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Dear All,

As we know that this following Miskowski's integral inequality is true for $1\leq p<\infty$

$ [\int_{S_1}|\int_{S_2}F(x,y)d\mu_1(x)|^pd\mu_2(y)dy]^{1/p} \leq \int_{S_2}[\int_{S_1}|F(x,y)|^pd\mu_2(y)]^{1/p}d\mu_1(x) $

I think that Minkowski's integral inequality is not right for the case $p=\infty$. And I am trying to find a counter-example when $p=\infty$ but I have not had luck. Does anyone know any counter example for that case OR if you know how to prove that it is still right. Thank you very much and I really appreciate your help.

Phi

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The wikipedia says that it IS true for $p=\infty" (en.wikipedia.org/wiki/Minkowski_inequality), and who are we to argue? –  Igor Rivin Dec 15 '11 at 21:30
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I'm confused. I see expressions of the form $[\ldots]^p$ in you formula. How should I interpret them when $p=\infty$? –  André Henriques Dec 15 '11 at 22:00
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The wikipedia says that the statement should be modified "in the obvious way" for $p=\infty.$ Presumably to replace the usual integrals by essential suprema (or garden variety suprema for continuous functions). –  Igor Rivin Dec 15 '11 at 22:09
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In any case, if you still have questions, I think they would belong better on math.stackexchange.com where you will find many people able and willing to clarify. –  Yemon Choi Dec 15 '11 at 22:19
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Phi, as Andre points out, what you wrote down doesn't make sense for $p=\infty$. There is an analogue for $p=\infty$: $$\bigg\|\int f(x,y)\; dy\bigg\|_\infty\le \int \big\|f(x,y)\big\| _\infty\; dy$$ This is not in every analysis book, but it is in Folland's Real Analysis (p. 194). –  B R Dec 15 '11 at 23:54
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closed as too localized by Igor Rivin, Yemon Choi, fedja, Bill Johnson, Andres Caicedo Dec 16 '11 at 6:24

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1 Answer

Presumably we want: $$ \sup_{y \in S_1} \left|\int_{S_2} F(x,y) d\mu_1(x)\right| \le \int_{S_2}\sup_{y \in S_1} |F(x,y)| d\mu_1(x) $$ where $\sup$ is actually essential supremum with respect to $\mu_2$ on $S_1$.

Notation a bit confusing, since here (unlike the wikipedia version) $\mu_1$ is on $S_2$ and $\mu_2$ is on $S_1$.

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