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Suppose $1<p<\infty$ and $f,g \in L^p (X,M,\mu)$. Where $||f||_P$ and $||g||_p$ are non zero, and $||f+g||_p = ||f||_p +||g||_p$ .

Proving that equality: $${f \over ||f||_p} = {g \over ||g||_p} \text{ }\mu -a.e.$$

What theorem is available for that prove? I can't find start point.

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See this –  leo Jun 14 '12 at 4:04
    
@leo: while t.b.'s answer there can be moved this way to get an answer, I don't think they are exactly dupes. –  Willie Wong Jun 14 '12 at 9:18
    
@WillieWong I just realized that this is about Minkowski's inequality. I have read the question too fast. Sorry about that. You are right this a good question in his own right. However, the conditions to equality in Minkowski's inequality are the same of Hölder's inequality. –  leo Jun 14 '12 at 16:11
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up vote 2 down vote accepted

The triangle inequality (Minkowski) is proven using Hölder. To show what you want, you can repeat that argument:

(note that $q(p-1)=p$)

$$ \|f\|_p+\|g\|_p=\|f+g\|_p=\left(\int|f+g|^p\right)^{1/p}=\left(\int|f+g|\,|f+g|^{p-1}\right)^{1/p}\leq\left(\int|f|\,|f+g|^{p-1}+\int|g|\,|f+g|^{p-1}\right)^{1/p} \leq\left(\|f\|_p \left(\int|f+g|^p\right)^{1/q}+\|g\|_p \left(\int|f+g|^p\right)^{1/q}\right)^{1/p} =\left((\|f\|_p+\|g\|_p)\left(\int|f+g|^p\right)^{1/q} \right)^{1/p} =\left(\|f+g\|_p \|f+g\|_p^{p/q} \right)^{1/p} =\left(\|f+g\|_p \|f+g\|_p^{p-1} \right)^{1/p} =\left( \|f+g\|_p^{p} \right)^{1/p} =\|f+g\|_p=\|f\|_p+\|g\|_p $$

In particular we have equality in the two Hölder inequalities we used in the middle (the second "$\leq$"). Equality in Hölder occurs only when the $p$ and $q$ powers of the two factors are linearly dependent. This means that there exist constants $\alpha,\beta$ such that $$ |f|^p=\alpha|f+g|^{q(p-1)},\ \ |g|^p=\beta|f+g|^{q(p-1)} $$ almost everywhere.

So $|g|^p=\gamma|f|^p$ a.e. for some constant $\gamma$, and then $|g|=\delta|f|$ a.e. for yet another constant. Then $$ \frac{|f|}{\|f\|_p}=\frac{|g|}{\|g\|_p}\ \ \ \text{a.e.} $$

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Perfect answer. Thanks @MartinArgerami –  Collins Jun 13 '12 at 5:19
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