📂Lebesgue Spaces

The Relationship between L1 Space and L2 Space

Definition

• $L^{1}$ Space

  A function $f$ is said to be (absolutely) integrable on the interval $[a,\ b]$ if it satisfies the following equation.

  $$ \int_{a}^b |f(x)| dx < \infty $$

  The set of functions that are integrable on the interval $[a,b]$ is denoted as $L^{1}(a,b)$.

  $$ L^{1}(a,b)= \left\{ f : \int_{-a}^{b} |f(x)| dx < \infty \right\} $$

• $L^{2}$ Space

  A function satisfying the following equation is said to be square-integrable.

  $$ \int_{a}^b |f(x)|^{2} dx < \infty $$

  The set of square-integrable functions on the interval $[a,b]$ is denoted as $L^{2}(a,b)$.

  $$ L^{2}(a,b) := \left\{ f : \int_{a}^b |f(x)|^{2} dx < \infty \right\} $$

Explanation¹

Unless specified otherwise, consider the interval as the entire real numbers $\mathbb{R}$.

$$ \begin{align*} L^{1} &= L^{1}(\mathbb{R})=\left\{ f : \int_{-\infty}^{\infty} |f(x)| dx < \infty \right\} \\ L^{2} &= L^{2}(\mathbb{R})=\left\{ f : \int_{-\infty}^{\infty} |f(x)|^{2} dx < \infty \right\} \end{align*} $$

At first glance, it may seem like there is an inclusion relationship between the spaces $L^{1}$ and $L^{2}$, but that is not the case at all.

$$ L^{1} \nsubseteq L^{2},\quad L^{2} \nsubseteq L^{1} $$

For example, consider the following function.

$$ \begin{align*} f(x) &= \begin{cases} x^{-\frac{2}{3}} & \mathrm{if}\ 0<x<1 \\ 0 & \mathrm{otherwise} \end{cases} \\ g(x) &= \begin{cases} x^{-\frac{2}{3}} & \mathrm{if}\ 1<x \\ 0 & \mathrm{otherwise} \end{cases} \end{align*} $$

Upon calculation, it is observed that $f$ is a $L^{1}$ function but not a $L^{2}$ function.

$$ \begin{align*} \int |f(x)|dx &= \int_{0}^1 x^{-\frac{2}{3}}dx=\left[ 3x^{\frac{1}{3}} \right]_{0}^1<\infty \\ \int |f(x)|^2dx &= \int_{0}^1 x^{-\frac{4}{3}}dx=\left[ -3x^{-\frac{1}{3}} \right]_{0}^1=\infty \end{align*} $$

Conversely, $g$ is a $L^{2}$ function but not a $L^{1}$ function.

$$ \begin{align*} \int |g(x)|dx &= \int_{1}^\infty x^{-\frac{2}{3}}dx=\left[ 3x^{\frac{1}{3}} \right]_{1}^\infty=\infty \\ \int |g(x)|^2dx &= \int_{1}^\infty x^{-\frac{4}{3}}dx=\left[ -3x^{-\frac{1}{3}} \right]_{1}^\infty<\infty \end{align*} $$

However, if the following condition is satisfied, a $L^{1}$ function can also be a $L^{2}$ function, or a $L^{2}$ function can also be a $L^{1}$ function. Moreover, if the integral interval is bounded, $L^{1} \subset L^{2}$ holds.

Theorem

(a) Let $f \in L^{1}$ and assume $f$ is bounded. Then, $f \in L^{2}$ is true.

(b) If $f \in L^{2}$ and $f$ is $0$ outside a finite interval, then $f \in L^{1}$ is true.

Proof

(a)

Assuming $f$ is bounded, a positive number $M$ exists.

$$ |f| \le M $$

Therefore, $|f|^{2} \le M|f|$ is true. Thus, the following is satisfied.

$$ \int |f|^2dx \le \int M|f|dx=M\int |f|dx <\infty $$

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(b)

By assumption, the following equation is true.

$$ \int |f|dx=\int_{a}^b|f|dx $$

Then, by the Cauchy-Schwarz inequality $| \langle x,\ y \rangle | \le \|x\| \|y\|$, the following is satisfied.

$$ \begin{align*} \int_{a}^b|f|dx =&\ \int_{a}^b1\cdot |f|dx \\ =&\ \langle 1 , |f| \rangle \\ \le& \| 1 \|_{2} \| |f| \|_{2} \\ =&\ (b-a)^{\frac{1}{2}}\left( \int_{a}^b|f|^{2} dx\right)^{\frac{1}{2}} \\ <& \infty \end{align*} $$

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