[Measure Theory] Lp Space
In this post, we introduce some spaces of functions called the $L^p$ spaces, or sometimes Lebesgue spaces that play a crucial role in analysis, and other applications such as statistics, physics, etc.
Definitions
Consider a measure space $(X, \mathcal{A}, \mu)$.
Define the $L^p$ norm of a measurable function $f$ by $$ \begin{aligned} \| f \|_p = \left[ \int | f(x) |^p \; \mu (dx)\right]^{1/p} \end{aligned} $$ for $1 \leq p < \infty$.
For $p = \infty$, define $$ \begin{aligned} \| f \|_\infty = \inf \left\{ M \geq 0 : \mu \left( \{ | f | > M \} \right) = 0 \right\} \end{aligned} $$ Note that if no such $M$ exists, then we define $\| f \|_\infty = \infty$.
There are another notions of supremum and infimum in the jargons of $L^p$ space, named essential supremum and infimum, and it is related to the definition of $L^\infty$.
Let $f$ be a real-valued measurable function. The essential supremum and essential infimum are defined by $$ \begin{aligned} \text{ess sup } f & = \inf \; \{ M : \mu (\{f > M\}) = 0 \} \\ \text{ess inf } f & = \sup \; \{ m : \mu (\{f < m\}) = 0 \} \\ \end{aligned} $$
In other words, the essential supremum is defined as the smallest almost everywhere upper bound of $f$, and the essential infimum is defined as the largest almost everywhere lower bound $f$. So, we always have
\[\begin{aligned} \text{ess sup } f & \leq \sup f \\ \text{ess inf } f & \geq \inf f \end{aligned}\]Also note that
\[\begin{aligned} \| f \|_\infty = \text{ess sup } | f | \end{aligned}\]Hence $| f(x) | \leq || f ||_\infty$ for $\mu$-almost every $x$.
Now we define the $L^p$ space:
For $1 \leq p \leq \infty$, the $L^p$ space is the set $$ \begin{aligned} \{ f: \| f \|_p < \infty \} \end{aligned} $$ We denote it by $L^p (X)$, $L^p (\mu)$.
Basic inequalities on $L^p$ space
The basics to the study of Lp spaces are inequalities. In this post, we discuss two inequalities: Hölder’s and Minkowsky inequality.
Hölder’s inequality
Including Hölder’s inequality, the condition $p^{-1} + q^{-1} = 1$ turns up frequently in $L^p$ theory, which is called conjugate exponent:
If $1 < p < \infty$, we define $q$ by $\frac{1}{p} + \frac{1}{q} = 1$. We call $q$ the conjugate exponent of $p$. If $p = 1$, set $q = \infty$. If $p = \infty$, set $q = 1$.
If $1 < p, q < \infty$, $\frac{1}{p} + \frac{1}{q} = 1$, and $f$ and $g$ are measurable, then $$ \begin{aligned} \int | fg | \; d\mu \leq \| f \|_p \| g \|_q. \end{aligned} $$ This also holds if $p = \infty$ and $q = 1$, or if $p = 1$ and $q = \infty$.
$\mathbf{Proof.}$
For the case of $p = \infty$ and $q = 1$, let \(M = \|f\|_\infty\), then $| f | \leq M$ a.e. and $\int | fg | \leq M \int | g |$. Thus the case $p=\infty$ and $q=1$, and similarly $p = 1$ and $q=\infty$ follow.
The main idea of the other parts is using $e^t$ is convex. In other words,
\[\begin{aligned} e^{\lambda a + (1- \lambda b)} \leq \lambda e^a + (1 - \lambda) e^b \end{aligned}\]for any $a \leq b$, $0 \leq \lambda \leq 1$. Then we set $\lambda = \frac{1}{p}$ and $1 - \lambda = \frac{1}{q}$ for the case $1 < p, q < \infty$.
With $F(x) = | f(x) | / ||f||_p$ and $G(x) = | g(x) | / ||g||_p$, we have
\[\begin{aligned} F(x) G(x) \leq \frac{F(x)^p}{p} + \frac{G(x)^q}{q} \end{aligned}\]by setting $e^a = (F(x))^{1/\lambda}$, $e^b = (G(x))^{1/(1-\lambda)}$. Note that this inequality also holds if $F = 0$ or $G = 0$.
