\documentclass[12pt,titlepage]{article}

\usepackage{amsmath}
\usepackage{mathrsfs}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{mathtools}
\usepackage{graphicx}
\usepackage{color}
\usepackage{ucs}
\usepackage[utf8x]{inputenc}
\usepackage{xparse}
\usepackage{hyperref}

%----Macros----------
%
% Unresolved issues:
%
%  \righttoleftarrow
%  \lefttorightarrow
%
%  \color{} with HTML colorspec
%  \bgcolor
%  \array with options (without options, it's equivalent to the matrix environment)

% Of the standard HTML named colors, white, black, red, green, blue and yellow
% are predefined in the color package. Here are the rest.
\definecolor{aqua}{rgb}{0, 1.0, 1.0}
\definecolor{fuschia}{rgb}{1.0, 0, 1.0}
\definecolor{gray}{rgb}{0.502, 0.502, 0.502}
\definecolor{lime}{rgb}{0, 1.0, 0}
\definecolor{maroon}{rgb}{0.502, 0, 0}
\definecolor{navy}{rgb}{0, 0, 0.502}
\definecolor{olive}{rgb}{0.502, 0.502, 0}
\definecolor{purple}{rgb}{0.502, 0, 0.502}
\definecolor{silver}{rgb}{0.753, 0.753, 0.753}
\definecolor{teal}{rgb}{0, 0.502, 0.502}

% Because of conflicts, \space and \mathop are converted to
% \itexspace and \operatorname during preprocessing.

% itex: \space{ht}{dp}{wd}
%
% Height and baseline depth measurements are in units of tenths of an ex while
% the width is measured in tenths of an em.
\makeatletter
\newdimen\itex@wd%
\newdimen\itex@dp%
\newdimen\itex@thd%
\def\itexspace#1#2#3{\itex@wd=#3em%
\itex@wd=0.1\itex@wd%
\itex@dp=#2ex%
\itex@dp=0.1\itex@dp%
\itex@thd=#1ex%
\itex@thd=0.1\itex@thd%
\advance\itex@thd\the\itex@dp%
\makebox[\the\itex@wd]{\rule[-\the\itex@dp]{0cm}{\the\itex@thd}}}
\makeatother

% \tensor and \multiscript
\makeatletter
\newif\if@sup
\newtoks\@sups
\def\append@sup#1{\edef\act{\noexpand\@sups={\the\@sups #1}}\act}%
\def\reset@sup{\@supfalse\@sups={}}%
\def\mk@scripts#1#2{\if #2/ \if@sup ^{\the\@sups}\fi \else%
  \ifx #1_ \if@sup ^{\the\@sups}\reset@sup \fi {}_{#2}%
  \else \append@sup#2 \@suptrue \fi%
  \expandafter\mk@scripts\fi}
\def\tensor#1#2{\reset@sup#1\mk@scripts#2_/}
\def\multiscripts#1#2#3{\reset@sup{}\mk@scripts#1_/#2%
  \reset@sup\mk@scripts#3_/}
\makeatother

% \slash
\makeatletter
\newbox\slashbox \setbox\slashbox=\hbox{$/$}
\def\itex@pslash#1{\setbox\@tempboxa=\hbox{$#1$}
  \@tempdima=0.5\wd\slashbox \advance\@tempdima 0.5\wd\@tempboxa
  \copy\slashbox \kern-\@tempdima \box\@tempboxa}
\def\slash{\protect\itex@pslash}
\makeatother

% math-mode versions of \rlap, etc
% from Alexander Perlis, "A complement to \smash, \llap, and lap"
%   http://math.arizona.edu/~aprl/publications/mathclap/
\def\clap#1{\hbox to 0pt{\hss#1\hss}}
\def\mathllap{\mathpalette\mathllapinternal}
\def\mathrlap{\mathpalette\mathrlapinternal}
\def\mathclap{\mathpalette\mathclapinternal}
\def\mathllapinternal#1#2{\llap{$\mathsurround=0pt#1{#2}$}}
\def\mathrlapinternal#1#2{\rlap{$\mathsurround=0pt#1{#2}$}}
\def\mathclapinternal#1#2{\clap{$\mathsurround=0pt#1{#2}$}}

