# Why are the mollifiers negative in this sequence?

I'm reading a proof that starts the following way:

Assume $$E$$ is open and $$u \geq 0$$ a.e. Given $$K \subset E$$ compact, let $$\psi\in C_0^{\infty}(\mathbb{R}^n)$$ satisfy: $$\psi = 1 \quad \text { on } \{x:\text{dist}(x,K) \leq \frac{1}{2}\text{dist}(\mathbb{R}^N-E,K)$$ $$\psi = 0 \quad \text { on } \mathbb{R}^N-E$$ and $$0 \leq \psi \leq 1$$. Let $$\phi_\epsilon(x)$$ be a family of mollifers with support $$\phi_\epsilon \subset B_\epsilon(0)$$ and set $$w_n(x) = u * \phi_{\epsilon_n}(x) = \int_{B_{\epsilon_n}}u(y)\phi_{\epsilon_n}(x-y)\,dy$$ where $$\epsilon_n \rightarrow 0$$ monotonically and $$\epsilon_0 < \text{dist}(\mathbb{R}^N-E,K)$$. So $$w_n \leq 0$$ on $$K$$.

Why is $$w_n \leq 0$$ on $$K$$? Also the proof switches notation from $$\psi$$ to $$\phi$$. I'm assuming this was in error, but perhaps there's something missing?

EDIT: $w_n$">

• Huh. From this, it seems like $\psi$ and the family of modifiers $\phi_\epsilon$ are probably different objects. But I agree with you that it seems strange for $w_n$ to be negative given that $u$ is positive and that mollifiers are typically positive as well... – C. Windolf Jun 4 at 16:12
• @C.Windolf Perhaps there's something I'm missing. It's pg 36, pt (iv). books.google.com/… – yoshi Jun 4 at 18:01

I would say that this is just a typo. It should read $$w_n \ge 0$$. This is also used at the end of the proof.
• if you can, can you comment on why $w_n$ is the desired sequence? (this is from the end of the proof) – yoshi Jun 5 at 15:32
• At the end of the proof, it is written "$u_n$ is the desired sequence". – gerw Jun 5 at 17:57
• sorry! that's a typo -- my question should read: "can you comment on why $u_n$ is the desired sequence?" – yoshi Jun 6 at 1:04
• The sequence $u_n$ has the desired properties (Lipschitz, $u_n \to u$ and $u \ge 0$ on $K$). Which of these properties is not clear to you? – gerw Jun 6 at 5:35
• I guess the nature of $w_n$ confuses me as $n \rightarrow \infty$. The mollifier's radius is shrinking to a point $u(x)$. $v_n(x)$ takes the value zero there. So it looks like in the limit, there's going to be a spike at $\lim_{n\rightarrow \infty}w_n (x)= u(x)$. Should the radius be shrinking? I've added a little picture to my question. – yoshi Jun 7 at 14:07