Prove the Part (a) and Part (b).

Question

Problem: Let $(X,\Sigma,\mu)$ be a measure space. Assume that $\mu$ is a finite measure (i.e. $\mu(X)< \infty$) and $ν<<\mu.$

• (a) Prove that Radon-Nikodym derivative $ f=\frac{d\nu}{d\mu} \in L^{p}(\mu)$ if and only if there exists a constant $c ≥ 0$ such that for every pair-wise disjoint countable partition $\{ E_{n}\}$ of $X$ we have $\sum\limits_{n=1}^{\infty} \frac{\nu(E_{n})^{p}}{\mu(E_{n})^{p-1}} ≤ c$ (where $1 ≤ p < 1$).

• (b) What is the relation between $c$ and $\Vert f\Vert_{p}$?

For Part (a) I need a hint.

Answer 1

For $1<p<\infty$, \begin{align*} \sum_{n=1}\dfrac{\nu(E_{n})^{p}}{\mu(E_{n})^{p-1}}&=\sum_{n=1}\dfrac{1}{\mu(E_{n})^{p-1}}\left(\int_{E_{n}}fd\mu\right)^{p}\\ &\leq\sum_{n=1}\dfrac{1}{\mu(E_{n})^{p-1}}\|f\|_{L^{p}(E_{n})}^{p}(\mu(E_{n}))^{p/q}\\ &=\sum_{n=1}\dfrac{1}{\mu(E_{n})^{p-1}}\|f\|_{L^{p}(E_{n})}^{p}(\mu(E_{n}))^{p-1}\\ &=\sum_{n=1}\|f\|_{L^{p}(E_{n})}^{p}\\ &=\|f\|_{L^{p}(X)}^{p}, \end{align*} where $q$ is the conjugate of $p$.

For the other direction. Let $E_{n}=\{n\leq f<n+1\}$, $n\in{\bf{Z}}$, and $S_{n}=\{2^{-(n+1)}\leq f<2^{-n}\}$, $n=0,1,...$, and $T_{n}=\{-2^{n}\leq f<-2^{-(n+1)}\}$, $n=0,1,...$, $N=\{f=0\}$. Then $(E_{n})$ is a disjoint partition of $X$, and $E_{0}=\displaystyle\bigcup_{k\geq 0}S_{k}\cup N$, $E_{-1}=\displaystyle\bigcup_{k\geq 0}T_{k}$.

For $n\geq 1$, we have \begin{align*} \int_{E_{n}}|f|^{p}d\mu&=\int_{E_{n}}f^{p}d\mu\\ &\leq(n+1)^{p}\mu(E_{n})\\ &\leq 2^{p}n^{p}\mu(E_{n})\\ &=2^{p}(n\mu(E_{n}))^{p}\cdot\dfrac{1}{\mu(E_{n})^{p-1}}\\ &\leq 2^{p}\left(\int_{E_{n}}fd\mu\right)^{p}\cdot\dfrac{1}{\mu(E_{n})^{p-1}}\\ &=2^{p}\cdot\dfrac{\nu(E_{n})^{p}}{\mu(E_{n})^{p-1}}, \end{align*} similarly, for $n\leq -2$, we have \begin{align*} \int_{E_{n}}|f|^{p}d\mu\leq2^{p}\cdot\dfrac{\nu(E_{n})^{p}}{\mu(E_{n})^{p-1}}. \end{align*} Now we investigate $n=0$. One has \begin{align*} \int_{E_{0}}|f|^{p}d\mu&=\int_{E_{0}-N}|f|^{p}d\mu\\ &=\sum_{k\geq 0}\int_{S_{k}}f^{p}d\mu\\ &\leq 2^{p}\sum_{k\geq 0}\dfrac{\nu(S_{k})^{p}}{\mu(S_{k})^{p-1}}, \end{align*} by the similar technique. Also, for $n=-1$, we have \begin{align*} \int_{E_{-1}}|f|^{p}d\mu\leq 2^{p}\sum_{k\geq 0}\dfrac{\nu(T_{k})^{p}}{\mu(T_{k})^{p-1}}. \end{align*} Now we use the assumption that \begin{align*} \sum_{n\geq 1}\dfrac{\nu(E_{n})^{p}}{\mu(E_{n})^{p-1}}+\sum_{n\leq -2}\dfrac{\nu(E_{n})^{p}}{\mu(E_{n})^{p-1}}+\sum_{k\geq 0}\dfrac{\nu(S_{k})^{p}}{\mu(S_{k})^{p-1}}+\sum_{k\geq 0}\dfrac{\nu(T_{k})^{p}}{\mu(T_{k})^{p-1}}\leq c \end{align*} to conclude that \begin{align*} \int_{X}|f|^{p}d\mu\leq 2^{p}c<\infty, \end{align*} note that $\nu(N)=0$ so that this term can be omitted.

Answer 2

The converse to the direction proven here by user284331 is similar. One may take some large $k \ge 1$ and have $E_j = \{|f| \in [\frac{j}{2^k},\frac{j+1}{2^k}]\}$ so that the Hölder inequality is basically tight on each of these $E_j$'s. We can then conclude (by taking $k$ larger and larger) that $||f||_{L^p(X)}^p \le c$. So then for part $b$, we get $c = ||f||_{L^p(X)}^p$.

Answer 3

How is my answer?

Since $f\in L^{p}(E_{n})$ then $\int_{E_{n}}\frac{(d\nu)^{p}}{(d\mu)^{p}}d\mu<\infty$ so $$\int_{E_{n}}\frac{(d\nu)^{p}}{(d\mu)^{p-1}}<\infty$$ $$\bigcup_{n}\int_{E_{n}}\frac{(d\nu)^{p}}{(d\mu)^{p-1}}<\infty$$ $$\sum_{n}\frac{(d\nu)^{p}}{(d\mu)^{p-1}}\mu(X)<\infty$$ $$\sum_{n}\frac{(d\nu)^{p}}{(d\mu)^{p-1}}<\infty$$