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Lemma Let $\Omega$ be a $\sigma$-finite measure space and $J \colon \Omega \to [0, \infty]$ a measurable function. If $1 < p < \infty$ and $F \ge 0$ then the following statements are equivalent:

  1. $\lVert J \rVert_p \le F$;
  2. $\forall g \in L^{p'}(\Omega), g \ge 0, \int_{\Omega} g^{p'}dx \le 1$ we have $\int_{\Omega}Jg\, dx \le F$, where $p^\prime$ is the conjugate of $p$.

This is a lemma from this ancient question(Infinite dimensional integral inequality), which originally came from a text of Hardy-Littlewood-Polya. The symbols in that text were too old to follow, and so I tried to prove it on my own. But I had some troubles.

The direction "$1\Rightarrow 2$" follows by Holder's inequality.

For the direction "$2\Rightarrow 1$", at first I wanted to use the Rieze representation theory, but later I found two flaws in my reasoning. And I can only prove a weaker version of this direction.

Let statement $2^\prime$ be that, $\forall g \in L^{p'}(\Omega), \int_{\Omega} |g|^{p'}dx \le 1$ we have $\int_{\Omega}Jg\, dx \le F$, and further, $J\in L^p.$

My proof for "$2^\prime\Rightarrow 1$":

By statement $2^\prime$, we know $\int_{\Omega}Jg\, dx$ is a continuous functional on $L^{p^\prime}(\Omega)$ and with a operator norm less than or equal to $F$. Because its norm is just $||J||_p$, we have the statement $1$.

My question is how to prove the direction "$2\Rightarrow 1$".

Thanks for help.

Sam Wong
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    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Feb 28 '20 at 09:54

1 Answers1

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Let $(\Omega,\mathcal A, \mu)$ be the measure space in question. Since $\mu$ is a $\sigma$-finite measure there exist $C_n$ measurable sets s.t. $\mu(C_n)<\infty$, $C_1\subset C_2 \subset \ldots$ and $\bigcup_n C_n = \Omega$.

First we will show that $\mu(\{J=\infty\})=0$. For arbitrary $K>0$ and $n$ define $r_{n,K}(\omega) := I(\{J> K\} \cap C_n) $. Then for any $q>1$ we have $\int r_{n,K}^q d\mu< \mu(C_n)<\infty$, hence one can define $s_{n,K}(\omega):= r_{n,K}(\omega)^{p-1} / (\int r_{n,K}^{p})^{(p-1)/p} = I(C_n\cap \{J>K\}) / \mu(C_n \cap \{J>K\})^{(p-1)/p}$. Since $\int s_{n,K}^{p'} d\mu =1$ and $r_{n,k} \ge 0$, we can use the assumptions $$ F\ge \int s_{n,K} J d\mu \ge K \mu(C_n\cap \{J>K\})^{1/p}, $$ hence $\mu(C_n\cap \{J>K\})\to 0$ as $K\to \infty$ for any $n$. Thus taking $n\to \infty$ we obtain $\mu(\{J=\infty\})=0$, using the continuity of the measure.

Next we will show that $J\in L^p$ and that $||J||_p\le F$ holds. Define $h_{n,K}:= J\cdot I(\{J\le K\} \cap C_n)$. Due to the fact that $\int h_{n,K} ^{q} \le K^{q} \mu(C_n)<\infty $ for any $q>1$, we can define $g_{n,K} := (h_{n,K}^{p-1}) / (\int h_{n,K}^{p}d\mu))^{(p-1)/p}$. Now $g_{n,K} \ge 0$ and $\int g_{n,K}^{p'} d\mu = 1$, hence the assumptions are satisfied, so $$ \int J g_{n,K} d\mu \le F $$ for any $n$ and $K$. Denote $h_{n,n}$ by $h_{n}$. Then $h_{n}\ge0$ and $h_n\le h_{n+1}$ hence by the Beppo-Levi theorem $$ \lim_n \int h_{n}^{q} d\mu \to \int \lim_n h_n^{q}d\mu $$ for any $q>1$. Let $g_n$ denote $g_{n,n}$. Using Fatou's lemma $$ \int \liminf_n g_{n,n} Jd\mu \le\liminf_n \int g_{n,n}J d\mu = ||J||_p \le F $$