% Renames \sqrt as \oldsqrt and redefine root to result in \sqrt[#1]{#2}
\let\oldroot\root
\def\root#1#2{\oldroot #1 \of{#2}}
\renewcommand{\sqrt}[2][]{\oldroot #1 \of{#2}}

% Manually declare the txfonts symbolsC font
\DeclareSymbolFont{symbolsC}{U}{txsyc}{m}{n}
\SetSymbolFont{symbolsC}{bold}{U}{txsyc}{bx}{n}
\DeclareFontSubstitution{U}{txsyc}{m}{n}

% Manually declare the stmaryrd font
\DeclareSymbolFont{stmry}{U}{stmry}{m}{n}
\SetSymbolFont{stmry}{bold}{U}{stmry}{b}{n}

% Manually declare the MnSymbolE font
\DeclareFontFamily{OMX}{MnSymbolE}{}
\DeclareSymbolFont{mnomx}{OMX}{MnSymbolE}{m}{n}
\SetSymbolFont{mnomx}{bold}{OMX}{MnSymbolE}{b}{n}
\DeclareFontShape{OMX}{MnSymbolE}{m}{n}{
    <-6>  MnSymbolE5
   <6-7>  MnSymbolE6
   <7-8>  MnSymbolE7
   <8-9>  MnSymbolE8
   <9-10> MnSymbolE9
  <10-12> MnSymbolE10
  <12->   MnSymbolE12}{}

% Declare specific arrows from txfonts without loading the full package
\makeatletter
\def\re@DeclareMathSymbol#1#2#3#4{%
    \let#1=\undefined
    \DeclareMathSymbol{#1}{#2}{#3}{#4}}
\re@DeclareMathSymbol{\neArrow}{\mathrel}{symbolsC}{116}
\re@DeclareMathSymbol{\neArr}{\mathrel}{symbolsC}{116}
\re@DeclareMathSymbol{\seArrow}{\mathrel}{symbolsC}{117}
\re@DeclareMathSymbol{\seArr}{\mathrel}{symbolsC}{117}
\re@DeclareMathSymbol{\nwArrow}{\mathrel}{symbolsC}{118}
\re@DeclareMathSymbol{\nwArr}{\mathrel}{symbolsC}{118}
\re@DeclareMathSymbol{\swArrow}{\mathrel}{symbolsC}{119}
\re@DeclareMathSymbol{\swArr}{\mathrel}{symbolsC}{119}
\re@DeclareMathSymbol{\nequiv}{\mathrel}{symbolsC}{46}
\re@DeclareMathSymbol{\Perp}{\mathrel}{symbolsC}{121}
\re@DeclareMathSymbol{\Vbar}{\mathrel}{symbolsC}{121}
\re@DeclareMathSymbol{\sslash}{\mathrel}{stmry}{12}
\re@DeclareMathSymbol{\bigsqcap}{\mathop}{stmry}{"64}
\re@DeclareMathSymbol{\biginterleave}{\mathop}{stmry}{"6}
\re@DeclareMathSymbol{\invamp}{\mathrel}{symbolsC}{77}
\re@DeclareMathSymbol{\parr}{\mathrel}{symbolsC}{77}
\makeatother

% \llangle, \rrangle, \lmoustache and \rmoustache from MnSymbolE
\makeatletter
\def\Decl@Mn@Delim#1#2#3#4{%
  \if\relax\noexpand#1%
    \let#1\undefined
  \fi
  \DeclareMathDelimiter{#1}{#2}{#3}{#4}{#3}{#4}}
\def\Decl@Mn@Open#1#2#3{\Decl@Mn@Delim{#1}{\mathopen}{#2}{#3}}
\def\Decl@Mn@Close#1#2#3{\Decl@Mn@Delim{#1}{\mathclose}{#2}{#3}}
\Decl@Mn@Open{\llangle}{mnomx}{'164}
\Decl@Mn@Close{\rrangle}{mnomx}{'171}
\Decl@Mn@Open{\lmoustache}{mnomx}{'245}
\Decl@Mn@Close{\rmoustache}{mnomx}{'244}
\makeatother

