# When $L_p = L_q$?

As we know that $L_p \subseteq L_q$ when $0 < p < q$ for probability measure, I was wondering when $L_p = L_q$ is true and why. Is it to impose some restriction on the domain space? Thanks!

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Seems to me that you only have equality when the support of the measure is finite. I don't have a proof of this, so I'm not adding it as answer. – kahen Nov 10 '10 at 11:57
The containment is backwards. If $p < q$ and the measure is finite, you have $L^q \subset L^p$. – Nate Eldredge Nov 10 '10 at 17:23

## 3 Answers

This is essentially Exercise 6.5 in Folland's Real Analysis. Let $(X,\mathcal{F},\mu)$ be a measure space, and let $$m = \inf\{\mu(A) : A \in \mathcal{F}, \mu(A) > 0\}$$ $$M = \sup\{\mu(A) : A \in \mathcal{F}, \mu(A) < \infty\}$$

For $0 < p < q < \infty$, it is a fact that $L^p(\mu) \subset L^q(\mu)$ iff $m > 0$, and $L^q(\mu) \subset L^p(\mu)$ iff $M < \infty$ (which in particular holds when $\mu$ is finite).

So for a finite measure $\mu$, it is necessary and sufficient that for some $m > 0$ every set either has measure 0 or has measure at least $m$. This is going to force your space to be "effectively" finite in some sense.

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Note that $L^p$ will even be finite dimensional. – scineram Mar 25 '11 at 10:29
+1. "So for a finite measure μ, it is necessary and sufficient that for some m>0 every set either has measure 0 or has measure at least m. This is going to force your space to be "effectively" finite in some sense." What is the necessary and sufficient condition for? – Tim Dec 28 '12 at 4:50
@Tim: Necessary and sufficient for having $L^p = L^q$ for some (or any) $p \ne q$. – Nate Eldredge Dec 28 '12 at 7:11
Thanks! Is the definition of $M$ allowing $\mu$ to take $\infty$ at some measurable subset, so $M < \infty$ doesn't necessarily mean $\mu$ is finite? Similarly, Is the definition of $m$ allowing $\mu$ to take zero at some measuable subset, so $m >0$ doesn't necessarily mean $\mu$ cannot take zero at some nonmepty measurable subset $A$? – Tim Dec 28 '12 at 13:11

Equality only when $p = q$. Suppose $p < q$, and let $r = \frac{p+q}{2}$.

Let $B_n$ be a sequence of disjoint measurable subsets of your probability space with the property that their measures $|B_n| = m_n \leq 2^{-n}$.

Let $\chi_n$ denote the characteristic function of $B_n$. Consider the sequence of functions

$$f_n = \sum_1^n \frac{1}{m_k^{1/r}} \chi_k$$

It is easy to check that this sequence is bounded in $L^p$ and Cauchy, so converges to a limit function $f$. But the sequence is unbounded in $L^q$. Hence the limit function $f$ is an element of $L^p$ that is not in $L^q$.

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Nice. Now I'm wondering if we can have equality with other (non-trivial!) measures. – Jonas Teuwen Nov 10 '10 at 12:38
sounds like the right idea.. but you assume that there exists such a sequence of disjoint sets of positive measure (which won't exist if the measure space is finite, where $L^p$ is $L^q$). Also you have to pass to a subsequence to ensure decreasing measure. – Zarrax Nov 10 '10 at 15:15
@Zaricuse: you are right on the first point of course. On the second: I only meant decrease in the eventual sense. If there exists a sequence of disjoint sets of positive measure $\lim |B_n|$ must go to zero. – Willie Wong Nov 10 '10 at 23:20
Ack, reading Nate's response made me realize that I forgot to check some signs and wrote down the wrong inequality. – Willie Wong Nov 10 '10 at 23:30
@WillieWong: Is "Equality only when p=q" wrong? I am reading the reply by Nate and the comment here math.stackexchange.com/questions/220945/…: "My bet (assuming p≠s): Only when these spaces are finite dimensional, i.e., when the total space is the finite union of atoms. – Harald Hanche-Olsen" – Tim Dec 28 '12 at 12:49

For probability measure $L_p \subset L_q$ . Equality is unreachable, see http://en.wikipedia.org/wiki/Lp_space.

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What about a discrete measure on a finite set? – scineram Mar 25 '11 at 13:15