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Let $(X,\mu)$ be a measure space. Suppose that $0 < p_{0} < p < p_{1} < \infty$ and $\frac{1}{p} = \frac{1-\theta}{p_{0}} + \frac{\theta}{p_{1}}$ for some $\theta \in (0,1)$. If $f \in L^{p_{0},\infty}(X,\mu) \cap L^{p_{1},\infty}(X,\mu)$, then $f \in L^{p,\infty}(X,\mu)$ and $$\left\|f\right\|_{L^{p,\infty}} \leq \left\|f\right\|_{L^{p_{0},\infty}}^{1-\theta}\left\|f\right\|_{L^{p_{1},\infty}}^{\theta}$$ It isn't hard to show this inequality for a constant greater than 1, using Holder's inequality for $L^{p,\infty}$ spaces (Grafakos Classical Fourier Analysis Exercise 1.1.15), but I can't figure out how to prove the stated inequality.

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  • $\begingroup$ Where is $p_1$? $0 < p_0 < p < p_1$? $\endgroup$
    – JT_NL
    Jul 28, 2011 at 20:15
  • $\begingroup$ It is late, but are you sure it is true? Let $f = 2 1_A$. Then $\|f\|_{p, \infty} = \mu(A)^{1/p}$. Hence we get $\mu(A)^{1/p} \leq \mu(A)$. Did I make a mistake somewhere? If not, this fails for $p = \frac12$ and $\mu(A) = 2$. $\endgroup$
    – JT_NL
    Jul 28, 2011 at 22:30
  • $\begingroup$ I'm not sure that the stated inequality is true, although Grafakos appears to claim that it's true by including it in his book. However, I don't understand how you get $\mu(A)^{\frac{1}{p}} \leq \mu(A)$. Don't we have $\left\|f\right\|_{L^{p_{0},\infty}} = \mu(A)^{\frac{1}{p_{0}}}$ and $\left\|f\right\|_{L^{p_{1},\infty}} = \mu(A)^{\frac{1}{p_{1}}}$. Hence, $$\left\|f\right\|_{L^{p_{0},\infty}}^{1-\theta}\left\|f\right\|_{L^{p_{1},\infty}}^{\theta} = \mu(A)^{\frac{1}{p}}$$ $\endgroup$ Jul 28, 2011 at 23:49
  • $\begingroup$ Right, I was mistaken (the sum is $1/p$ not $1$). I will see if I can find a solution, part a) was easy (I have done it some years ago) but with the Hölder inequality for weak $L^p$ you get an additional constant... $\endgroup$
    – JT_NL
    Jul 29, 2011 at 12:25

1 Answer 1

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WLOG assume $f \geq 0$.

$$ \|f\|_{p,\infty}^p := \sup_{t > 0} t^p \mu\{ f > t\} $$

Now, fix $\lambda,\eta \in (0,1)$, you have

$$ t^p \mu\{ f > t\} = t^{\eta p} \left(\mu\{f > t\}\right)^\lambda \cdot t^{(1-\eta)p}\left( \mu\{f > t\}\right)^{(1-\lambda)} $$

or

$$ = \left( t^{\eta p / \lambda} \mu\{f > t\}\right)^{\lambda} \cdot \left( t^{(1-\eta) p /(1-\lambda)} \mu\{f > t\}\right)^{1-\lambda} $$

so taking the sup of the terms inside the parentheses

$$ \| f \|_{p,\infty}^p \leq \| f \|_{\eta p/\lambda,\infty}^{\frac{\eta p}{\lambda}\cdot\lambda}\| f\|_{(1-\eta)p/(1-\lambda),\infty}^{\frac{(1-\eta)p}{1-\lambda}\cdot(1-\lambda)} $$

or

$$ \| f \|_{p,\infty} \leq \| f \|_{\eta p / \lambda,\infty}^\eta \| f \|_{(1-\eta)p/(1-\lambda),\infty}^{1-\eta} $$

Now let $p_0 = \eta p / \lambda$ and $p_1 = (1-\eta)p / (1-\lambda)$. It is simple to check that

$$ \frac{\eta}{p_0} + \frac{1-\eta}{p_1} = \frac{\lambda}{p} + \frac{1-\lambda}{p} = \frac{1}{p} $$

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    $\begingroup$ Hah, it's that easy. Nice +1. $\endgroup$
    – JT_NL
    Jul 29, 2011 at 15:52
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    $\begingroup$ @Jonas: The natural next question to ask is what can be said about general Lorentz spaces $L^{p,s}$. Using the same method you can prove something that allows to interpolate both in $p$ and in $s$, but I am not sure what the constant should be. $\endgroup$ Jul 29, 2011 at 16:20
  • $\begingroup$ @Willie: Many Thanks! $\endgroup$ Jul 29, 2011 at 19:15

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