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Added minkowski's inequality for integrals.

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This commit is contained in:
Bokuan Li
2026-09-19 11:28:31 -04:00
parent 8b93c445fa
commit c05faf4a77
3 changed files with 97 additions and 6 deletions
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src/fa/lp/definition.tex
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\begin{theorem}[Hölder's Inequality]
\label{theorem:holder}
Let $(X, \cm, \mu)$ be a measure space, $E, F$ be a normed vector spaces, $p, q \in [1, \infty]$ be Hölder conjugates, then for any $f \in \mathcal{L}^p(X; E)$ and $g \in \mathcal{L}^q(X; F)$,
Let $(X, \cm, \mu)$ be a measure space and $p, q \in [1, \infty]$ be Hölder conjugates, then for any $f, g: \in L^+(X)$,
\[
\int \norm{f}_E \norm{g}_F d\mu \le \norm{f}_{L^p(X; E)}\norm{g}_{L^q(X; F)}
\int fg d\mu \le \norm{f}_{L^p(X; \real)}\norm{g}_{L^q(X; \real)}
\]
\end{theorem}
\begin{proof}[Proof, {{\cite[Theorem 6.2]{Folland}}}. ]
First suppose that $p = 1$ and $q = \infty$. In this case,
\[
\int \norm{f}_E \norm{g}_F d\mu \le \norm{g}_{L^\infty(X; F)}\int \norm{f}_Ed\mu = \norm{f}_{L^1(X; E)}\norm{g}_{L^\infty(X; F)}
\int fg d\mu \le \norm{g}_{L^\infty(X; \real)}\int fd\mu = \norm{f}_{L^1(X; \real)}\norm{g}_{L^\infty(X; \real)}
\]
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 \hyperref[Young's inequality]{lemma:young-inequality},
Now suppose that $p, q \in (1, \infty)$ are Hölder conjugates, and assume without loss of generality that $\norm{f}_{L^p(X; \real)} = \norm{g}_{L^q(X; \real)} = 1$. In which case, by \hyperref[Young's inequality]{lemma:young-inequality},
\[
\int \norm{f}_E \norm{g}_F d\mu \le \int \frac{\norm{f}_E^p}{p} + \frac{\norm{g}_F^q}{q} d\mu = \frac{1}{p}\int \norm{f}_E d\mu + \frac{1}{q}\int \norm{g}_F^q d\mu = 1
\int fg d\mu \le \int \frac{f^p}{p} + \frac{g^q}{q} d\mu = \frac{1}{p}\int f d\mu + \frac{1}{q}\int g d\mu = 1
\]
\end{proof}
\begin{theorem}[Minkowski's Inequality, {{\cite[6.5]{Folland}}}]
src/fa/lp/ineq.tex
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\end{align*}
\end{proof}
\begin{theorem}[Minkowski's Inequality for Integrals]
\label{theorem:minkowski-integral}
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:
\begin{enumerate}
\item If $E = \real$ and $f \ge 0$, then for any $p \in [1, \infty)$,
\[
\braks{\int_{X}\paren{\int_Y f(x, y)\nu(dy)}^{p}\mu(dx)}^{1/p} \le \int_{Y}\norm{f(\cdot, y)}_{L^p(X, \mu; E)} \nu(dy)
\]
\item For any $p \in [1, \infty]$, if
\begin{enumerate}[label=(\alph*)]
\item For $\nu$-almost every $y \in Y$, $f(\cdot, y) \in L^p(\mu; E)$.
\item The mapping $y \mapsto \norm{f(\cdot, y)}_{L^p(\nu; E)}$ is in $L^1(\mu; \real)$.
