📂Lebesgue Spaces

Lp Spaces, Lebesgue Spaces

Definition^{1 2 3}

Let $\Omega \subset \mathbb{R}^{n}$ be an open set, and $p$ be a positive real number.

For all measurable functions $f$ defined on $\Omega$, define 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 is briefly denoted as $L^{p}$. Typically, textbooks on functional analysis describe it as above, while measure theory and real analysis textbooks describe it as follows.

Given a measure space $(X, \mathcal{E}, \mu)$, for measurable functions $f$ defined on $X$, define 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 the measure. It is simply denoted as $L^{p}(\mu), L^{p}(X)$ and so on.

Properties

1. $L^{p}$ is a vector space.
2. For $1 \le p \le \infty$, $L^{p}$ is a normed space.
3. $L^{p}$ is a complete space.
4. 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

2. If $p \lt 1$, then $\left\| \cdot \right\|_{p}$ does not satisfy the triangle inequality and is not a norm. However, if $p = \infty$, then $L^{p}$ space becomes a normed space.

A complete normed vector space is specifically called a Banach space. Therefore, $L^{p}$ space is a Banach space. $L^{p}$ is especially important as a space where the Hölder inequality and the Minkowski inequality hold.

A vector space on which an inner product is defined is called an inner product space. A complete inner product space is specifically called a Hilbert space. For $L^{2}$ 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, $L^{2}$ space is a Hilbert space.

4. Let’s focus on the condition $\mu (E) < \infty$. If the integration range is not bounded, then $L^{1} (E)$ and $L^{2} (E)$ do not have any inclusion relation. For certain conditions that $1 \le p \lt q \lt r$ meets, ${u \in L^{p} \cap L^{r} \implies u \in L^{q}}$ can also be satisfied.

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 $\| \cdot \|_{p}$ becomes the norm of $L^{p}$ space. (When $0<p<1$, it does not become a norm.) It is obvious that $\| f \|_{p} \ge 0$, and it is also obvious that $\| f \|_{p}=0 \iff f=0$. It can also be shown that for $c \in \mathbb{C}$, $\| cf \|_{p} = \left| c \right| \left\| f \right\|_{p}$ holds 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 $f,g \in L^{p}$, $\left\| f + g \right\|_{p} \le \| f \|_{p} + \| g \|_{p}$ also holds, and this is known as the Minkowski inequality.

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3.

Strategy: Almost everything is solved by Fatou’s lemma.

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Given a Cauchy sequence $f_{n}$, a subsequence $f_{n_{k}}$ that satisfies $\left\| f_{n} - f_{n_{k}} \right\|_{p} < \dfrac{1}{2^{k}}$ can be found. For all $k \in \mathbb{N}$,

$$\begin{align*} g_{k} :=& \sum_{i=1}^{k} \left| f_{n_{i+1}} - f_{n_{i}} \right| \\ g :=& \lim_{k \to \infty} g_{k} = \sum_{i=1}^{\infty} \left| f_{n_{i+1}} - f_{n_{i}} \right| \end{align*}$$

by defining it, the triangle inequality provides

$$\left\| g_{k} \right\|_{p} \le \sum_{i}^{k} \dfrac{1}{2^{i}} < 1$$

Fatou’s Lemma

For a sequence of non-negative measurable functions $\left\{ f_{n} \right\}$,

$$\int \left( \liminf_{n \to \infty} f_{n} \right) d \mu \le \liminf_{n \to \infty} \int f_{n} d \mu$$

According to 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 in $L^{p}$, $L^{p}$ is a complete space.

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4.

Strategy: Showing the inequality $|f(x)|^{p} \le 1 + |f(x)|^{q}$ is sufficient; the rest follows from the properties of Lebesgue integration.

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Let’s assume $f \in L^{q}$. Then the following equation holds.

$$\begin{align*} | f(x) | \le 1 \implies& |f(x) |^{p} \le 1 \\ 1 \le |f(x)| \implies& |f(x)|^{p} \le |f(x)|^{q} \end{align*}$$

Hence, whether $| f(x) |$ is larger or smaller than $1$, the following holds.

$$|f(x)|^{p} \le 1 + |f(x)|^{q}$$

Taking the Lebesgue integral $\displaystyle \int_{E} d \mu$ results in

$$\int_{E} |f|^{p} d \mu \le \int_{E} 1 d \mu + \int_{E} |f|^{q} d \mu = m(E) + \int_{E} |f|^{q} d \mu < \infty$$

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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See Also

• $L^{1}$ space
• $L^{2}$ space
• Hilbert space