For the case that $|| f ||_p \neq 0$ and $|| g ||_p \neq 0$,
\[\begin{aligned} \int FG \; d\mu \leq \int \frac{F(x)^p}{p} \; d\mu + \int \frac{G(x)^q}{q} \; d\mu = \frac{\| F \|_p^p}{p} + \frac{\| G \|_q^q}{q} = \frac{1}{p} + \frac{1}{q} = 1. \end{aligned}\] \[\tag*{$\blacksquare$}\]Note that when $p = q = 2$, this inequality is the Cauchy-Schwartz inequality.
Minkowsky’s inequality
$\mathbf{Proof.}$
If $p = 1$ or $a = 0$, then the inequality holds. Thus let’s assume $p > 1$ and $a > 0$.
Set $f (x) = 2^{p-1} + 2^{p-1} x^p - (1+x)^p$ with $x = b / a$. Note that $f(0) > 0$, $f(1) = 0$, and $\lim _{x \to \infty} f(x) = \infty$.
Then
\[\begin{aligned} f^\prime (x) & = 2^{p - 1} p x^{p - 1} - p (1 + x)^{p-1} \\ & = p [ (2x)^{p-1} - (1 + x)^{p-1}] \end{aligned}\]Since $f^\prime < 0$ for $1 > x \geq 0$ and $f^\prime > 0$ for $x > 1$, $f$ takes its minimum at $x = 1$ and, hence $f(x) \geq 0$ for $x \geq 0$.
\[\tag*{$\blacksquare$}\]If $1 \leq p \leq \infty$ and $f, g$ are measurable, then $$ \begin{aligned} \| f + g \|_p \leq \| f \|_p + \| g \|_p. \end{aligned} $$
$\mathbf{Proof.}$
If $p = 1$, we have the results from the fact that
\[\begin{aligned} | (f + g) (x) | \leq | f(x) | + | g(x) | \end{aligned}\]Consider $p = \infty$. If \(\| f \|_\infty = \infty\) or \(\| g \|_\infty = \infty\), then the inequality holds trivially. If both are finite, then
\[\begin{aligned} \| f + g \|_\infty = \text{ ess sup } | f + g | \leq \| f \|_\infty + \| g \|_\infty \end{aligned}\]as \(\| f \|_\infty\) is a.e. upper bound of $| f |$ and \(\| g \|_\infty\) is a.e. upper bound of $| g |$.
Consider $1 < p < \infty$. If \(\| f \|_p\) or \(\| g \|_p\) is infinite, the result is obvious, so we may assume both are finite.
Let $a = | f |$ and $b = | g |$. From the lemma and integrating both side, we have
\[\begin{aligned} \int | f + g |^p \; d\mu \leq \int ( |f| + |g| )^p \; d\mu \leq \int 2^{p-1} |f|^p + 2^{p-1} |g|^p \; d\mu. \end{aligned}\]Therefore we have \(\| f + g \|_p < \infty\). Clearly we may assume \(\| f + g \|_p > 0\). Note that
\[\begin{aligned} | f + g |^p \leq |f| | f + g |^{p-1} + |g| | f + g |^{p-1}. \end{aligned}\]By Hölder’s inequality with $q = (1 - \frac{1}{p})^{-1}$, i.e. $(p-1)q = 1$, we obtain
\[\begin{aligned} \| f + g \|_p^p = \int | f + g |^p \leq \; & \| f \|_p \left(\int | f + g |^{(p - 1)q} \right)^{1/q} \\ & + \| g \|_p \left(\int | f + g |^{(p - 1)q} \right)^{1/q} \\ = \; & (\| f \|_p + \| g \|_p) \cdot \| f + g \|_p^{p/q} \end{aligned}\]It implies that
\[\begin{aligned} \| f + g \|_p^{p(1-\frac{1}{q})} = \| f + g \|_p \leq \| f \|_p + \| g \|_p \end{aligned}\] \[\tag*{$\blacksquare$}\]Completeness
In this section, we show that the $L^p$ space is complete. Recall that the definition of completeness is following:
Every Cauchy sequence of points in $X$ has a limit that is also in $X$.