% Widecheck
\makeatletter
\DeclareRobustCommand\widecheck[1]{{\mathpalette\@widecheck{#1}}}
\def\@widecheck#1#2{%
    \setbox\z@\hbox{\m@th$#1#2$}%
    \setbox\tw@\hbox{\m@th$#1%
       \widehat{%
          \vrule\@width\z@\@height\ht\z@
          \vrule\@height\z@\@width\wd\z@}$}%
    \dp\tw@-\ht\z@
    \@tempdima\ht\z@ \advance\@tempdima2\ht\tw@ \divide\@tempdima\thr@@
    \setbox\tw@\hbox{%
       \raise\@tempdima\hbox{\scalebox{1}[-1]{\lower\@tempdima\box
\tw@}}}%
    {\ooalign{\box\tw@ \cr \box\z@}}}
\makeatother

% \mathraisebox{voffset}[height][depth]{something}
\makeatletter
\NewDocumentCommand\mathraisebox{moom}{%
\IfNoValueTF{#2}{\def\@temp##1##2{\raisebox{#1}{$\m@th##1##2$}}}{%
\IfNoValueTF{#3}{\def\@temp##1##2{\raisebox{#1}[#2]{$\m@th##1##2$}}%
}{\def\@temp##1##2{\raisebox{#1}[#2][#3]{$\m@th##1##2$}}}}%
\mathpalette\@temp{#4}}
\makeatletter

% udots (taken from yhmath)
\makeatletter
\def\udots{\mathinner{\mkern2mu\raise\p@\hbox{.}
\mkern2mu\raise4\p@\hbox{.}\mkern1mu
\raise7\p@\vbox{\kern7\p@\hbox{.}}\mkern1mu}}
\makeatother

%% Fix array
\newcommand{\itexarray}[1]{\begin{matrix}#1\end{matrix}}
%% \itexnum is a noop
\newcommand{\itexnum}[1]{#1}

%% Renaming existing commands
\newcommand{\underoverset}[3]{\underset{#1}{\overset{#2}{#3}}}
\newcommand{\widevec}{\overrightarrow}
\newcommand{\darr}{\downarrow}
\newcommand{\nearr}{\nearrow}
\newcommand{\nwarr}{\nwarrow}
\newcommand{\searr}{\searrow}
\newcommand{\swarr}{\swarrow}
\newcommand{\curvearrowbotright}{\curvearrowright}
\newcommand{\uparr}{\uparrow}
\newcommand{\downuparrow}{\updownarrow}
\newcommand{\duparr}{\updownarrow}
\newcommand{\updarr}{\updownarrow}
\newcommand{\gt}{>}
\newcommand{\lt}{<}
\newcommand{\map}{\mapsto}
\newcommand{\embedsin}{\hookrightarrow}
\newcommand{\Alpha}{A}
\newcommand{\Beta}{B}
\newcommand{\Zeta}{Z}
\newcommand{\Eta}{H}
\newcommand{\Iota}{I}
\newcommand{\Kappa}{K}
\newcommand{\Mu}{M}
\newcommand{\Nu}{N}
\newcommand{\Rho}{P}
\newcommand{\Tau}{T}
\newcommand{\Upsi}{\Upsilon}
\newcommand{\omicron}{o}
\newcommand{\lang}{\langle}
\newcommand{\rang}{\rangle}
\newcommand{\Union}{\bigcup}
\newcommand{\Intersection}{\bigcap}
\newcommand{\Oplus}{\bigoplus}
\newcommand{\Otimes}{\bigotimes}
\newcommand{\Wedge}{\bigwedge}
\newcommand{\Vee}{\bigvee}
\newcommand{\coproduct}{\coprod}
\newcommand{\product}{\prod}
\newcommand{\closure}{\overline}
\newcommand{\integral}{\int}
\newcommand{\doubleintegral}{\iint}
\newcommand{\tripleintegral}{\iiint}
\newcommand{\quadrupleintegral}{\iiiint}
\newcommand{\conint}{\oint}
\newcommand{\contourintegral}{\oint}
\newcommand{\infinity}{\infty}
\newcommand{\bottom}{\bot}
\newcommand{\minusb}{\boxminus}
\newcommand{\plusb}{\boxplus}
\newcommand{\timesb}{\boxtimes}
\newcommand{\intersection}{\cap}
\newcommand{\union}{\cup}
\newcommand{\Del}{\nabla}
\newcommand{\odash}{\circleddash}
\newcommand{\negspace}{\!}
\newcommand{\widebar}{\overline}
\newcommand{\textsize}{\normalsize}
\renewcommand{\scriptsize}{\scriptstyle}
\newcommand{\scriptscriptsize}{\scriptscriptstyle}
\newcommand{\mathfr}{\mathfrak}
\newcommand{\statusline}[2]{#2}
\newcommand{\tooltip}[2]{#2}
\newcommand{\toggle}[2]{#2}