Mick
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  • Hi Mick, I think there may be two problems in this proof. The first one is that we don't know if $J\in L^p$, i.e. we don't know if $||J||_p \lt \infty$. If $||J||_p = \infty$, then your $g$ is not well defined. The second one is that, even though your $g$ is well defined, it may not be nonnegative, which is required in the statement $2$. – Sam Wong Feb 28 '20 at 09:47
  • And actually this reasoning is a part of proof of the Riesz representation theorem, more specifically, the isometry part. – Sam Wong Feb 28 '20 at 09:51
  • True, I skipped the $L^p$ part. One could choose $g = I(J = \infty \mbox{ on a compact } C)/ \lambda(C)$. Then $\int g \le 1$. I think you can show that $|| J ||_p$ cannot be $\infty$ using this, and choosing bigger and bigger compacts, exploiting the $\sigma$-finite property of $\Omega$. How could $g$ be negative? $J$ is nonnegative by assumption and so is its norm (whether or not it is finite). – Mick Feb 28 '20 at 10:26
  • I meant to write $\int g ^ {p'} \le 1$. – Mick Feb 28 '20 at 10:35
  • Oh yeah, sorry I forgot the assumption on $J$, which indeed made $g$ nonnegative. I still have problem in understanding the $L^p$ part. Did you define $g$ be constant $1/\lambda(C)$ on the set $C$? There may be another problem that $J$ is finite every where and so the measure $\lambda(C)$ is zero, with $||J||_p =\infty.$ – Sam Wong Feb 28 '20 at 10:45
  • That is true. I wanted to get rid of $x /\infty$ type stuff. Let $K>0$ and $n\in \mathbb N$, define $$ g_{K,n} := \big(J\cdot I(J\le K\ on\ cl(B(0,n)) / \int_{cl(B(0,n)\cap{J\le K}} J \big)^{p-1} $$ then $g\ge 0$ and $\int g^{p'} =1$. Taking the limit $K\to \infty$ and $n\to \infty$, this should work... – Mick Feb 28 '20 at 11:04
  • Yes, now $g_{K,n}$ are well defined for every $K,n$, but I still can't see why this contradicts the $\sigma$-finiteness of $\Omega$. Can you elaborate it a bit as an edit in your answer? Thank you so much. – Sam Wong Feb 28 '20 at 11:37
  • I edited my answer. Should be ok now, let know if something still doesn't add up... – Mick Feb 28 '20 at 14:09
  • It makes perfect sense to me. I think if you also mention the increasing monotonicity of ${h_{n,n}}$, it may be better because when taking limit in the denominator of $g_{n,n}$, we also need to quote the monotone convergence theorem (i.e. your B-L theorem). Also, I want to ask if there is a small typo in this inequalities $F\ge \int s_{n,K} J d\mu \ge K \mu(C_n\cap {J>K}).$ I think the rightmost term should be $K [\mu(C_n\cap {J>K})]^{\frac{1}{p}}$. However, even if it was a typo, your proof still perfectly works. Thank you so much :) – Sam Wong Feb 28 '20 at 15:52
  • Yes, that was a typo, it's corrected now. I also added some details on the integral of the $h_{n,n}'s$ – Mick Feb 28 '20 at 16:05
  • Errr, it may be a little bit redundant, but shall we need to specify that $g_{n,K}$ are defined a.e. (i.e. on $\Omega \setminus {J=\infty}$). Because that's the reason why we proved $\mu ({J=\infty})=0$ to avoid the case of $\infty \over \infty$. Thanks. – Sam Wong Feb 28 '20 at 16:06
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    Update: it is sufficient to use Fatou's lemma. Corrected in the answer. Yes you are right, we might want to leave out the set ${J =\infty}$ from $\Omega$ to make it absolutely precise. – Mick Feb 28 '20 at 17:28
  • I don't understand how you get the last equality. I understand that $h_n$ increases to $J$ (taking $J$ to be finite everywhere by the first part), so $Jh_n^{p-1}$ increases to $J^p$. But I don't see what exactly happens to $Jg_n$. I have the following: $$\int J^p = \int \liminf Jh_n^{p-1}\leq \liminf\int Jh_n^{p-1}\leq\liminf F(\int h_n^p)^{1-1/p}$$ where the last inequality comes from the hypothesis (by moving the normalizing factor to the right side). But I don't see any bounds on the right side. – red whisker Nov 04 '21 at 13:35