A normed vector space that is complete with respect to the norm metric is called a Banach space. And the following is a useful criterion for completeness of a normed vector space.
$\mathbf{Proof.}$
Suppose $\chi$ is complete and \(\sum_{n=1}^\infty \| x_n \| < \infty\). Let $S_N = \sum_{i=1}^N x_n$. Then for $N > M$, we have
\[\begin{aligned} \| S_N - S_M \| \leq \sum_{n = M + 1}^N \| x_n \| \to 0 \text{ as } M, N \to \infty, \end{aligned}\]thus the sequence \(\{ S_N \}\) is Cauchy and hence convergent.
Conversely, suppose that every absolute convergent series convergent, and let \(\{ x_n \}\) be a Cauchy sequence. We can choose $n_1 < n_2 < \cdots$ such that \(\| x_n - x_m \| < 2^{-j}\) for $m, n \geq n_j$. Let $y_1 = x_{n_1}$ and $y_j = x_{n_j} - x_{n_{j-1}}$ for $j > 1$. Then $\sum_{j=1}^k y_j = x_{n_k}$ and
\[\begin{aligned} \sum_{j=1}^\infty \| y_j \| \leq \| y_1 \| + \sum_{j=1}^\infty 2^{-j} = \| y_1 \| + 1 < \infty, \end{aligned}\]thus it is convergent;
\[\begin{aligned} \lim x_{n_k} = \sum_{j=1}^\infty y_j \end{aligned}\]exists. But since \(\{ x_n \}\) is Cauchy, we can show that \(\{ x_n \}\) converges to the same limit as \(\{ x_{n_k} \}\):
Suppose \(\{ x_{n_k} \}\) converges to $x$ and let any $\varepsilon > 0$ be given. Then, as \(\{ x_n \}\) is Cauchy, there exists $N > 0$ such that for all $n, m > N$, we have
\[\begin{aligned} \| x_n - x_m \| < \frac{\varepsilon}{2} \end{aligned}\]Also, there exists $M > 0$ such that for all $n_k > M$, we have
\[\begin{aligned} \| x_{n_k} - x \| < \frac{\varepsilon}{2} \end{aligned}\]Let \(K = \max \{N,M\}\). For all $n, m, n_k > K$, we have
\[\begin{aligned} \| x_{n} - x \| \leq \| x_{n} - x_{n_k} \| + \| x_{n_k} - x \| < \varepsilon. \end{aligned}\] \[\tag*{$\blacksquare$}\]Recall our definition of $L^p$ norm, \(\{ f: \| f \|_p < \infty \}\). Thus, it is possible to utilize the previous theorem to prove the completeness of $L^p$ space.
If $1 \leq p \leq \infty$, then $L^p$ is complete.
$\mathbf{Proof.}$
By the previous theorem, we only need to show that every absolutely convergent series in $L^p$ converges in $L^p$.
Step 1. $p < \infty$
Suppose \(\{f_k\} \in L^p\) and \(\sum_{k=1}^\infty \| f_k \|_p < \infty\). Set \(B = \sum_{k=1}^\infty \| f_k \|_p < \infty\).
Let $G_n = \sum_{k=1}^n | f_k |$ and $G(x) = \sum_{k=1}^\infty | f_k | (x)$. In other words, $\lim _{n \to \infty} G_n (x) = G(x)$.
Then, for each $n \in \mathbb{N}$, we have
\[\begin{aligned} \| G_n \|_p \leq \sum_{k=1}^n \| f_k \|_p \leq B. \end{aligned}\]By the monotone convergence theorem,
\[\begin{aligned} \int G^p \; d\mu & = \int \lim _{n \to \infty} G_n^p \; d\mu \\ & = \lim _{n \to \infty} \int G_n^p \; d\mu \\ & = \lim _{n \to \infty} \| G_n \|_p^p \leq B^p \end{aligned}\]Hence, $G \in L^p$, and in particular, $G < \infty$ a.e. Thus, $\sum_{k=1}^\infty f_k$ converges a.e. from the definition $G = \sum_{k=1}^\infty | f_k |$. Set $F = \sum_{k=1}^\infty f_k$. Note that $| F | \leq \sum_{k=1}^\infty | f_k | = G$. Since $G \in L^p$, $F \in L^p$.
