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/Part 3: Functional Analysis/Chapter 16: $L^{p}$ Spaces

16.2 $L^{p}$ Inequalities

Theorem 16.2.1 (Markov’s Inequality).label Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, and $f: X \to E$ be a Borel measurable function, then

  1. (1)

    For any $\alpha > 0$,

    \[\mu\bracs{\norm{f}_E \ge \alpha}\le \frac{1}{\alpha}\norm{f}_{L^1(X; E)}\]

  2. (2)

    For $\alpha > 0$ and strictly increasing function $\phi: [0, \infty) \to [0, \infty)$,

    \[\mu\bracs{\norm{f}_E \ge \alpha}\le \frac{1}{\phi(\alpha)}\norm{\phi \circ |f|}_{L^1(X; E)}\]

  3. (3)

    For any $\alpha > 0$,

    \[\mu\bracs{\norm{f}_E \ge \alpha}\le \frac{1}{\alpha^{p}}\norm{f}_{L^p(X; E)}^{p}\]

Proof. (1): For any $\alpha > 0$,

\begin{align*}\mu\bracs{\norm{f}_E \ge \alpha}&= \int_{\bracs{\norm{f}_E \ge \alpha}}d\mu \le \frac{1}{\alpha}\int |\norm{f}_{E}d\mu = \frac{\norm{f}_{L^1(X; E)}}{\alpha}\end{align*}

(2): By Lemma 26.3.3, $\phi: [0, \infty) \to [0, \infty)$ is Borel measurable. Since $\phi$ is strictly increasing, $\bracs{\norm{f}_E \ge \alpha}= \bracs{\phi \circ \norm{f}_E \ge \phi(\alpha)}$, and the result holds by applying (1) to $\phi \circ \norm{f}_{E}$.

(3): Since $x \mapsto x^{p}$ is strictly increasing on $[0, \infty)$, applying (2) to $\phi(x) = x^{p}$ yields the desired result.$\square$

Lemma 16.2.2.label Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, $p \in [1, \infty)$, and $f, g \in L^{p}(X; E)$, then

\[|\norm{f}_{L^p(X; E)}^{p} - \norm{g}_{L^p(X; E)}^{p}| \le p\norm{f - g}_{L^p(X; E)}(\norm{f}_{L^p(X; E)}\vee \norm{g}_{L^p(X; E)})^{p-1}\]

Proof. By Lemma 3.2.2,

\begin{align*}|\norm{f}_{L^p(X; E)}^{p} - \norm{g}_{L^p(X; E)}^{p}|&\le p|\norm{f}_{L^p(X; E)}- \norm{g}_{L^p(X; E)}|\\&\times (\norm{f}_{L^p(X; E)}\vee \norm{g}_{L^p(X; E)})^{p-1}\\&\le p\norm{f - g}_{L^p(X; E)}(\norm{f}_{L^p(X; E)}\vee \norm{g}_{L^p(X; E)})^{p-1}\end{align*}

$\square$

Theorem 16.2.3 (Minkowski’s Inequality for Integrals).label Let $(X, \cm, \mu)$ and $(Y, \cn, \nu)$ be $\sigma$-finite measure spaces, $E$ be a Banach space, and $f: X \times Y \to E$ be $(\cm \otimes \cn)$-strongly measurable. Then:

  1. (1)

    If $E = \real$ and $f \ge 0$, then for any $p \in [1, \infty)$,

    \[\braks{\int_{X}\paren{\int_Y f(x, y)\nu(dy)}^{p}\mu(dx)}^{1/p}\le \int_{Y}\norm{f(\cdot, y)}_{L^p(X, \mu; E)}\nu(dy)\]

  2. (2)

    For any $p \in [1, \infty]$, if

    1. (a)

      For $\nu$-almost every $y \in Y$, $f(\cdot, y) \in L^{p}(\mu; E)$.

    2. (b)

      The mapping $y \mapsto \norm{f(\cdot, y)}_{L^p(\nu; E)}$ is in $L^{1}(\mu; \real)$.


    then the mapping $x \mapsto \int_{Y} f(x, y)\nu(dy)$ is in $L^{p}(\mu; E)$, with

    \[\norm{\int_Y f(\cdot, y)\nu(dy)}_{L^p(X, \mu; E)}\le \int_{Y}\norm{f(\cdot, y)}_{L^p(X, \mu; E)}\nu(dy)\]

Proof, [Theorem 6.19, Fol99]. (1): If $p = 1$, then by Tonelli’s Theorem,

\[\int_{X}\int_{Y} f(x, y) \nu(dy)\mu(dx) = \int_{Y}\int_{X} f(x, y)\mu(dx)\nu(dy)\]

Now suppose that $p \in (1, \infty)$, assume without loss of generality that $\int_{Y}\norm{f(\cdot, y)}_{L^p(X, \mu; E)}\nu(dy) < \infty$. Let $q$ be the Hölder conjugate of $p$ and $\phi \in L^{q}(\mu; \real) \cap L^{+}(\mu)$ with $\phi \ge 0$, then by Tonelli’s Theorem,

\begin{align*}\int_{X} \int_{Y} f(x, y)\nu(dy) \phi(x) \mu(dx)&= \int_{X}\int_{Y} f(x, y)\phi(x)\nu(dy) \mu(dx) \\&= \int_{Y}\int_{X} f(x, y) \phi(x) \mu(dx) \nu(dy)\end{align*}

By Hölder’s Inequality,

\[\int_{Y}\int_{X} f(x, y) \phi(x) \mu(dx) \nu(dy) \le \norm{\phi}_{L^q(\mu; \real)}\int_{Y} \norm{f(\cdot, y)}_{L^p(\mu; \real)}\nu(dy)\]

As $f \ge 0$ and the above holds for all $\phi \in L^{q}(\mu; \real) \cap L^{+}(\mu)$, Theorem 16.3.3 implies that $x \mapsto \int_{Y} f(x, y)\nu(dy) \in L^{p}(\mu)$ with

\[\braks{\int_{X}\paren{\int_Y f(x, y)\nu(dy)}^{p}\mu(dx)}^{1/p}\le \int_{Y}\norm{f(\cdot, y)}_{L^p(X, \mu; E)}\nu(dy)\]

$\square$

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