How to prove the following without using Hölder's inequality :
$$ \|f\|_{p} = \sup_{\|g\|_q =1} \int |fg| d\mu ; \frac{1}{p} + \frac{1}{q} =1$$
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4Why do you want to avoid Hölder's inequality? Without using Hölder's inequality, I'd say you prove it just like you prove Hölder's inequality, it is, after all, Hölder's inequality in disguise. – Daniel Fischer Jun 28 '14 at 13:26
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@DanielFischer: The wikipedia page http://en.wikipedia.org/wiki/Minkowski_inequality says that it is a general caseof Minkowski to prove which holder is used. So, I thought there must be a different way to prove. – Anonymous Jun 28 '14 at 15:16
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Well, the equality says in particular that $\int \lvert fg\rvert,d\mu \leqslant \lVert f\rVert_p \cdot \lVert g\rVert_q$ for all $g$ with $\lVert g\rVert_q = 1$. Homogeneity gives the inequality for all $\lVert g\rVert_q\neq 0$, and the case $\lVert g\rVert_q = 0$ is evident. So there you have Hölder. Additionally, the equality here says that you can come arbitrarily close to equality in Hölder's inequality, but the usual proofs of Hölder's inequality also yield the equality in case $\lvert g\rvert^q = \lvert f\rvert^p$ (a.e.). – Daniel Fischer Jun 28 '14 at 15:49
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I'll try to clarify what Wikipedia says:
The Minkowski inequality is the triangle inequality in $L^p(S)$. In fact, it is a special case of the more general fact $$\|f\|_p = \sup_{\|g\|_q = 1} \int |fg| d\mu, \qquad 1/p + 1/q = 1 \tag{*}$$ where it is easy to see that the right-hand side satisfies the triangular inequality.
Indeed, once you have (*), Minkowski's inequality easily follows:
$$\|f+h\|_p = \sup_{\|g\|_q = 1} \int |(f+h)g| d\mu
\le \sup_{\|g\|_q = 1} \int |fg| d\mu+\sup_{\|g\|_q = 1} \int |hg| d\mu
=\|f\|_p+\|h\|_p $$
As Daniel Fischer said, (*) is a statement that contains Hölder's inequality. If you are proving (*), you are proving Hölder's inequality along the way.
If you want a proof of Minkowski's inequality without Hölder's inequality, see here.
