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I am reading wikipedia's entry on Holder's inequality https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality#Counting_measure

There are quite a few variations of Holder's inequality "applied to different measures" - a phrase quite new to me. I have no idea what this notion of a measure is, when I look up, for example, the counting measure, it says that the counting measure $\nu: S \to [0, \infty]$, is the size of a set $A$. https://www.ma.utexas.edu/users/gordanz/notes/measures.pdf

Can someone please clarify what it means to "apply an inequality to different measures". I am not a mathematician, what is a simple way to understand this idea? Must we specify which measure we are applying an inequality against?

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    $\begingroup$ If you don't know what a measure is, it's hard to see how one can write an answer that would be helpful. Understanding what measures are and how to use them is typically a half-year course at the advanced undergraduate or graduate level. $\endgroup$ – Nate Eldredge Nov 19 '15 at 3:27
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Hölder's inequality is an inequality involving integrals. You may be familiar with ordinary integrals of functions $f:\mathbb{R}\to\mathbb{R}$, but integrals can be defined in much more general contexts as well. A "measure" on a set $S$ is a gadget that allows you to define integrals of functions $f:S\to\mathbb{R}$ (the measure on $\mathbb{R}$ that gives ordinary integration is called "Lebesgue measure"). Hölder's inequality is valid when talking about the integrals obtained by any measure in this way. So "applying Hölder's inequality to a measure" just means using Hölder's inequality where the integrals in the inequality are integrals defined using that measure.

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  • $\begingroup$ This is exactly my question: what is the process that give birth to Holder's inequality. Is this inequality the result of of a single unifying inequality to different contexts, or do different contexts have their own holder inequality but no unifying holder inequality. This is a very natural question, given that for example, Holder give birth to Cauchy Schwartz for p = q = 2. $\endgroup$ – Carlos - the Mongoose - Danger Nov 19 '15 at 3:40
  • $\begingroup$ I'm not sure what you mean by a "single unifying inequality". It is possible to give a single proof of Hölder's inequality that works for all measures simultaneously, if that's what you mean. $\endgroup$ – Eric Wofsey Nov 19 '15 at 3:42
  • $\begingroup$ In your answer you have specified that the unifying guy here is the one involving integrals. Defining the integral with respect to different measures give rises to different Holder inequality (different in the sense it is defined wrt to different measures). And yes it would be interesting to see that proof Eric! $\endgroup$ – Carlos - the Mongoose - Danger Nov 19 '15 at 3:44
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Holder's inequality is a very general result concerning very general integrals in an arbitrary measure space. This is probably the most remarkable thing about Holder's inequality, and why it is so useful.

Even in the statement of the theorem (as stated in the wikipedia link) the result only requires that we have a measure space $(S,\Sigma,\mu)$. All they mean by "applied to different measures" is that we can explore some meaningful implications and say explicitly what the inequality means if we are given a priori certain popular measure spaces (e.g., counting measure, Lebesgue measure, etc...).

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  • $\begingroup$ Thanks, it is not obvious to me as an engineer why we need to specify it. I just use it. If I am given a bunch of functions, I apply the one that has integral sign. If I am given a vector, I use the one with the counting measure. I never think about the underlying "measurable space" but then again I don't do proofs I just crank out results. Good to have this clarified. $\endgroup$ – Carlos - the Mongoose - Danger Nov 19 '15 at 3:34

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