Talk:Hölder's inequality
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What about the generalized inequality of Hölder?
I have removed the following nonsense
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[edit] Generalizations
Hölder's inequality can be generalized to lessen requirements on its two parameters. While p ≥ 1 and q ≥ 1, with 1/p + 1/q = 1, one has in terms of any two positive numbers, r > 0 and s > 0:
provided only that the integrability conditions can be generalized as well, namely that f is in Lr(S) and g is in Ls(S).
The latter inequality can be derived from Hölder's inequality applied to where . (Igny 19:59, 7 April 2006 (UTC))
I can understand that this generalization can be considered irrelevant, but I don't see why it is nonsense
- In particular, because you can not lessen requirements on the parameters by introducing different parameters.(Igny 20:33, 7 November 2006 (UTC))
- Ok. You mean it is not a generalization, but a consequence or corollary. Right?
- Yes, an unimportant corollary. (Igny 16:55, 30 November 2006 (UTC))
- Thanks for the clarification. My interest in this was only that I corrected the wrong 'proof' that was originally given for that fact on this page, so I wanted to know if I had made a mistake. At the time I did not feel comfortable totally removing something that was put by someone else. However I totally agree with you on the decision. (GBlanchard 10:44, 4 December 2006 (UTC))
- Yes, an unimportant corollary. (Igny 16:55, 30 November 2006 (UTC))
[edit] does it hold in case of infinite norm?
Following some discussion on Talk:Minkowski inequality I wonder if the inequality can still be proved if we allow one of and to be infinite. Or perhaps we can even allow both f and g to have infinite norm? --MarSch 13:29, 2 October 2007 (UTC)
- Okay, yeah, R.e.b. is right when he says it holds trivially. Still it would be nice to mention for completeness and because the proof of the Minkowski inequality uses this. --MarSch 16:59, 2 October 2007 (UTC)
[edit] L^infty and L^1 case
Does Holder hold if p = 1 and q = infinity? The statement on the page seems to be in the affirmative, but the proof via Young's Inequality doesn't cover this case. Lavaka (talk) 17:21, 15 January 2008 (UTC)
- yeah, after thinking about it, it's obvious for this case, since
- .
- The proof should at least mention this case though. Lavaka (talk) 17:43, 15 January 2008 (UTC)
[edit] Removed inequality
I have removed the inequality which seems trivial and has nothing to to with Holder's.
- One can also apply the inequality to a set of a set of numbers aij, where i is in the interval [1,m] and j is in the interval [1,n]. In this case,
Basically, for and m>1, because of the convexity of m-th root, we have
Or I am missing something? (Igny (talk) 19:05, 30 January 2008 (UTC))
Now, when I thought a bit about that, I think the contributor meant the following inequality (
which follows from generalized Holder's inequality. (Igny (talk) 02:56, 1 February 2008 (UTC))