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Consider $1. Let's define the space: $L_{V}^{p}(-1,1)=\left \{ f:(-1,1)\rightarrow \mathbb{R}:\int_{-1}^{1}\left | f(x) \right |^{p}V(x)dx<\infty \right \}$

Consider the norm: $\left \| f \right \|_{L_{V}^{p}}=(\int_{-1}^{1}\left | f(x) \right |^{p}V(x)dx)^{\frac{1}{p}}$

Consider $W\subset L_{V}^{p}(-1,1)$ to be a finite dimensional subspace. For a given $f$ in $L_{V}^{p}(-1,1)$, we define the minimizer $m$ in $W$ such that: $\left \| f(x)-m(x) \right \|_{L_{V}^{p}}= min\left \| f(x)-q(x) \right \|_{L_{V}^{p}}$ for all $q$ in $W$.

We need to show that $m$ satisfies the following:

$\int_{-1}^{1}\left | f(x)-m(x) \right |^{p-2}(f(x)-m(x))q(x)V(x)dx=0$ for all $q\in W$

Any help is ppreciated?

1 Answers 1

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Consider, for each $q\in W$, the functions $J_q:\mathbb{R} \to \mathbb{R}$ given by $ J_q(t)=\int_{-1}^1 |f(x)-m(x)+tq(x)|^pV(x)dx $ and note that your conclusion is equivalent to $J_q'(0)=0$ for all $q$, but your hypothesis imply that $J_q$ has a minimum at $0$, and the result follows.

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    @M.Krov: Just notice that $g_t:=-m+tq\in W$ since the latter is a vector space. By the definition of $m$ we then have $J_q(0)=\| f-m\|_{L_V^p}^p \leq \| f-g_t\|_{L^p_V}^p =J_q(t)$2012-12-02