📂Lebesgue Spaces

L infinity space

Definitions¹

• Let $\Omega \subset \mathbb{R}^{n}$ be called an open set. For a measurable function $u$ on $\Omega$, if there exists a constant $K$ that satisfies the following condition for $u$, then $u$ is said to be essentially bounded on $\Omega$.

  $$ \left| u(x) \right| \le K \text{ a.e. on } \Omega $$

  Here, $\text{a.e.}$ means almost everywhere.

• The supremum of such $K$ is called the essential supremum of $\left| u \right|$ and is denoted as follows.

  $$ \underset{x\in \Omega}{\text{ess sup}}\left| u(x) \right| := \inf \left\{ K : \left| u(x) \right| \le K \text{ a.e. on } \Omega \right\} $$

• The set of all functions $u$ that are essentially bounded on $\Omega$ is defined as $L^{\infty}(\Omega)$.

  $$ L^{\infty}(\Omega) := \left\{ u : u \text{ is essentially bounded on } \Omega \right\} $$

Explanation

$L^{\infty}$ space is read as L infinity space.

‘Saying almost everywhere bounded’ essentially means ‘frankly bounded’, ‘putting it plainly bounded’, so it makes sense to say ’essentially bounded’. Especially in the space $L^{p}$, since it’s about integration, if it’s bounded almost everywhere, it’s essentially bounded literally.

Meanwhile, the $L^{\infty}$ space becomes a norm space, and the defined essential supremum can be used as is.

$$ \left\| u \right\|_{\infty} = \underset{x\in \Omega}{\text{ess sup}}\left| u(x) \right|, \quad u \in L^{\infty}(\Omega) $$

It is not difficult to verify this actually being a norm. The important fact is, as you can guess from the notation, that this really is the same as the limit of $\left\| u \right\|_{p}$. If for $p \lt \infty$, $u \in L^{\infty} \cap L^{p}$ then

$$ \left\| u \right\|_{\infty} = \lim \limits_{p \to \infty} \left\| u \right\|_{p} $$

Moreover, the Hölder’s inequality and its corollaries that were valid for $1 \lt p, p^{\prime} \lt \infty$ are extended to hold for $p = 1, p^{\prime} = \infty$ and $p = \infty, p^{\prime} = 1$ respectively.

Hölder’s Inequality

Consider two constants $1 \le p \le \infty, 1 \le p^{\prime} \le \infty$ satisfying the following expression.

$$ \dfrac{1}{p}+\dfrac{1}{p^{\prime}} = 1 \left(\text{or } p^{\prime} = \frac{p}{p-1} \right) $$

If $u \in L^p(\Omega)$, $v\in L^{p^{\prime}}(\Omega)$ then $uv \in L^1(\Omega)$ and the inequality below holds.

$$ \| uv \|_{1} = \int_{\Omega} |u(x)v(x)| dx \le \| u \|_{p} \| v \|_{p^{\prime}} $$