Orthogonal projection on subspace Let $\Omega$ be a measure space and let $h : \Omega → [0, +∞)$ be a measurable function. 
Let$$K = \{u ∈ L^2(\Omega);\ |u(x)| ≤ h(x)\ a.e. on\ \Omega\}.$$
Check that K is a non-empty closed convex set in $H = L^2(\Omega)$. Determine $P_K$.


*

*$K$ non-empty: I do not have a clue

*$K$ $\textbf{closed}$: I would take a sequence $(f_n)_{n\geq 1}\subset K$ such that $f_n\to f$ in $L^2$ and show that $|f(x)| ≤ h(x)\ a.e$ but I do not see how this is implied by the onvergence.

*determine $P_K$: take $f\in H$, then $<f-P_kf,v>=0, \forall v\in K$, so I would write: $<f-P_kf,v>=\int_\Omega (f(x)-P_kf(x))\cdot v(x)dx=0$ and try to solve but I do not get anywhere. 
Any help?

1) $u(x)\equiv0$ is in $K$.
2) If $f_n\to f$ in $L^2$, there is a subsequence $f_{n_k}$ converging to $f$ almost everywhere. Since $|f_n(x)|\le h(x)$ a.e, it follows that $|f(x)|\le h(x)$ a.e. and $f\in K$.
3) The condition you state for $P_Kf$ is incorrect (it would be OK if $K$ where a subspace.) It should be
$$
\langle f-P_kf,u-P_Kf\rangle\le0\quad\forall u\in K.
$$
$$
P_Kf(x)=\begin{cases}
f(x) & \text{if }|f(x)|\le h(x),\\
h(x) & \text{if }f(x)> h(x),\\
-h(x) & \text{if }f(x)< - h(x).
\end{cases}
$$
