Given $f \in L^2(\Omega)$, set $M=\{u\in L^2(\Omega): u\ge f\}$. Is $M$ a half space of $L^2(\Omega)$ and can we describe the projection of an arbitrary element $g\in L^2(\Omega)$ onto $M$?
2026-03-26 04:49:30.1774500570
Projection onto a set
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The answer is $\max \{f,g\}$. Proof: note that $\|g-\max \{f,g\}\|_2^{2}=\int_{g \leq f} (g-f)^{2}$. For any $u \in M$ we have $\|g-u\|_2^{2}\geq \int_{g \leq f} (g-u)^{2}$ and just observe that $|g-u| =u-g \geq f-g =|f-g|$ on $\{g \leq f\}$ so $\|g-u\|_2^{2}\geq \int_{g \leq f} (g-f)^{2}=\|g-\max \{f,g\}\|_2^{2}$.