an inequality in the proof of properties of mollifier does not hold for $p=\infty$

Question

This is part of the proof of mollifier properties in Evans PDE, which has been posted numerous times...

In the above, the inequality does not hold for $p=\infty$, why? (in the case of $U=\mathbb{R}$ I think it does.

Could anyone provide a contradictory proof or similar? Thanks

Answer 1

Let $f \in L^\infty_{loc}(U)$, and we take $f$ such that its norm is given by its supremum instead of its essential supremum.
Now, we have for $x\in U_\epsilon$ $$|f^\varepsilon(x)| \le \int_{B(0,\varepsilon)} | \eta_\varepsilon(y) f(x-y) | \textrm{d}y \le \sup_{z \in B(x, \varepsilon) }|f(z)|\int_{B(0,\varepsilon)} | \eta_\varepsilon(y)| \textrm{d}y = \sup_{z \in B(x, \varepsilon) }|f(z)|$$ The first inequality is taking the absolute value inside the integral. The last equality is using $\eta_\varepsilon(y) \ge 0$ and $\int_{B(0,\varepsilon)} \eta_\varepsilon(y) \textrm{d}y = 1$.

Now, we have $V \subset \subset W \subset \subset U$, so there exists a compact $K$ such that $V \subset K \subset W$. Furthermore, $W$ is open, thus disjoint with it boundary $\partial W$, which is closed. As $K \subset W$, we also have $K \cap \partial W = \emptyset$.
So we have compact set $K$ and closed set $\partial W$ which are disjoint. So by this post there exists a $\varepsilon >0$ such that $\textrm{dist}(K , \partial W) > \varepsilon$.

Now, consider $H = \{ x \in \mathbb{R}^n \mid x \in K \textrm{ or } \textrm{dist}(x,K) < \frac{\varepsilon}{2}\}$.
Then if $x \in H \backslash W$, we have $\textrm{dist}(x,K) < \frac{\varepsilon}{2}$. Furthermore, we have $x \not\in \partial W = \bar{W} \backslash W$, so $x \not\in \bar{W}$. So we can find a closed set $G$ around $x$ such that $G \cap W = \emptyset$ and $G$ has non-empty diameter. But then there must lie a $z$ on the line between $y \in K$, such that $\| x-y\| = \textrm{dist}(x,K)$, and $x$ such that $z \in \partial W$. This would imply that $\textrm{dist}(K,\partial W) \le \textrm{dist}(K,z) \le \textrm{dist}(K,x) \le \frac{\varepsilon}{2}$. Which contradicts the minimum distance between $K$ and $\partial W$.
So we have $H \subset W$.

But then we have $$\sup_{x \in V} |f^\frac{\varepsilon}{2}(x) \le \sup_{x \in K} | f^\frac{\varepsilon}{2}(x) | \le \sup_{x \in H} | f(x) | \le \sup_{x \in W} | f(x)|$$

Which show that the inequality also holds for $p = \infty$.