$\mathbf{Proof.}$
Thus by the dominated convergence theorem, we have
\[\begin{aligned} \left\| F - \sum_{k=1}^n f_k \right\|_p^p = \int \left| F - \sum_{k=1}^n f_k \right|^p \to \int \lim _{n \to \infty} \left| F - \sum_{k=1}^n f_k \right|^p = 0 \end{aligned}\]as $n \to \infty$.
\[\tag*{$\blacksquare$}\]Step 2. $p = \infty$
Suppose \(\{ f_k \} \subset L^\infty\) and \(\sum_{k=1}^\infty \| f_k \|_\infty = B < \infty\).
Let $G_n = \sum_{k=1}^n | f_k |$ and $G = \sum_{k=1}^\infty | f_k |$. Then, $G (x) < \infty$ a.e. Hence, $\sum_{k=1}^\infty f_k$ converges, say to $F$, a.e.
We have
\[\begin{aligned} \left\| F - \sum_{k=1}^n f_k \right\|_\infty = \left\| \sum_{k=n+1}^\infty f_k \right\|_\infty \leq \sum_{k=n+1}^\infty \| f_k \|_\infty \to 0 \end{aligned}\]as $n \to \infty$.
\[\tag*{$\blacksquare$}\]Dense Subspaces
In the same way that $\mathbb{Q}$ is dense in $\mathbb{R}$, the collection of simple functions is dense in $L^p$ for $1 \leq p \leq \infty$.
$\mathbf{Proof.}$
Step 1. $p < \infty$
Let $f \in L^p (X)$. Assume that $f$ is non-negative. Then, there exists a sequence of non-negative simple functions \(\{ s_n \}\) increasing to $f$. Note that $s_n \in L^p (X)$ as $s_n \leq f$ and $| f |_p < \infty$. Also,
\[\begin{aligned} | f - s_n |^p \leq | f |^p \in L^1 \end{aligned}\]Thus, by dominated convergence theorem, we have
\[\begin{aligned} \lim _{n \to \infty} \| f - s_n \|_p = 0. \end{aligned}\]Now let $f \in L^p(X)$ but not necessarily non-negative. Write $f = f^+ - f^-$. Then, again there exist sequences of non-negative simple functions \(\{ s^+_n \}\) and \(\{ s^-_n \}\) increasing to $f^+$ and $f^-$ and $| f^+ - s^+_n |_p \to 0$ and $| f^- - s^-_n |_p \to 0$, respectively. Then
\[\begin{aligned} \| f - s_n \|_p \leq \| f^+ - s^+_n \|_p + \| f^- - s^-_n \|_p \to 0 \end{aligned}\]as $n \to \infty$ where $s_n = s^+_n - s^-_n$.
Step 2. $p = \infty$
Let $f \in L^\infty (X)$. In other words, $f$ is bounded. Assume that $f$ is non-negative. Then, there exists a sequence of non-negative simple functions \(\{ s_n \}\) increasing to $f$. Since $f$ is bounded, $s_n$ increases to $f$ uniformly.
Formally, let any $\varepsilon > 0$ be given. Then there exists $n_0 \in \mathbb{N}$ such that $| f - s_n | < \varepsilon$ whenever $n \geq n_0$ for any $x \in X$. That is,
\[\begin{aligned} \| f - s_n \|_\infty \leq \varepsilon \end{aligned}\]for all $n \geq n_0$. Hence \(\lim _{n \to \infty} \| f - s_n \|_\infty = 0\). With this result, we can show the result for general $f \in L^\infty (X)$ as $1 \leq p < \infty$ case.
\[\tag*{$\blacksquare$}\]Reference
[1] Richard F. Bass, Real Analysis for Graduate Students, Version 4.3
[2] Folland, Gerald B. Real analysis: modern techniques and their applications. Vol. 40. John Wiley & Sons, 1999.
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