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

We define 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 $W$.

I 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 $q\in W$

I have no idea how to prove the above identity based on the information given in the problem. Any ideas?

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up vote 2 down vote accepted

Hint: Since $m$ is the minimizer of $\|f - m\|_p$, for any $q \in W$, $$ \frac{d}{dt} \|f - m - tq\|_p^p = 0 $$ (why?) Then try and evaluate this expression.

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Can you, please, explain why is the fact that $m$ is a minimizer implies $\frac{d}{dt}\left \| f-m-tq \right \|_{p}^{p}=0$ ? – M.Krov Dec 2 '12 at 3:19
For any $t \in \mathbb{R}$, $m + tq$ is in $W$, so you know $||f - m||_p \leq ||f - m - tq||_p$. Some tools from calculus (of a single real variable) can help here. – Pot Dec 2 '12 at 12:28

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