Lp Space, Lebesgue Space
Definition1 2 3
Let $\Omega \subset \mathbb{R}^{n}$ be an open set and let $p$ be a positive real number.
For all measurable functions $f$ defined on $\Omega$, define the set $L^{p}(\Omega)$ as follows.
$$ L^{p}(\Omega) := \left\{ f : \int_{\Omega} \left| f(x) \right|^{p} dx < \infty \right\} $$
This is called the Lp space or Lebesgue space, and it is also written simply as $L^{p}$. Functional analysis textbooks usually state it as above, while measure theory and real analysis textbooks state it as follows.
Suppose a measure space $(X, \mathcal{E}, \mu)$ is given. For a measurable function $f$ defined on $X$, define the set $L^{p}(X, \mathcal{E},\mu)$ as follows.
$$ L^{p}(X, \mathcal{E}, \mu) := \left\{ f : \int \left| f \right|^{p} d \mu < \infty \right\} $$
Here, $\mu$ is a measure. It is written simply as $L^{p}(\mu), L^{p}(X)$, and so on.
Properties
- $L^{p}$ is a vector space.
- For $1 \le p \le \infty$, $L^{p}$ is a normed space.
- $L^{p}$ is a complete space.
- For $E\subset X$, if $1 \le p \le q \le \infty$ and $\mu (E) < \infty \implies L^{q} (E) \subset L^{p} (E)$
Explanation
In 2., if $p \lt 1$, then $\left\| \cdot \right\|_{p}$ does not satisfy the triangle inequality and thus fails to be a norm. On the other hand, when $p = \infty$, the $L^{p}$ space is a normed space.
A normed vector space that is complete is specifically called a Banach space. Therefore, the $L^{p}$ space is a Banach space. $L^{p}$ is particularly important as a space in which the Hölder inequality and the Minkowski inequality hold.
A vector space equipped with an inner product is called an inner product space. An inner product space that is complete is specifically called a Hilbert space. In the case of the $L^{2}$ space, an inner product can be defined as follows.
$$ \left( \int |f(x)|^2 dx\right)^{\frac{1}{2}} = \left( \int f(x)\overline{f(x)}dx \right) ^{\frac{1}{2}} = \langle f,f \rangle ^{\frac{1}{2}} $$
Therefore, the $L^{2}$ space is a Hilbert space.
In 4., note the condition $\mu (E) < \infty$. If the domain of integration is not bounded, then $L^{1} (E)$ and $L^{2} (E)$ have no inclusion relation whatsoever. For $1 \le p \lt q \lt r$ satisfying certain conditions, ${u \in L^{p} \cap L^{r} \implies u \in L^{q}}$ may also hold.
Proof
2.
For $1\le p <\infty$, define $\| \cdot \|_{p}$ as follows.
$$ \left\| f \right\|_{p} := \left( \int_{\Omega} \left| f(x) \right|^{p} dx \right)^{1/p},\quad f\in L^{p}(\Omega) $$
Then $\| \cdot \|_{p}$ is a norm on the $L^{p}$ space. (When $0<p<1$, it is not a norm.) By definition, $\| f \|_{p} \ge 0$ is obvious, and $\| f \|_{p}=0 \iff f=0$ is also obvious. That $\| cf \|_{p} = \left| c \right| \left\| f \right\|_{p}$ holds for $c \in \mathbb{C}$ can be shown as follows.
$$ \begin{align*} \left\| cf \right\|_{p} =& \left( \int_{\Omega} \left| cf(x) \right|^{p} dx \right)^{1/p} \\ =& \left( \left| c \right|^{p} \int_{\Omega} \left| f(x) \right|^{p} dx \right)^{1/p} \\ =& \left| c \right| \left( \int_{\Omega} \left| f(x) \right|^{p} dx \right)^{1/p} \\ =& \left| c \right| \left\| f \right\|_{p} \end{align*} $$
For $f,g \in L^{p}$, $\left\| f + g \right\|_{p} \le \| f \|_{p} + \| g \|_{p}$ holds as well, and this is known by the name of the Minkowski inequality.
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3.
Strategy: Almost everything is settled by Fatou’s lemma.
Given a Cauchy sequence $f_{n}$, one can find a subsequence $f_{n_{k}}$ satisfying $\left\| f_{n} - f_{n_{k}} \right\|_{p} < \dfrac{1}{2^{k}}$. If, for all $k \in \mathbb{N}$, we define
$$ \begin{align*} g_{k} :=& \sum_{i=1}^{k} \left| f_{n_{i+1}} - f_{n_{i}} \right| \\ g :=& \lim_{k \to \infty} g_{k} = \sum_{i=1}^{\infty} \left| f_{n_{i+1}} - f_{n_{i}} \right| \end{align*} $$
then by the triangle inequality,
$$ \left\| g_{k} \right\|_{p} \le \sum_{i}^{k} \dfrac{1}{2^{i}} < 1 $$
For a sequence $\left\{ f_{n} \right\}$ of non-negative measurable functions,
$$ \int \left( \liminf_{n \to \infty} f_{n} \right) d \mu \le \liminf_{n \to \infty} \int f_{n} d \mu $$
By Fatou’s lemma,
$$ \left\| g \right\|_{p}^{p} \le \int \lim_{n \to \infty} g_{k}^{p} d \mu \le \liminf_{k \to \infty} \int g_{k}^{p} d \mu \le 1 $$
Since $g$ is finite almost everywhere,
$$ f_{n_{k}} = f_{n_{1}}(x) + \sum_{i=1}^{ k } \left[ f_{n_{i}} (x) - f_{n_{i-1}} (x) \right] $$
converges almost everywhere. If we define $f := \lim\limits_{k \to \infty} f_{n_{k}}$, then by Fatou’s lemma,
$$ \left\| f - f_{m} \right\|_{p} = \int |f - f_{m}|^{p} d \mu \le \liminf_{k \to \infty} \int | f_{n_{k}} - f_{m}|^{p} d \mu \le \varepsilon^{p} $$
Therefore, $f - f_{m} \in L^{p}$, and $f = f_{m} + (f - f_{m} ) \in L^{p}$. Since every Cauchy sequence in $L^{p}$ converges to an element of $L^{p}$, $L^{p}$ is a complete space.
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4.
Strategy: It suffices to show the inequality $|f(x)|^{p} \le 1 + |f(x)|^{q}$; the rest of the proof is finished by the properties of the Lebesgue integral.
Let $f \in L^{q}$. Then the following holds.
$$ \begin{align*} | f(x) | \le 1 \implies& |f(x) |^{p} \le 1 \\ 1 \le |f(x)| \implies& |f(x)|^{p} \le |f(x)|^{q} \end{align*} $$
Therefore, whether $| f(x) |$ is greater or less than $1$, the following holds.
$$ |f(x)|^{p} \le 1 + |f(x)|^{q} $$
Taking the Lebesgue integral $\displaystyle \int_{E} d \mu$ gives the following.
$$ \int_{E} |f|^{p} d \mu \le \int_{E} 1 d \mu + \int_{E} |f|^{q} d \mu = m(E) + \int_{E} |f|^{q} d \mu < \infty $$
Since $m(E) < \infty$ and $\displaystyle \int_{E} |f|^{q} d \mu < \infty$, the following holds.
$$ \int_{E} |f|^{p} d \mu < \infty $$
In other words, since $f \in L^{q} \implies f \in L^{p}$,
$$ L^{q} (E) \subset L^{p} (E) $$
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