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Consider a probability measure $\mu$ on a set $X$. Let $p,q \in (1, \infty)$, $f \in L^{pq} \cap L^1$ (so also $f\in L^p \cap L^q$) by non-negative. Can we say anything about the relationship between

$$\int f^{qp} d\mu + \left(\int f d\mu\right)^{qp}$$


$$\left(\int f^p d\mu \right)^q + \left(\int f^q d\mu \right)^p ?$$

In other words, is there an inequality saying that one of these two quantities is greater than or equal to the other under certain circumstances? It seems as if there should be a way to deduce something like this from Hölders/Jensens inequalities, but I have been unable to do so.

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I corrected the spelling of "Hölder". If you can't type the "ö" character, then "oe" is an acceptable substitute. Thus "Hölder" is the same as "Hoelder", but "Holder" is a different spelling. – Michael Hardy Jun 8 '12 at 21:36
If the space has finite measure, and in particular if it's a probability space, and $pq>1$, then $f\in L^{pq}$ implies $f\in L^1$, i.e. $L^{pq}$ is in that case a subset of $L^1$. So there's no need to write "$L^{pq}\cap L^1$". – Michael Hardy Jun 8 '12 at 22:28
Since $pq$ is the largest exponent here, you can make $\int f^{qp}\,d\mu$ huge while keeping the other three as small as you wish. On the other hand, direct application of Holder's inequality gives a bound for each of the terms in the second sum in terms of $\int f^{qp}\,d\mu$. – user31373 Jun 9 '12 at 1:14

Let $I(f):=\int f^{pq}d\mu+\left(\int f\mu\right)^{pq}$ and $J(f):=\left(\int f^pd\mu\right)^q+\left(\int f^qd\mu\right)^p$. Since, by Jensen inequality, $$\left(\int f^pd\mu\right)^q\leq \int f^{pq}d\mu\quad \mbox{and}\quad\left(\int f^qd\mu\right)^p\leq \int f^{pq}d\mu,$$ we have $J(f)\leq 2I(f)$ for any $f\in L^{pq}$.

But in general, we can't find $C$ universal (i.e. which doesn't depend on $f$ but only on $p$ and $q$) such that $I(f)\leq C\cdot J(f)$. Indeed, if it was the case, then in particular, for $f=\chi_A$, where a is measurable, then $$\mu(A)\leq C(\mu(A)^p+\mu(A)^q)$$ hence $\frac 1C\leq \mu(A)^{p-1}+\mu(A)^{q-1}$. If we can find a sequence $\{A_n\}$ of measurable sets such that $\mu(A_n)>0$ for all $n$ and $\mu(A_n)\to 0$ we get a contradiction. (for example when the space has not $\mu$-atoms)

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Is it possible to get $J(f) \leq I(f)$, without the 2? – user15464 Jun 9 '12 at 11:57

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