Talk:Hölder's inequality

Generalization

What about the generalized inequality of Hölder?

I have removed the following nonsense

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:

$\| f ~ g \|_{\frac{1}{1/r + 1/s}} \le \| f \|_{r} ~ \| g \|_{s},$

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 $(f^{\theta}, g^\theta)\,; p = r/\theta , q = s/\theta ,$ where $\theta = (r^{-1} + s^{-1})^{-1}\,$.

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))

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 $\|f\|_p$ and $\|g\|_q$ 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)

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
$\int |fg| \le \int |f| \cdot |g| \le \int \|f\|_\infty \cdot |g| = \|f\|_\infty \cdot \|g\|_1$.
The proof should at least mention this case though. Lavaka (talk) 17:43, 15 January 2008 (UTC)

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,
$\prod_{i=1}^{m}\left(\sum_{j=1}^{n}a_{i_{j}}\right)\ge\left(\sum_{j=1}^{n}\sqrt[m]{\prod_{i=1}^{m}a_{i_{j}}}\right)^{m}$

Basically, for $a_{ij}\geq 0$ and m>1, because of the convexity of m-th root, we have

$\prod_i \sum_j (\cdot) \geq \sum_j\prod_i(\cdot)\geq \left[\sum_j\left(\prod_i(\cdot)\right)^{1/m}\right]^m$

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 ($1\leq p_i\leq \infty, \sum_i (1/p_i)=1.$

$\sum_{j=1}^{n}\prod_{i=1}^m a_{ij}\leq \prod_{i=1}^{m}\left(\sum_{j=1}^{n} a_{ij}^{p_i}\right)^{1/p_i}$

which follows from generalized Holder's inequality. (Igny (talk) 02:56, 1 February 2008 (UTC))

Converse inequality, extremal equality?

I added this "converse inequality" that was not exactly given by the article, and that is useful for the duality of Lp spaces. Bdmy (talk) 23:22, 2 November 2008 (UTC)

I was not very happy with the name I gave to the section, and changed it to "Extremal equality". If someone comes with a better suggestion...
I also made the text more precise, and mentioned the duality $L^p$ - $L^q$. Bdmy (talk) 09:34, 3 November 2008 (UTC)

What about f and g are complex valued functions?

I think in that case, we should use the conjugate of g. —Preceding unsigned comment added by 201.223.189.127 (talk) 06:41, 16 December 2008 (UTC)

No. Why adding a useless "bar" in these inequalities? You may apply the given inequality to $\overline g$, if you need to. Bdmy (talk) 12:25, 16 December 2008 (UTC)

Inner product form

I added this section to the page. The distinction is almost trivial, but it thoroughly confused me when I first saw the inequality, since some people use Wikipedia's convention and others use the inner product version, and nobody seems to mention the difference. Saying that Cauchy-Schwarz follows with p=q=2 doesn't help (since this seems to hint that ||fg||1=|(f,g)| ). I imagine that this is what confused the person (201.223.189.127) who started the section just above this one on the talk page. If someone thinks that this should be integrated into the article in a different way then fair enough, but I feel strongly that there should at least be a mention of it. Quietbritishjim (talk) 13:57, 2 May 2009 (UTC)

Used to prove Minkowski's Inequality?

The page mentions that Holder's inequality is used to prove Minkowski's inequality. This is true for the proof given on the Minkowski inequality Wikipedia page, but it doesn't seem necessary at all; as best I can tell, the Minkowski Inequality follows easily from the convexity of $t^p$ on the interval $[0,\infty)$ for $1 \leq p < \infty$. Royden's Real Analysis (a standard textbook) seems to agree with me, and in fact it proves Minkowski before it proves Holder. Is there some subtlety I'm missing? Obscureeconomist (talk) 03:00, 6 September 2010 (UTC)