<Home>Qualifying Exams><Analysis>Fall 2019 - Problem 3

Fall 2019 - Problem 3

Lp spaces

Consider a measure space $\p{X, \mathcal{X}}$ with $\sigma$ -finite measure $\mu$ and let $p \in \p{1, \infty}$ . Let $L^{p,\infty}$ be the set of measurable $\func{f}{X}{\R}$ with $\br{f}_p = \sup_{t>0} t\mu\p{\set{\abs{f} > t}}^{1/p}$ finite. Let

\norm{f}_{p,\infty} = \sup_{\stackrel{E \in \mathcal{X}}{\mu\p{E} \in \p{0,\infty}}} \frac{1}{\mu\p{E}^{1-\frac{1}{p}}} \int_E \abs{f} \,\diff\mu.

Prove that there exist $c_1, c_2 \in \p{0, \infty}$ ---which may depend on $p$ and $\mu$ ---such that

\forall f \in L^{p,\infty}, \quad c_1\br{f}_p \leq \norm{f}_{p,\infty} \leq c_2\br{f}_p.

Solution.

If $f = 0$ , then the inequalities hold trivially. Otherwise, there exists some $t > 0$ such that $\mu\p{\set{\abs{f} > t}} > 0$ . Then

\begin{aligned} \frac{1}{\mu\p{\set{\abs{f} > t}}^{1-\frac{1}{p}}} \int_{\set{\abs{f} > t}} \abs{f} \,\diff\mu &\geq \frac{t\mu\p{\set{\abs{f} > t}}}{\mu\p{\set{\abs{f} > t}}^{1-\frac{1}{p}}} \\ &= t\mu\p{\set{\abs{f} > t}}^{1/p}. \end{aligned}

Taking the supremum over $t$ and then over positive $\mu$ -measure sets, we see that

\br{f}_p \leq \norm{f}_{p,\infty}.

Conversely, let $E \in \mathcal{X}$ be with $\mu\p{E} \in \p{0, \infty}$ . Then

\begin{aligned} \frac{1}{\mu\p{E}^{1-\frac{1}{p}}} \int_E \abs{f} \,\diff\mu &= \frac{1}{\mu\p{E}^{1-\frac{1}{p}}} \int_0^\infty \mu\p{\set{x \in E \mid \abs{f\p{x} > t}}} \,\diff{t} \\ &\leq \frac{1}{\mu\p{E}^{1-\frac{1}{p}}} \int_0^\infty \min\,\set{\mu\p{E}, \mu\p{\set{\abs{f} > t}}} \,\diff{t} \\ &\leq \frac{1}{\mu\p{E}^{1-\frac{1}{p}}} \int_0^\infty \min\,\set{\mu\p{E}, \frac{\br{f}_p^p}{t^p}} \,\diff{t} \\ &= \frac{1}{\mu\p{E}^{1-\frac{1}{p}}} \p{\int_{\set{\frac{\br{f}_p^p}{t^p} \leq \mu\p{E}}} \frac{\br{f}_p^p}{t^p} \,\diff{t} + \int_{\set{\frac{\br{f}_p^p}{t^p} \geq \mu\p{E}}} \mu\p{E} \,\diff{t}} \\ &= \frac{1}{\mu\p{E}^{1-\frac{1}{p}}} \p{\int_\lambda^\infty \frac{\br{f}_p^p}{t^p} \,\diff{t} + \int_0^\lambda \mu\p{E} \,\diff{t}} && \p{\lambda = \frac{\br{f}_p}{\mu\p{E}^{1/p}}} \\ &= \frac{1}{\mu\p{E}^{1-\frac{1}{p}}} \p{\frac{\lambda^{1-p}\br{f}_p^p}{p - 1} + \lambda\mu\p{E}} \\ &= \frac{1}{\mu\p{E}^{1-\frac{1}{p}}} \p{\frac{\mu\p{E}^{1-\frac{1}{p}}\br{f}_p}{p - 1} + \mu\p{E}^{1-\frac{1}{p}}\br{f}_p} \\ &= \p{\frac{p}{p - 1}} \br{f}_p, \end{aligned}

which completes the proof.