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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?

mastro
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1 Answers1

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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 $$ \langle f-P_kf,u-P_Kf\rangle\le0\quad\forall u\in K. $$ $$ P_Kf(x)=\begin{cases} f(x) & \text{if }|f(x)|\le h(x),\\ h(x) & \text{if }f(x)> h(x),\\ -h(x) & \text{if }f(x)< - h(x). \end{cases} $$

Julián Aguirre
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  • Thank you very much! Is the condition you are referring to: $<f−Pkf,v>=0,∀v∈K$? Isn't K a subspace because if I sum two elements in $K$, then the result can be outside $K$? And how do you prove formally that the projection is what you wrote? – mastro Jan 16 '15 at 16:33
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    $K$ is not a subspace: $|u(x)|\le h(x)$ and $|v(x)|\le h(x)$ do not imply that $|u(x)+v(x)|\le h(x)$ or $|c,u(x)|\le h(x)$ for all constants $c$. As for the proof, I leave it to you; you should do part of your homework yourself. – Julián Aguirre Jan 16 '15 at 16:58
  • Thank you again. For the proof regarding the projection, I would like to do it by myself but I do not get what technique I should use to prove it formally. – mastro Jan 18 '15 at 17:21
  • Let $u$ be any function in $K$ and show that $\langle f-P_kf,u-P_Kf\rangle\le0$. – Julián Aguirre Jan 19 '15 at 09:53