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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.

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    $\begingroup$ I think this is a beautiful approach. $\endgroup$
    – Kenny Wong
    Commented Apr 29, 2023 at 22:01
  • $\begingroup$ @KennyWong Thank you so much for your verification! $\endgroup$
    – Analyst
    Commented Apr 29, 2023 at 22:38

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