Suppose $1 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.
$L^p$ measurable functions equality
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0@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. – 2012-06-14
1 Answers
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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0Perfect answer. Thanks @MartinArgerami – 2012-06-13