\end{enumerate}
then the mapping $x \mapsto \int_Y f(x, y)\nu(dy)$ is in $L^p(\mu; E)$, with
\[
\norm{\int_Y f(\cdot, y)\nu(dy)}_{L^p(X, \mu; E)} \le \int_{Y}\norm{f(\cdot, y)}_{L^p(X, \mu; E)} \nu(dy)
\]
\end{enumerate}
\end{theorem}
\begin{proof}[Proof, {{\cite[Theorem 6.19]{Folland}}}. ]
(1): If $p = 1$, then by \hyperref[Tonelli's Theorem]{theorem:fubini-tonelli},
\[
\int_X\int_Y f(x, y) \nu(dy)\mu(dx) = \int_Y\int_X f(x, y)\mu(dx)\nu(dy)
\]
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 \hyperref[Tonelli's Theorem]{theorem:fubini-tonelli},
\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 \hyperref[Hölder's Inequality]{theorem:holder},
\[
\int_Y\int_X f(x, y) \phi(x) \mu(dx) \nu(dy) \le \norm{\phi}_{L^q(\mu; \real)}\int_Y \norm{f(\cdot, y)}_{L^p(\mu; \real)} \nu(dy)
\]
As $f \ge 0$ and the above holds for all $\phi \in L^q(\mu; \real) \cap L^+(\mu)$, \autoref{theorem:lp-dual-function} implies that $x \mapsto \int_Y f(x, y)\nu(dy) \in L^p(\mu)$ with
\[
\braks{\int_{X}\paren{\int_Y f(x, y)\nu(dy)}^{p}\mu(dx)}^{1/p} \le \int_{Y}\norm{f(\cdot, y)}_{L^p(X, \mu; E)} \nu(dy)
\]
\end{proof}
src/fa/lp/weak.tex
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\section{Lorentz Spaces}
\label{section:weak-lp}
\begin{definition}[Distribution Function]
\label{definition:weak-lp-distribution}
Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, and $f \in L^0(X; E)$ be strongly measurable, then the \textbf{distribution function} of $f$ is the mapping
\[
\lambda_f: (0, \infty) \to [0, \infty] \quad \alpha \mapsto \mu\bracsn{\norm{f}_E > \alpha}
\]
\end{definition}
\begin{proposition}
\label{proposition:weak-lp-distribution-properties}
Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, and $f, g: X \to E$ be strongly measurable, then:
\begin{enumerate}
\item $\lambda_f$ is non-increasing and right-continuous.
\item If $|f| \le |g|$, then $\lambda_f \le \lambda_g$.
\item For each $\alpha > 0$ and $t \in (0, 1)$, $\lambda_{f + g}(\alpha) \le \lambda_f(t\alpha) + \lambda_g((1 - t)\alpha)$.
\item For any strongly measurable functions $\seq{f_n} \subset E^X$, if $\norm{f_n}_E \upto \norm{f}_{E}$ pointwise, then $\lambda_{f_n} \upto \lambda_f$ pointwise.
\end{enumerate}
\end{proposition}
\begin{proof}
(1): Let $\alpha > 0$ and $\seq{\alpha_n} \subset (\alpha, \infty)$ such that $\alpha_n \downto \alpha$, then
\[
\mu\bracsn{\norm{f}_E > \alpha} = \mu\paren{\bigcup_{n \in \natp}\bracsn{\norm{f}_E > \alpha_n}} = \limv{n}\mu\bracsn{\norm{f}_E > \alpha}
\]
by \hyperref[continuity from below]{proposition:measure-properties}.
(2): For each $\alpha > 0$ and $t \in (0, 1)$,
\[
\bracsn{\norm{f + g}_E > \alpha} \subset \bracsn{\norm{f}_E > t\alpha} \cup \bracsn{\norm{g}_E > (1-t)\alpha}
\]
(3): For each $\alpha > 0$, $\bracsn{\norm{f}_E > \alpha} =
Assume without loss of generality that $\fF \subset 2^{L^0(X; [0, \infty))}$ and $f \in L^0(X; [0, \infty))$. Fix $\alpha > 0$, then for each $t \in (0, 1)$,
\[
\lambda_{f}(\alpha) \le \lambda_{f - g}(t\alpha) + \lambda_{g}((1 - t)\alpha)
\]
\end{proof}