Talk:Minkowski inequality

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proof is incomplete[edit]

The proof is incomplete as one of the conditions to use Hoelder's inequality is in this case that or equivalently that . To avoid reasoning circularly this would have to be proved without using the Minkowski inequality. Anybody knows how to do this in a nice way? --MarSch 13:58, 1 October 2007 (UTC)

This condition is not necessary for the use of Holder's inequality (it is only used if one wants to prove that both sides of Holder's inequality are finite). R.e.b. 15:20, 1 October 2007 (UTC)
The current proof of Hölder's inequality uses this finiteness condition to justify, in effect, division (normalization). So if what you say is correct then the problem has merely shifted from here to there. --MarSch 13:23, 2 October 2007 (UTC)
Holder's inequality holds trivially if one of the functions has infinite norm. R.e.b. 14:59, 2 October 2007 (UTC)

You are correct of course. --MarSch 16:58, 2 October 2007 (UTC)

The division by at the end of the proof also requires that finiteness of has been proven. --MarSch 10:54, 3 October 2007 (UTC)

Okay, I think we're all good now. Thanks R.e.b.. --MarSch 12:19, 4 October 2007 (UTC)

The proof is due to Hardy and Littlewood[edit]

I believe so, but I don't have a source confirming it at hand. So, I'm putting this here instead of the article. -- Taku (talk) 00:12, 2 January 2009 (UTC)

Comments for Minkowski's integral inequality.[edit]

Since the integral form is for some obscure reason often not covered in elementary textbooks, I find that the section concerning the integral form ought to be put in a completely precise form. In that regard, the conditions stated are insufficient. Both measures are usually assumed sigma-finite. The proof of the integral form is more or less the same as the non-integral version (with an extra limit argument to assure finiteness), but relies on the Fubini-Tonelli Theorem (at the point where sum and integration are interchanged), which is where the sigma-finite condition enters. Note that the standard counter example to the Fubini-Tonelli Theorem for non-sigma-finite measures is also a counter example for Minkowski's Integral Inequality for non-sigma-finite measures: Take with f(x,y)=1 if x=y and zero otherwise, and put Lebesgue meaure in one coordinate and counting measure in the other. The sigma-finite condition on both measure spaces is thus not superfluous.

Also, I find the following form of Minkowski's Integral Inequality enlightening, as it shows in a pretty neat fashion the relation between the 'normal' Minkowski inequality and the integral version (generalising by considering integrals instead of sums):

Note that the standard Minkowski Inequality can be put in the form


Now we see that the integral form can equivalently be cast in the form


where the dot means that the norm is calculated with respect to the dotted variable. Isdatmaths (talk) 09:21, 14 February 2013 (UTC)