Consider a smooth solution that satisfies the elliptic equation $$-\Delta u=(u-1)(u+1)\ \text{ in }\ \{|x|<1\},\ \ u=f(x)\ \text{ on }\ |x|=1,$$ where $f$ is a continuous function such that $-1<f(x)<1$ on $|x|=1$. Prove that $-1<u\le 1$ in $|x|<1$.
This question comes from the qualify exam of UCLA.
I can only prove that $u\geq-1$: if $u(x_0)=\min_{|x|\le 1}u<-1$, then $|x_0|<1$ and $\Delta u(x_0)\geq 0$, which leads into a contradiction with our original equation. But the similar argument fails for the upper bound of $u$, and I really don't know how to exclude the possibility of $\min_{|x|\le 1}u=-1$.
Any help would be appreciated.