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/Part 3: Functional Analysis/Chapter 12: $L^{p}$ Spaces

12.1 Basic Properties

Definition 12.1.1 ($\mathcal{L}^{p}$ Spaces). Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, $f: X \to E$ be strongly measurable, and $p \in [1, \infty)$, then $f$ is $p$-integrable if

\[\norm{f}_{L^p(X; E)}= \norm{f}_{L^p(\mu; E)}= \norm{f}_{L^p(X, \cm, \mu; E)}= \braks{\int \norm{f}_E^p d\mu}^{1/p}< \infty\]

The set $\mathcal{L}^{p}(X; E) = \mathcal{L}^{p}(\mu; E) = \mathcal{L}^{p}(X, \cm, \mu; E)$ is the space of all $p$-integrable functions on $X$.

Definition 12.1.2 (Essential Supremum). Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, and $f: X \to E$ be strongly measurable, then $f$ is essentially bounded if

\[\norm{f}_{L^\infty(X; E)}= \norm{f}_{L^\infty(\mu; E)}= \norm{f}_{L^\infty(X, \cm, \mu; E)}= \inf\bracs{\alpha \ge 0|\mu(\bracs{f > \alpha}) = 0}< \infty\]

In which case, $\norm{f}_{L^\infty(X; E)}$ is the essential supremum of $f$.

Definition 12.1.3 (Hölder conjugates). Let $p, q \in (1, \infty)$, then $p$ and $q$ are Hölder conjugates if

\[\frac{1}{p}+ \frac{1}{q}= 1\]

Lemma 12.1.4. Let $p, q \in (1, \infty)$ be Hölder conjugates, then:

  1. $q = p/(p - 1)$.

  2. $p = q(p - 1)$.

Theorem 12.1.5 (Hölder’s Inequality, [6.2, Fol99]). Let $(X, \cm, \mu)$ be a measure space, $E, F$ be a normed vector spaces, $p, q \in [1, \infty]$. If $p, q$ are Hölder conjugates or if $p = 1$ and $q = \infty$, then for any $f \in \mathcal{L}^{p}(X; E)$ and $g \in \mathcal{L}^{q}(X; F)$,

\[\int \norm{f}_{E} \norm{g}_{F} d\mu \le \norm{f}_{L^p(X; E)}\norm{g}_{L^q(X; F)}\]

Proof. First suppose that $p = 1$ and $q = \infty$. In this case,

\[\int \norm{f}_{E} \norm{g}_{F} d\mu \le \norm{g}_{L^\infty(X; F)}\int \norm{f}_{E}d\mu = \norm{f}_{L^1(X; E)}\norm{g}_{L^\infty(X; F)}\]

Now suppose that $p, q \in (1, \infty)$ are Hölder conjugates. Assume without loss of generality that $\norm{f}_{L^p(X; E)}= \norm{g}_{L^q(X; F)}= 1$. By Young’s inequality,

\[\int \norm{f}_{E} \norm{g}_{F} d\mu \le \int \frac{\norm{f}_{E}^{p}}{p}+ \frac{\norm{g}_{F}^{q}}{q}d\mu = \frac{1}{p}\int \norm{f}_{E} d\mu + \frac{1}{q}\int \norm{g}_{F}^{q} d\mu = 1\]
$\square$

Theorem 12.1.6 (Minkowski’s Inequality, [6.5, Fol99]). Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, $p \in [1, \infty]$, and $f, g \in \mathcal{L}^{p}(X; E)$, then

\[\norm{f + g}_{L^p(X; E)}\le \norm{f}_{L^p(X; E)}+ \norm{g}_{L^p(X; E)}\]

Proof. If $p = 1$, then the theorem holds directly.

If $p = \infty$, then for any $\alpha > \norm{f}_{L^\infty(X; E)}$ and $\beta > \norm{g}_{L^\infty(X; E)}$, $\alpha + \beta \ge f + g$ $\mu$-almost everywhere, so

\[\norm{f + g}_{L^\infty(X; E)}\le \norm{f}_{L^\infty(X; E)}+ \norm{g}_{L^\infty(X; E)}\]

Now suppose that $p \in (1, \infty)$, then $p = q(p - 1)$, and

\begin{align*}\norm{f + g}_{E}^{p}&\le (\norm{f}_{E} + \norm{g}_{E})\norm{f + g}_{E}^{p - 1}\\ \int\norm{f + g}_{E}^{p}d\mu&\le (\norm{f}_{L^p(X; E)}+ \norm{g}_{L^p(X; E)})\braks{\int \norm{f + g}_E^{q(p - 1)}d\mu}^{1/q}\\ \int\norm{f + g}_{E}^{p}d\mu&\le (\norm{f}_{L^p(X; E)}+ \norm{g}_{L^p(X; E)})\braks{\int \norm{f + g}_E^{p}d\mu}^{1/q}\\ \braks{\int\norm{f + g}_E^pd\mu}^{1/p}&\le \norm{f}_{L^p(X; E)}+ \norm{g}_{L^p(X; E)}\end{align*}
$\square$

Definition 12.1.7 ($L^{p}$ Space). Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, and $p \in [1, \infty]$, then $\norm{\cdot}_{L^p(X; E)}$ is a seminorm on $\mathcal{L}^{p}(X; E)$. The quotient

\[L^{p}(X, \cm, \mu; E) = \mathcal{L}^{p}(X, \cm, \mu; E)/\bracs{f|f = 0\text{ a.e.}}\]

is a normed vector space, known as the $E$-valued $L^{p}$ space on $(X, \cm, \mu)$.

Proof. By Minkowski’s Inequality.$\square$

Proposition 12.1.8. Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, and $p \in [1, \infty)$, then $\Sigma(X, \cm; E) \cap L^{p}(X; E)$ is dense in $L^{p}(X; E)$.

Proof. Let $f \in L^{p}(X; E)$. By Definition 17.1.1, there exists $\seq{f_n}\subset \Sigma(X, \cm; E)$ such that $\norm{f_n}_{E} \le \norm{f}_{E}$ and $\norm{f_n - f}_{E} \to 0$ pointwise as $n \to \infty$.

For each $n \in \nat$, $\norm{f_n}_{E} \le \norm{f}_{E}$, so $\norm{f_n}_{L^p(X; E)}\le \norm{f}_{L^p(X; E)}< \infty$, and $\norm{f_n - f}_{E} \le 2\norm{f}_{E}$. By the Dominated Convergence Theorem,

\[\limv{n}\int \norm{f_n - f}_{E}^{p} d\mu = \int \limv{n}\norm{f_n - f}_{E}^{p} d\mu = 0\]

Therefore $\norm{f_n - f}_{L^p(X; E)}\to 0$ as $n \to \infty$.$\square$

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