I'm trying to solve one exercise from Brezis "Let $A$ be a measure space and let $h : A \rightarrow [0, +\infty)$ be a measurable function. Let $K=\{u ∈ L^2(A); \lvert u(x)\rvert \le h(x)\ \text{a.e. on} \ A\}$.
Check that $K$ is a nonempty closed convex set in $H = L^2(A)$. Determine $P_K$."
The first part is OK. The problem is $P_K$, the projection. I tried to find a good candidate for the projection, but so far nothing. Any ideas?
If $f \in L^2(A)$, then you want to find $k_0\in K$ such that $$\|f-k_0\| \le \|f-k\|,\;\;\; k\in K.$$ If you can find $k_0\in K$ such that $|f(x)-k_0(x)| \le |f(x)-k(x)|$ a.e. for all $k\in K$, then the above must hold. A little thought gives $$ \max\{|f(x)|-u(x),0\}\le |f(x)-k(x)|,\;\;\; x \in A,\;\; k\in K. $$ Therefore, if you can find $k_0\in K$ such that $$ \max\{|f(x)|-u(x),0\} = |f(x)-k_0(x)|, $$ then it must be true that $\|f-k_0\| \le \|f-k\|$ for all $k\in K$.
