14.1 Basic Properties

Definition 14.1.1 ($\mathcal{L}^{p}$ Spaces).label 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 14.1.2 (Essential Supremum).label 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 14.1.3 (Hölder conjugates).label Let $p, q \in (1, \infty)$, then $p$ and $q$ are Hölder conjugates if

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

Lemma 14.1.4.label Let $p, q \in (1, \infty)$ be Hölder conjugates, then:

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

Theorem 14.1.5 (Hölder’s Inequality, [6.2, Fol99]).label 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 14.1.6 (Minkowski’s Inequality, [6.5, Fol99]).label Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector 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 14.1.7 ($L^{p}$ Space).label Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector 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 14.1.8.label Let $(X, \cm, \mu)$ be a measure space, $E$ be a Banach space over $K \in \RC$, $p \in [1, \infty)$, $\seq{f_n}\subset L^{p}(X; E)$, and $f \in L^{p}(X; E)$. If

(a)
$f_{n} \to f$ strongly pointwise.
(b)
There exists $g \in L^{p}(X) \cap L^{+}(X)$ such that $\norm{f_n}_{E} \le g$ for all $n \in \natp$.

then $f_{n} \to f$ in $L^{p}(X; E)$.

Proof. By assumptions $a$ and $b$, $\norm{f_n - f}_{E} \le 2g$ for all $n \in \natp$. Since $\seq{f_n}\subset L^{p}(X; E)$, $f \in L^{p}(X; E)$, and $g \in L^{p}(X)$, $\norm{f_n - f}_{E}^{p}, g^{p} \in L^{1}(X)$ for all $n \in \natp$. By the Dominated Convergence Theorem,

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

$\square$

Proposition 14.1.9.label Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector 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 23.1.1, there exists $\seq{f_n}\subset \Sigma(X, \cm; E)$ such that $\norm{f_n}_{E} \le \norm{f}_{E}$ and $\norm{f_n - f}_{E} \to 0$ strongly pointwise as $n \to \infty$. By the Dominated Convergence Theorem, $f_{n} \to f$ in $L^{p}(X; E)$.$\square$

Theorem 14.1.10 ([III.6.5, SW99]).label Let $(X, \cm, \mu)$ be a measure space and $E$ be a Banach space over $K \in \RC$, then the map $L^{1}(X; K) \td{\otimes}_{\mu} E \to L^{1}(X; E)$ defined by extending

\[L^{1}(X; K) \times E \to L^{1}(X; E) \quad f \otimes x \mapsto x \cdot f\]

is an isometric isomorphism.

Proof. By (U) of the tensor product, the given map admits a unique extension

\[M: L^{1}(X; K) \otimes E \to L^{1}(X; E) \quad \sum_{j = 1}^{n} f_{j} \otimes x_{j} \mapsto \sum_{j = 1}^{n} x_{j} \cdot f_{j}\]

Restricting $M$ to the simple functions yields a linear isomorphism

\[M: [L^{1}(X; K) \cap \Sigma(X; K)] \otimes E \to L^{1}(X; E) \cap \Sigma(X; E)\]

For any $\phi \in L^{1}(X; E) \cap \Sigma(X; E)$, write

\[\phi = \sum_{y \in \phi(X) \setminus \bracs{0}}y \cdot \one_{\bracs{\phi = y}}= M\braks{\sum_{y \in \phi(X) \setminus \bracs{0}}\one_{\bracs{\phi = y}} \otimes y}\]

then

\[\normn{M^{-1}\phi}_{L^1(X; K) \otimes E}\le \sum_{y \in \phi(X) \setminus \bracs{0}}\norm{y}_{E} \cdot \mu\bracs{\phi = y}= \int \norm{\phi}_{E} d\mu = \norm{\phi}_{L^1(X; E)}\]

On the other hand, for any representation $M^{-1}\phi = \sum_{j = 1}^{n} a_{j} \one_{A_j}$,

\[\normn{\phi}_{L^1(X; E)}\le \sum_{j = 1}^{n} \norm{a_j}_{E} \mu(A_{j}) = \sum_{j = 1}^{n} \norm{a_j}_{E} \normn{\one_{A_j}}_{L^1(X; K)}\]

As this holds for all such representations, $\normn{\phi}_{L^1(X; E)}= \normn{M^{-1}\phi}_{L^1(X; K) \otimes E}$. Therefore $M$ restricted to $[L^{1}(X; K) \cap \Sigma(X; K)] \otimes E$ is an isometry. By Proposition 14.1.9, $[L^{1}(X; K) \cap \Sigma(X; K)] \otimes E$ is dense in $L^{1}(X; K) \widehat{\otimes}_{\pi} E$, and $L^{1}(X; E) \cap \Sigma(X; E)$ is dense in $E$. By the Linear Extension Theorem, $M$ extends uniquely into the given map on $L^{1}(X; K) \otimes E$, which then extends into an isometry $L^{1}(X; K) \otimes E \to L^{1}(X; E)$.$\square$

Theorem 14.1.11 (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 $\alpha > 0$,
\[\mu\bracs{|f| \ge \alpha}\le \frac{1}{\alpha}\norm{f}_{L^1(X; E)}\]
(2)
For $\alpha > 0$ and strictly increasing function $\phi: [0, \infty) \to [0, \infty)$,
\[\mu\bracs{|f| \ge \alpha}\le \frac{1}{\phi(\alpha)}\norm{\phi \circ |f|}_{L^1(X; E)}\]
(3)
For any $\alpha > 0$,
\[\mu\bracs{|f| \ge \alpha}\le \frac{1}{\alpha^{p}}\norm{f}_{L^p(X; E)}^{p}\]

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

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

(2): By Lemma 21.3.3, $\phi: [0, \infty) \to [0, \infty)$ is Borel measurable. Since $\phi$ is strictly increasing, $\bracs{|f| \ge \alpha}= \bracs{\phi \circ |f| \ge \phi(\alpha)}$, and the result holds by applying (1) to $\phi \circ |f|$.

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

Proposition 14.1.12.label Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, $p \in [1, \infty]$, $\seq{f_n}\subset L^{p}(X; E)$, and $f \in L^{p}(X; E)$ such that $f_{n} \to f$ in $L^{p}$, then $f_{n} \to f$ in measure.

Proof. Let $\eps > 0$. If $p < \infty$, then by Markov’s Inequality,

\[\limv{n}\mu\bracs{|f_n - f| \ge \eps}\le \limv{n}\frac{1}{\eps^{p}}\norm{f_n - f}_{L^p(X; E)}^{p} = 0\]

If $p = \infty$, then there exists $N \in \natp$ such that $\norm{f_n - f}_{L^\infty(X; E)}< \eps$ for all $n \ge N$. In which case, $\mu\bracs{|f_n - f| \ge \eps}= 0$ for all $n \ge N$.$\square$

Jerry's Digital Garden

14.1 Basic Properties

Direct References

Jerry's Digital Garden

Direct References