Why are the mollifiers negative in this sequence?

Question

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: $w_n$">

Answer 1

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.