# Determine the orthogonal projection of $f$ onto $K := \{u \in H : |u| \le h \text{ a.e.}\}$

Let $$(\Omega, \mathcal F, \mu)$$ be a $$\sigma$$-finite measure space. Let $$h:\Omega \to [0, \infty)$$ be $$\mu$$-measurable. Let $$H:= L^2(\Omega)$$ and $$K := \{u \in H : |u| \le h \text{ a.e.}\}.$$

Then $$K$$ is non-empty closed convex subset of $$H$$. I'm trying to solve below exercise

Fix $$f \in H$$ and let $$u$$ be the orthogonal projection of $$f$$ onto $$K$$. Determine $$u$$.

Could you verify my below attempt?

Let $$\Omega_1 := \{ |f| \le h \}, \Omega_2 := \{ f < -h \}$$ and $$\Omega_3 := \{ f > h \}$$. We define $$u$$ by $$u (\omega) := \begin{cases} f(\omega) &\text{if} \quad \omega \in \Omega_1, \\ -h(\omega) &\text{if} \quad \omega \in \Omega_2, \\ h(\omega) &\text{if} \quad \omega \in \Omega_3. \end{cases}$$

Then $$u \in K$$. Fix $$v \in K$$. It suffices to prove that $$\int_\Omega (f-u)(v-u) \le 0$$. First, $$\int_{\Omega_1} (f-u)(v-u) =0$$. We have $$f+h<0$$ on $$\Omega_2$$ and $$v+h \ge 0$$, so $$\int_{\Omega_2} (f-u)(v-u) = \int_{\Omega_1} (f+h)(v+h) \le 0.$$

We have $$f-h>0$$ on $$\Omega_3$$ and $$v-h \le 0$$, so $$\int_{\Omega_3} (f-u)(v-u) = \int_{\Omega_1} (f-h)(v-h) \le 0.$$

This completes the proof.

• I think this is a beautiful approach. Commented Apr 29, 2023 at 22:01
• @KennyWong Thank you so much for your verification! Commented Apr 29, 2023 at 22:38