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The following question should be part of the questions I recently asked here Prove or disprove a claim related to $L^p$ space

If $g \in L^p(\Omega, \lambda)$ where $\Omega$ is a bounded subset of $\mathbb{R^n}$, $p>1$ and $\lambda$ is the Lebesgue measure. By Holder's inequality, we know that for any measurable set $E \subset \Omega$, $$\int_E |g| d \lambda \le ||g||_p \lambda (E)^{\frac{p-1}{p}}$$.

Now the question is, if there exits a constant C, such that for any measurable set $E \subset \Omega$, $$\int_E |g| d \lambda \le C \lambda (E)^{\frac{p-1}{p}}$$ Does this imply $g \in L^p(\Omega, \lambda)$?

Here is my partial work. I tried to use the the duality of $L^q$ space by contradiction or the fact that simple functions are dense to prove this characterization, but I cannot control the constant. I think I need a powerful elementary inequality. Or maybe this characterization is not true.

Any comments would be appreciated. Thanks!

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student
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3 Answers3

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The condition does not imply that $g\in L^p$.

Consider $g(x)=x^{-1/p}$ on $[0,1]$. Obviously, $g(x)\not\in L^p([0,1])$.

Let $E_\alpha=\{x:x\le\alpha^{-p}\}$, then $|E_\alpha|=\alpha^{-p}$ and $$ \begin{align} \int_{E_\alpha}g(x)\,\mathrm{d}x &=\frac{p}{p-1}\alpha^{1-p}\\ &=\frac{p}{p-1}|E_\alpha|^{\frac{p-1}p} \end{align} $$ Since $g(x)$ is smaller on any other set of measure $\alpha^{-p}$, we have $$ \int_Eg(x)\,\mathrm{d}x\le\frac{p}{p-1}|E|^{\frac{p-1}p} $$

robjohn
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  • Thanks! Could you take a look at a related problem here http://math.stackexchange.com/questions/1221758/prove-or-disprove-a-claim-related-to-lp-space ? – student Apr 06 '15 at 00:07
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Indeed, let $h=\Sigma_ia_i\chi_{E_i}$ be a simple function with $E_i \subset \Omega, E_i \cap E_j = \emptyset$ and $||h||_{\frac{p}{p-1}}=1$. $$\int gh= \Sigma_ia_i\int _{E_i} g \le C \Sigma_i a_i |E_i|^{\frac{p-1}{p}}=C||h||_{\frac{p}{p-1}}=C $$

Hence by duality and the fact that simple functions are dense, we have $g \in L^p$ and $||g||_p \le C$.

Remark: I think you made a mistake when calculating $||h||_{\frac{p}{p-1}}$. I made a mistake at the very beginning... You partial work is actually in the right direction!

  • Since $$|h|{\frac{p}{p-1}}=\left(\sum_ia_i^{\frac{p}{p-1}}|E_i|\right)^{\frac{p-1}p}\ge\sum_ia_i|E_i|^{\frac{p-1}p}\left(\sum_i|E_i|\right)^{-\frac1p}$$ I don't believe there is a way to bound $|h|{\frac{p}{p-1}}$ with $\sum\limits_ia_i|E_i|^{\frac{p-1}p}$. – robjohn Apr 03 '15 at 09:22
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Later my friends proved that the condition in the question holds if and only if $g$ is weak $L^p$.

Here is the proof:

If $g$ satisfies the condition in the question, then let $E=\{|g|>\mu\}$, then we have $$\mu |E| \le \int_E |g| \le C|E|^{1-1/p}$$Hence $\mu |E|^{1/p} \le C$, thus $g$ is weak $L^p$.

If $g$ is weak $L^p$, then $\int_E |g|=\int_0^{\infty} |E \cap \{|g|>\mu\} d \mu \le t|E| + \int_t^{\infty}\frac{C}{\mu ^p} d \mu \le t|E|+Ct^{1-p}, \forall t>0$. Hence the condition follows by letting $t=|E|^{-1/p}$.

In particular, the proof shows there exists $g \in L^{p,w}-L^p$ such that $g$ satisfies the condition in the question.

student
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