Theorem 15.2.1 (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)
For any $\alpha > 0$,
\[\mu\bracs{\norm{f}_E \ge \alpha}\le \frac{1}{\alpha}\norm{f}_{L^1(X; E)}\] - (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)
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 25.3.3, $\phi: [0, \infty) \to [0, \infty)$ 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.$\square$
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