Proiezioni negli $$L^p$$

Problema. Sia $$f \in L^p ([-1,1])$$, con $$p \in [1, +\infty)$$, e consideriamo $Y = \{h \in L^p ([-1,1]) \, : \, h \text{ è pari}\}.$$$Y$$ è un sottospazio chiuso. Mostrare che $$g(x) = (f(x)+f(-x))/2$$ è tale che $\min_{h \in Y} \|f - h\|_p = \|f - g \|_p.$
E' vero anche per $$p = \infty$$?
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Re: Proiezioni negli $$L^p$$

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For $$f \in L^p([-1,1])$$ we want to solve the minimization problem $0 \le \ell= \inf_{ h \in Y } \|f - h \|_p = \inf_{ \tilde{h} \in Y } \|f_o - \tilde{h} \|_p$ where we set $$\tilde{h}= h - f_e$$ and $$f_e$$, $$f_o$$ are respectively the (essentially unique) even and odd parts of $$f$$ defined by $f_e (x) = \frac{f(x) + f(-x)}{2} \quad \text{and} \quad f_o (x) = \frac{f(x) - f(-x)}{2}$almost everywhere. By Theorem 7' page 76 of Lax's book (we switch the role of $$Y$$ and $$Y^{\bot}$$ with respect to the formulation of the theorem, but in this case the operation is harmless because $$Y$$ is closed and therefore $$(Y^{\bot})^{\bot} =Y$$) we have that (using duality arguments) $\ell = \sup_{l \in Y^{\bot}, \|l\| \le 1} |l(f_o)| = \sup_{g \in Y^{\bot}, \|g\|_q \le 1} \left| \int_{-1}^1 f_o g \, d \mu \right| \le \|f_o\|_p$ where the supremum is actually reached by $$\bar{g} = f_o |f_o|^{p-2} / \|f_o\|_p ^{p-1} \in L^q ([-1,1]) \cap Y^{\bot}$$, which is an odd function (indeed $$Y^{\bot}$$ contains all the odd functions belonging to $$L^q$$); here $$q$$ is the Hölder conjugate exponent of $$p$$. We conclude noticing that $\left| \int_{-1}^1 f_o \bar{g} \, d \mu \right| = \|f_o\|_p$ and therefore $\inf_{ \tilde{h} \in Y } \|f_o - \tilde{h} \|_p = \| f_o \|_p.$Now, for $$p \in (1, +\infty)$$ the spaces $$L^p ([-1,1])$$ are uniformly convex and since $$Y$$ is a linear closed subspace the minimizer exists and it is unique. It follows that $$\tilde{h}=0$$ a.e. which implies $$h = f_e$$ almost everywhere. For $$p=1$$ the infimum is reached for sure for $$\tilde{h} = 0$$, but it not necessarily unique.
consumami
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