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I have been staring at this for hours. I cannot figure out how to prove the following from Folland, problem 6.38. Show that: $$f \in L^p \iff \sum_{k=-\infty}^{+\infty}2^{kp}\lambda_f(2^k)<\infty$$

Where $\lambda_f(2^k)=\mu \left\{x: |f(x)|>2^k \right\}$.

Some guidance would be greatly appreciated. Thank you.

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1 Answer 1

Hint: Consider the function $g = \sum_{k\in\mathbb{Z}} 2^k \chi_{\left\{x: |f(x)|>2^k\right\}}$, and show that $|f|\leq g\leq 2|f|$.


Added: in the interest of thoroughness, I'll expound. For any $p\geq 1$ we may define $$g_p = \sum_{k\in\mathbb{Z}} 2^{pk} \chi_{\left\{x: |f(x)|>2^k\right\}},$$ and note that whenever $f(x)\neq 0$, and $n\in\mathbb{Z}$ such that $2^n\leq|f(x)|<2^{n+1}$, then we have $$g_p(x) = \sum_{m=-n}^\infty\left(\frac{1}{2^p}\right)^{m} = 2^{(n+1)p}\frac{1}{2^p-1},$$ hence (everywhere, including such $x$ as $f(x)=0$) $$\frac{1}{2^p-1}|f(x)|^p \leq g_p(x) \leq \frac{2^p}{2^p-1}|f(x)|^p.$$ Finally, this implies that $$f\in L^p\iff g_p\in L^1\iff \sum_{k\in\mathbb{Z}} 2^{pk} \mu(\left\{x: |f(x)|>2^k\right\})<\infty.$$

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