Lp norm and distribution function with dyadic series

By | May 27, 2015

Proposition. For any measurable function f:X\rightarrow\mathbb{C} and 0<p<\infty we have

    \[ \norm fp^{p}\sim_{p}\sum_{n\in\mathbb{Z}}\lambda_{f}\left(2^{n}\right)2^{np}. \]

Here X\sim_{p}Y means there are constants c_{p},C_{p}>0 depending only on p so that c_{p}Y\le X\le C_{p}Y.

Proof. For each n\in\mathbb{Z}, define

    \[ E_{n}=\left\{ x\in X:\left|f\left(x\right)\right|\ge2^{n}\right\} . \]

Then \lambda\left(2^{n}\right)=\mu\left(E_{n}\right) for all n\in\mathbb{Z}. Also, E_{n}\supset E_{n+1} for all n. Define A_{n}=E_{n}\setminus E_{n+1}. Then \bigcup A_{n}=\bigcup E_{n}=X and \left\{ A_{n}\right\} is a family of disjoint sets in X. Then we have

    \begin{align*} \norm fp^{p} & =\int_{X}\left|f\right|^{p}d\mu\\  & =\int_{\bigcup_{n\in\mathbb{Z}}A_{n}}\left|f\right|^{p}d\mu\\  & =\sum_{n\in\mathbb{Z}}\int_{A_{n}}\left|f\right|^{p}d\mu. \end{align*}

Note that x\in A_{n} if and only if 2^{n}\le\left|f\left(x\right)\right|<2^{n+1} for all n\in\mathbb{Z}.

For the upper bound, we have

    \begin{align*} \norm fp^{p} & \le\sum_{n\in\mathbb{Z}}\left(2^{n+1}\right)^{p}\int_{A_{n}}d\mu\\  & =\sum_{n\in\mathbb{Z}}\left(2^{n+1}\right)^{p}\mu\left(A_{n}\right)\\  & =\sum_{n\in\mathbb{Z}}\left(2^{n+1}\right)^{p}\left(\lambda_{f}\left(2^{n}\right)-\lambda_{f}\left(2^{n+1}\right)\right)\\  & \le\sum_{n\in\mathbb{Z}}\left(2^{n+1}\right)^{p}\lambda_{f}\left(2^{n}\right)\\  & \le C_{p}\sum_{n\in\mathbb{Z}}\lambda_{f}\left(2^{n}\right)2^{np} \end{align*}

where C_{p}=2^{p}.

For the lower bound, we have

    \begin{align*} \norm fp^{p} & =\sum_{n\in\mathbb{Z}}\int_{A_{n}}\left|f\right|^{p}d\mu\\  & \ge\sum_{n\in\mathbb{Z}}\left(2^{np}\right)\mu\left(A_{n}\right)\\  & =\sum_{n\in\mathbb{Z}}2^{np}\left(\lambda_{f}\left(2^{n}\right)-\lambda_{f}\left(2^{n+1}\right)\right). \end{align*}

So

    \[ \sum_{n\in\mathbb{Z}}2^{np}\lambda_{f}\left(2^{n}\right)\le\norm fp^{p}+\sum_{n\in\mathbb{Z}}\lambda_{f}\left(2^{n+1}\right)2^{np}. \]

Hence,

    \[ \sum_{n\in\mathbb{Z}}\left(1-\frac{1}{2^{p}}\right)\lambda_{f}\left(2^{n}\right)2^{np}\le\norm fp^{p}. \]

Take c_{p}=1-\frac{1}{2^{p}}>0. Then we finally get \norm fp^{p}\sim_{p}\sum_{n\in\mathbb{Z}}\lambda_{f}\left(2^{n}\right)2^{np}.

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