% Theorem Environments
\theoremstyle{plain}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\newtheorem{prop}{Proposition}
\newtheorem{cor}{Corollary}
\newtheorem*{utheorem}{Theorem}
\newtheorem*{ulemma}{Lemma}
\newtheorem*{uprop}{Proposition}
\newtheorem*{ucor}{Corollary}
\theoremstyle{definition}
\newtheorem{defn}{Definition}
\newtheorem{example}{Example}
\newtheorem*{udefn}{Definition}
\newtheorem*{uexample}{Example}
\theoremstyle{remark}
\newtheorem{remark}{Remark}
\newtheorem{note}{Note}
\newtheorem*{uremark}{Remark}
\newtheorem*{unote}{Note}

%-------------------------------------------------------------------

\begin{document}

%-------------------------------------------------------------------


\section*{Minkowski&#39;s inequality}

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

\hypertarget{analysis}{}\paragraph*{{Analysis}}\label{analysis}

[[!include analysis - contents]]

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

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{proofs}{Proofs}\dotfill \pageref*{proofs} \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}

What is commonly known as \emph{Minkowski's inequality} is the statement that the [[p-norm]] ${\Vert f\Vert_p} \coloneqq \root{p}{\int_X {\vert f\vert^p} d\mu}$ on [[Lebesgue spaces]] indeed satisfies the [[triangle inequality]].

\hypertarget{proofs}{}\subsection*{{Proofs}}\label{proofs}

Our proof of Minkowski's inequality is broken down into a few 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 set|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}
Since conditions 2. and 3. are pretty obviously equivalent, we just prove 1. and 3. are equivalent. Condition 1. implies condition 3. easily: if $\|u\| = 1 = \|v\|$ 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\|} \\
 & = & t + (1-t) = 1.}
\end{displaymath}
Now we 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'\|}} & = & \left(\frac{{\|v\|}}{{\|v\|}+{\|v'\|}}\right)\frac{v}{{\|v\|}} + \left(\frac{{\|v'\|}}{{\|v\|}+{\|v'\|}}\right)\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}, the Minkowski 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.

\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}
The function $g: x \mapsto {|x|}$ is convex, and for $1 \lt p$ the function $f: t \mapsto t^p$ for $t \geq 0$ is monotone increasing and convex, by the first and second derivative tests. Thus $g(t x + (1-t)y) \leq t g(x) + (1-t)g(y)$ and then

\begin{displaymath}
\itexarray{
f(g(t x + (1-t)y)) & \leq & f(t g(x) + (1-t)g(y)) \\
 & \leq & t f(g(x)) + (1-t)f(g(y))
}
\end{displaymath}
so $f \circ g: x \mapsto {|x|^p}$ is convex.

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

\begin{displaymath}
\int_X {|t u + (1-t)v|^p} \,d\mu \leq \int_X 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}
Another commonly seen proof of Minkowski's inequality derives it with the help of [[Hölder's inequality]]; see there for some commentary on this. But this is probably not the first thing one would think of unless one knows the trick, whereas the alternative proof given above seems geometrically motivated and fairly simple.

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

\begin{itemize}%
\item [[Hölder's inequality]]

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

\begin{itemize}%
\item [[Todd Trimble]], \emph{[[toddtrimble:p-norms]]}


\item Wikipedia, \emph{\href{https://en.wikipedia.org/wiki/Minkowski_inequality}{Minowski's inequality}}


\item [[eom]], \emph{\href{https://www.encyclopediaofmath.org/index.php/Minkowski_inequality}{Minkowski inequality}}



\end{itemize}
[[!redirects Minkowski inequality]]

[[!redirects Minkowski inequality]] [[!redirects Minkowski's inequality]] [[!redirects Minkowski's inequality]] [[!redirects Minkowski's inequality]] [[!redirects Minkowski/`s inequality]]

[[!redirects Minkowki's inequality]] [[!redirects Minkowki's inequality]]



\end{document}