Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.


share|improve this question

closed as too localized by Igor Rivin, Yemon Choi, fedja, Bill Johnson, Andres Caicedo Dec 16 '11 at 6:24

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

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
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
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
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
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

1 Answer 1

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 $S_1$.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.