1
$\begingroup$

Let $K$ be a compact set in $\mathbb R^n$.

Suppose we have a Radon measure $\sigma$ and denote the Lebesgue measure by $m$.

Suppose we have $f\in L^p_\sigma(K):=\{g \text{ is measurable}\,\vert \,\int_K \vert g \vert^p \,d\sigma <\infty \}$.

I want something like $\int_K \vert f \vert^p \,dm \le C \int_K \vert f \vert^p \,d\sigma$ for any $f\in L^p_\sigma(K)$, where $C$ is some positive constant. Is this possible to hold? And if it holds, can you give me a hint on the proof of it?

Or alternatively, does this weaker claim holds? Weaker Claim: $\int_K \vert f \vert \,dm \le C \int_K \vert f \vert^p \,d\sigma$ for any $f\in L^p_\sigma(K)$.


My attempt: I have considered the Riesz–Markov–Kakutani representation theorem (which relates Radon measures to Riemann-Stieltjes integrals), but our function $f$ may neither be continuous nor have compact support. So this theorem may not apply. I am quite confused now. Thanks for any help.

$\endgroup$

1 Answer 1

1
$\begingroup$

Take $\sigma = \delta_0$ and assume that $0∈ K$. Then for any $f∈ C^0$, $$ \int_K |f|^p \,\delta_0(\mathrm d x) = |f(0)|^p $$ However, take $f_n(x) = \min(n,|x|^{-1/2p})$. Then $$ \int_K |f_n|^p \,\delta_0(\mathrm d x) = n^p \underset{n\to\infty}{\longrightarrow} \infty $$ but $$ \int_K |f_n|^p \,\mathrm d x \underset{n\to\infty}{\longrightarrow} \int_K |x|^{-1/2} \,\mathrm d x < \infty \\ \int_K |f_n| \,\mathrm d x \underset{n\to\infty}{\longrightarrow} \int_K |x|^{-1/2p} \,\mathrm d x < \infty $$ It proves that the inequality you are looking for does not hold in general, even your "weaker" version. The only possibility is when $\sigma$ is absolutely continuous with respect to the Lebesgue measure (or at least its positive part, you can always add a general Radon measure as a negative part). Indeed, if your inequality is true, then for any smooth function compactly test function $\varphi∈ C^\infty_c(K)$, $$ \langle \sigma,\varphi\rangle \leq C\, \langle 1,\varphi\rangle $$ which is equivalent to say that $\sigma ≤ C$ in the sense of distributions, and so $\sigma$ is absolutely continuous with respect to the Lebesgue measure, and its associated function $g∈ L^\infty$. Reciprocally, if $σ(\mathrm d x) = g(x)\,\mathrm d x$ for some $g∈ L^\infty$, then $$ \int_K |f(x)|^p\, \sigma(\mathrm d x) = \int_K |f(x)|^p\, g(x)\,\mathrm d x ≤ \|g\|_{L^\infty} \,\int_K |f(x)|^p\,\mathrm d x $$


Remark: I used the notation $\sigma(\mathrm d x)$ instead of $\mathrm d \sigma$ to avoid confusions with the Stieltjes integrals notation, as explained in my answer here.

$\endgroup$
4
  • $\begingroup$ Hi LL, thanks for the answer. I have three questions regarding to the "weaker version" part in your answer. Q1: How do we deduce $\langle \sigma,\varphi\rangle \leq C\, \langle 1,\varphi\rangle$ from the "weak claim"? Q2: I know the Radon-Nikodym theorem but I don't know the result that $\int_K |f(x)|^p\, \sigma(\mathrm d x) = \int_K |f(x)|^p\, g(x)\,\mathrm d x$. Can you give me a hint to prove it or a reference? Q3: The Radon measure $\sigma$ is positive (where $\sigma$ is not a signed measure). What do you mean by its positive part? Thank you so much. $\endgroup$
    – Sam Wong
    Commented Aug 23, 2021 at 9:17
  • $\begingroup$ Q1: You can just take $f = \varphi^{1/p}$ if $\varphi$ is positive (and decompose $φ = \varphi_+ - \varphi_-$ in the general case). Q2: Yes, the Radon-Nykodym theorem tells you that the identity holds when $|f|^p$ is the indicator of a measurable set. For general Lebesgue measurable functions, it follows from the fact that one can approximate measurable functions by simple functions. (Another point of view is saying that $σ = g d x$ in the sense of measures and so in the sense of distributions, and then use the density of smooth functions) $\endgroup$
    – LL 3.14
    Commented Aug 23, 2021 at 15:52
  • $\begingroup$ Q3: If $\sigma$ is positive, then forget about my remark ;) $\endgroup$
    – LL 3.14
    Commented Aug 23, 2021 at 15:53
  • $\begingroup$ Hi LL, thanks for the comments. I understand Q2 now. But I am still confused with the Q1. Even though we take $f=\varphi^{1\over p}$, shouldn't we get from the "weak claim" something like $\langle 1,\varphi^{1\over p}\rangle \leq C\, \langle \sigma,\varphi\rangle$? Can you write out some more steps about how to derive the inequality? Thank you so much. $\endgroup$
    – Sam Wong
    Commented Aug 24, 2021 at 3:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .