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\section*{Lebesgue space}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{functional_analysis}{}\paragraph*{{Functional analysis}}\label{functional_analysis}

[[!include functional analysis - contents]]

\hypertarget{integration_theory}{}\paragraph*{{Integration theory}}\label{integration_theory}

[[!include integration theory - contents]]

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{Definition}{Definitions}\dotfill \pageref*{Definition} \linebreak
\noindent\hyperlink{MinkowskiInequality}{Minkowski's inequality}\dotfill \pageref*{MinkowskiInequality} \linebreak
\noindent\hyperlink{related_concepts}{Related concepts}\dotfill \pageref*{related_concepts} \linebreak
\noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

The term `Lebesgue space' can stand for two distinct notions: one is the general notion of [[measure space]] (compare the \href{http://eom.springer.de/L/l057910.htm}{Springer Encyclopaedia of Mathematics}) and another is the notion of $L^p$ space (or $L_p$ space). Here we discuss the latter.

Lebesgue spaces $L^p$ in this sense are [[normed vector spaces]] of [[functions]] on a [[measure space]], equipped with the suitable version of the [[p-norm]].

Beware that sometimes the notation `$L_p$' is used as a synonym for $L^p$; sometimes it is used to mean $L^{1/p}$.

\hypertarget{Definition}{}\subsection*{{Definitions}}\label{Definition}

If $1 \leq p \lt \infty$ is a [[real number]] and $(\Omega,\mu)$ is a [[measure space]], one considers the \textbf{$L^p$ space} $L_p(\Omega)$, which is the [[vector space]] of [[equivalence classes]] of those [[measurable function|measurable]] (complex- or real-valued) functions $f\colon \Omega \to \mathbb{K}$ whose (absolute values of) $p$th powers are integrable, in that the integral

\begin{displaymath}
{\|f\|_p} 
    \coloneqq \left(\int_\Omega {|f|^p} \,d\mu\right)^{1/p} 
  \lt \infty
\end{displaymath}
exists. Two such are taken to be equivalent, $f \sim g$, if ${\|f-g\|_p} = 0$. For $p = 2$ this is the space $L^2$ of [[square integrable functions]].

On these spaces $L^p(X)$ of equivalence classes of $p$-power integrable functions, the function ${\|f\|_p}$ satisfies the [[triangle inequality]] (due to [[Minkowski's inequality]], see \hyperlink{MinkowskiInequality}{below}) and hence defines a [[norm]], the \emph{[[p-norm]]}, making them [[normed vector spaces]].

The $L^p$ spaces are examples of [[Banach spaces]]; they are continuous analogues of $l^p$ spaces of $p$-summable series. (Indeed, $l^p(S)$, for $S$ a [[set]], is simply $L^p(S)$ if $S$ is equipped with [[counting measure]].)

For fixed $f$, the norm ${\|f\|_p}$ is continuous in $p$. Accordingly, for $p = \infty$, one may take the limit of ${\|f\|}_p$ as $p \to \infty$. However, this turns out to be the same as the [[essential supremum]] norm $\|f\|_\infty$. Therefore, $L^\infty(\Omega)$ makes sense as long as $\Omega$ is a [[measurable space]] equipped with a family of [[null sets]] (or [[full sets]]); the measure $\mu$ is otherwise irrelevant.

For $0 \leq p \lt 1$, one can modify the definition to make $L^p$ into an [[F-space]] (but not a Banach space). See the definitions at [[p-norm]].

\hypertarget{MinkowskiInequality}{}\subsection*{{Minkowski's inequality}}\label{MinkowskiInequality}

We offer here a proof that ${\|f\|_p}$ indeed defines a norm in the case $1 \lt p \lt \infty$, in that it satisfies the [[triangle inequality]]. This is usually known as \emph{[[Minkowski's inequality]]}.

(The cases $p = 1$ and $p = \infty$ follow by continuity and are easy to check from first principles.)

The most usual textbook proofs involve a clever application of [[Hölder's inequality]]; the following proof is more straightforwardly geometric. All functions $f$ may be assumed to be real- or complex-valued.

\begin{theorem}
\label{}\hypertarget{}{}
Suppose $1 \leq p \leq \infty$, and suppose $\Omega$ is a [[measure space]] with [[measure]] $\mu$. Then the function ${|(-)|_p}\colon L^p(\Omega, \mu) \to \mathbb{R}$ defined by

\begin{displaymath}
{\|f\|_p} \coloneqq (\int_\Omega {|f|^p} \,d\mu)^{1/p}
\end{displaymath}
defines a [[norm]].

\end{theorem}
One must verify three things:

\begin{enumerate}%
\item Separation axiom: ${\|f\|_p} = 0$ implies $f = 0$.


\item Scaling axiom: ${\|t f\|}_p = {|t|} \, {\|f\|_p}$.


\item Triangle inequality: ${\|f + g\|_p} \leq {\|f\|_p} + {\|g\|_p}$.



\end{enumerate}
The first two properties are obvious, so it remains to prove the last, which is also called \textbf{[[Minkowski's inequality]]}.

Our proof of Minkowski's inequality is broken down into a series of simple lemmas. The plan is to boil it down to two things: the scaling axiom, and convexity of the function $x \mapsto {|x|^p}$ (as a function from real or complex numbers to nonnegative real numbers).

First, some generalities. Let $V$ be a (real or complex) vector space equipped with a function ${\|(-)\|}\colon V \to [0, \infty]$ that satisfies the scaling axiom: ${\|t v\|} = {|t|} \, {\|v\|}$ for all scalars $t$, and the separation axiom: ${\|v\|} = 0$ implies $v = 0$. As usual, we define the [[unit ball]] in $V$ to be $\{v \in V \;|\; {\|v\|} \leq 1\}.$

\begin{lemma}
\label{conditions}\hypertarget{conditions}{}
Given that the scaling and separation axioms hold, the following conditions are equivalent:

\begin{enumerate}%
\item The triangle inequality is satisfied.
\item The unit ball is convex.
\item If ${\|u\|} = {\|v\|} = 1$, then ${\|t u + (1-t)v\|} \leq 1$ for all $t \in [0, 1]$.

\end{enumerate}
\end{lemma}
\begin{proof}
Condition 1. implies condition 2. easily: if $u$ and $v$ are in the unit ball and $0 \leq t \leq 1$, we have

\begin{displaymath}
\itexarray{
{\|t u + (1-t)v\|} & \leq & {\|t u\|} + {\|(1-t)v\|} \\
 & = & t {\|u\|} + (1-t) {\|v\|} \\
 & \leq & t + (1-t) = 1.}
\end{displaymath}
Now 2. implies 3. trivially, so it remains to prove that 3. implies 1. Suppose ${\|v\|}, {\|v'\|} \in (0, \infty)$. Let $u = \frac{v}{{\|v\|}}$ and $u' = \frac{v'}{{\|v'\|}}$ be the associated unit vectors. Then

\begin{displaymath}
\itexarray{
\frac{v + v'}{{\|v\|}+{\|v'\|}} & = & (\frac{{\|v\|}}{{\|v\|}+{\|v'\|}})\frac{v}{{\|v\|}} + (\frac{{\|v'\|}}{{\|v\|}+{\|v'\|}})\frac{v'}{{\|v'\|}} \\
 & = & t u + (1-t)u'}
\end{displaymath}
where $t = \frac{{\|v\|}}{{\|v\|} + {\|v'\|}}$. If condition 3. holds, then

\begin{displaymath}
{\|t u + (1-t)u'\|} \leq 1
\end{displaymath}
but by the scaling axiom, this is the same as saying

\begin{displaymath}
\frac{{\|v + v'\|}}{{\|v\|} + {\|v'\|}} \leq 1,
\end{displaymath}
which is the triangle inequality.

\end{proof}
Consider now $L^p$ with its $p$-norm ${\|f\|} = {|f|_p}$. By Lemma \ref{conditions}, this inequality is equivalent to

\begin{itemize}%
\item \textbf{Condition 4:} If ${|u|_{p}^{p}} = 1$ and ${|v|_{p}^{p}} = 1$, then ${|t u + (1-t)v|_{p}^{p}} \leq 1$ whenever $0 \leq t \leq 1$.

\end{itemize}
This allows us to remove the cumbersome exponent $1/p$ in the definition of the $p$-norm.

The next two lemmas may be proven by elementary calculus; we omit the proofs. (But you can also see the \href{http://ncatlab.org/toddtrimble/published/p-norms}{full details}.)

\begin{lemma}
\label{2deriv}\hypertarget{2deriv}{}
Let $\alpha, \beta$ be two complex numbers, and define

\begin{displaymath}
\gamma(t) = {|\alpha + \beta t|^p}
\end{displaymath}
for \emph{real} $t$. Then $\gamma''(t)$ is nonnegative.

\end{lemma}
\begin{lemma}
\label{convex}\hypertarget{convex}{}
Define $\phi\colon \mathbb{C} \to \mathbb{R}$ by $\phi(x) = |x|^p$. Then $\phi$ is convex, i.e., for all $x, y$,

\begin{displaymath}
{|t x + (1-t)y|^p} \leq t{|x|^p} + (1-t){|y|^p}
\end{displaymath}
for all $t \in [0, 1]$.

\end{lemma}
\begin{proof}
Let $u$ and $v$ be unit vectors in $L^p$. By condition 4, it suffices to show that ${|t u + (1-t)v|_p} \leq 1$ for all $t \in [0, 1]$. But

\begin{displaymath}
\int_\Omega {|t u + (1-t)v|^p} \,d\mu \leq \int_\Omega t{|u|}^p + (1-t){|v|}^p \,d\mu
\end{displaymath}
by Lemma \ref{convex}. Using $\int {|u|^p} = 1 = \int {|v|^p}$, we are done.

\end{proof}
\hypertarget{related_concepts}{}\subsection*{{Related concepts}}\label{related_concepts}

\begin{itemize}%
\item [[square integrable function]]


\item [[canonical Hilbert space of half-densities]]


\item [[tempered distribution]]



\end{itemize}
\hypertarget{references}{}\subsection*{{References}}\label{references}

Named after [[Henri Lebesgue]].

\begin{itemize}%
\item W. Rudin, \emph{Functional analysis}, McGraw Hill 1991.


\item L. C. Evans, \emph{Partial differential equations}, Amer. Math. Soc. 1998.


\item Wikipedia (English): \href{http://en.wikipedia.org/wiki/Lp_space}{Lp space}



\end{itemize}
category: analysis

[[!redirects Lebesgue space]] [[!redirects Lebesgue spaces]]

[[!redirects L-p space]] [[!redirects L-p spaces]] [[!redirects L-p-space]] [[!redirects L-p-spaces]] [[!redirects L{\tt \symbol{94}}p space]] [[!redirects L{\tt \symbol{94}}p spaces]] [[!redirects L\_p space]] [[!redirects L\_p spaces]]

[[!redirects L-1 space]] [[!redirects L-1 spaces]] [[!redirects L-1-space]] [[!redirects L-1-spaces]] [[!redirects L{\tt \symbol{94}}1 space]] [[!redirects L{\tt \symbol{94}}1 spaces]] [[!redirects L\_1 space]] [[!redirects L\_1 spaces]]

[[!redirects L-2 space]] [[!redirects L-2 spaces]] [[!redirects L-2-space]] [[!redirects L-2-spaces]] [[!redirects L{\tt \symbol{94}}2 space]] [[!redirects L{\tt \symbol{94}}2 spaces]] [[!redirects L\_2 space]] [[!redirects L\_2 spaces]]



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