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This question is from Folland 6.38, Show that $f \in L^p $ iff $\sum_{k=-\infty}^ {\infty} 2^{pk} \mu \{{x: |f(x)|>2^{k}}\} \lt \infty$

  1. If $f \in L^p $, I applied the Chebyshev's inequality
  2. But for the other direction, I don't know how to begin.

Any advice would be appreciated. Thanks

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2 Answers

up vote 4 down vote accepted

On $E_k = \{ 2^k < \left| f(x) \right| \leq 2^{k+1} \}$, we have

$$ 2^{pk} < \left|f(x)\right|^{p} \leq 2^{p(k+1)}.$$

Thus integrating on $E_k$ we have

$$ 2^{pk} \mu (E_k) < \int_{E_k} \left|f(x)\right|^{p} \, d\mu \leq 2^{p(k+1)} \mu(E_k).$$

Summing through $k$, we obtain

$$ \sum_{k=-\infty}^{\infty} 2^{pk} \mu (E_k) < \int \left|f(x)\right|^{p} \, d\mu \leq 2^{p} \sum_{k=-\infty}^{\infty} 2^{pk} \mu(E_k).$$

Finally, note that

$$ \sum_{j=-\infty}^{k} 2^{pj} = \frac{2^{kp}}{1 - 2^{-p}}. $$

Thus we have

$$ \begin{align*} \sum_{k=-\infty}^{\infty} 2^{pk} \mu (E_k) &= (1 - 2^{-p}) \sum_{k=-\infty}^{\infty} \sum_{j=-\infty}^{k} 2^{jp} \mu (E_k) \\ &= (1 - 2^{-p}) \sum_{j=-\infty}^{\infty} \sum_{k=j}^{\infty} 2^{jp} \mu (E_k) \\ &= (1 - 2^{-p}) \sum_{j=-\infty}^{\infty} 2^{jp} \mu \{ 2^j < \left| f(x) \right| \}. \end{align*}, $$

from which the conclusion follows.

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Note that $$\int|f|^pd\mu=p\int_0^\infty t^{p-1}\mu\{x:|f(x)|>t\}dt$$ and $$\int_{2^k}^{2^{k+1}}t^{p-1}\mu\{x:|f(x)|>t\}dt\le 2^{(k+1)p}\mu\{x:|f(x)|>2^k\}.$$

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