. $$ Observe that we have "control" over the second and third term of this expression: $g^{\epsilon}$ converges uniformly to $g$ on $V$ and $ \|g-f \|_{L^{p}(V)} < \delta.$ However, we don't know a priori if the term $\|f^{\epsilon}-g^{\epsilon} \|_{L^{p}(V)}$ will tend to zero. Nevertheless, if we use the previous estimate, $$\| f^{\epsilon}-g^{\epsilon} \|_{L^{p}(V)} \leq \| f-g \|_{L^{p}(V)} $$ Then it becomes clear that we also have control over the first term. I consider that the use of the set $W$ is merely to give a convenient way to estimate the last inequality.
Finally, we do need to choose $\epsilon$ small enough to be sure that each convolution is well defined. Recall that when we are defining the convolution of a function $f: U \longrightarrow \mathbb{R}$ we actually define it on the set $U^{\epsilon} = \{ x \in U \hspace{0.1 cm} | \hspace{0.1 cm} dist(x,\partial U)> \epsilon\}$ the thing is that $$f^{\epsilon}(x) = \int_{U} \eta_{\epsilon}(x-y)f(y)dy = \int_{B(0,\epsilon)} \eta_{\epsilon}(y)\color{red}{f(x-y)}dy \hspace{1 cm} (x \in U^{\epsilon})$$ and we need to be sure that the term in red is well defined, i.e. that $x-y \in U$ for any $y \in B(0,\epsilon)$. In the specific context of the proof, you may have something like this:
So if $\epsilon$ is not small enough, you can not even guarantee that the convolution is well defined.