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Let $K:= \{ u \in L^2(\Omega) : |u| \leq 1\}$ where the absolute value means the $L^2$ norm. Consider the Hilbert space projection operator $P:L^2(\Omega) \to K$, which satisfies by definition $$(Px-x, Px-a) \leq 0$$ for all $a \in K$.

I read in here that:

  1. If $|x| > 1$, then $Px = \frac{x}{|x|}$

  2. If $|x| = 1$, and $w \in L^2$ and $t > 0$ (can be chosen sufficiently small) is such that $|x+tw| \leq (1+t^2|w|^2)^{\frac 12}$, then $P(x+tw) = \frac{x+tw}{\sup(1, |x+tw|)}$.

I don't see why either of these are true. I tried plugging these expressions into the definition, but I wasn't able to prove it. ANy help please.

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    Part 1 holds in any Hilbert space. Do you know that $Px$ is the closest point in $K$ to $x$? This fact, together with the equality case of the triangle inequality, leads to (1). – 2017-01-07
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    $Px = x / \max(1, \|x\|)$ for any $x$. You can easily obtain this by applying the method of multipliers to the problem "minimize $\|u-x\|^2$ subject to $\|u\|^2 = \langle u,u\rangle \le 1$". – 2017-01-08
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    Thanks @dohmatob . Useful to know – 2017-01-08

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