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Viewing Version
3
of
'proof of Darboux's theorem'
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| Title of object: |
proof of Darboux's theorem |
| Canonical Name: |
ProofOfDarbouxsTheorem |
| Type: |
Proof |
| Created on: |
2002-06-06 03:56:10 |
| Modified on: |
2005-02-18 08:04:25 |
| Classification: |
msc:26A06 |
Revision comment (for changes between this and next version):
| Changes for correction #13919 ('Locally Increasing at a'). |
Preamble:
% this is the default PlanetMath preamble. as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.
% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
% used for TeXing text within eps files
%\usepackage{psfrag}
% need this for including graphics (\includegraphics)
%\usepackage{graphicx}
% for neatly defining theorems and propositions
%\usepackage{amsthm}
% making logically defined graphics
%\usepackage{xypic}
% there are many more packages, add them here as you need them
% define commands here
\newcommand{\Prob}[2]{\mathbb{P}_{#1}\left\{#2\right\}}
\newcommand{\norm}[1]{\left\|#1\right\|}
% Some sets
\newcommand{\Nats}{\mathbb{N}}
\newcommand{\Ints}{\mathbb{Z}}
\newcommand{\Reals}{\mathbb{R}}
\newcommand{\Complex}{\mathbb{C}} |
Content:
Without loss of generality we migth and shall assume $f'_{+}(a)>t>f'_{-}(b)$. Let $g(x):=f(x)-tx$. Then $g'(x)=f'(x)-t$, $g'_{+}(a)>0>g'_{-}(b)$, and we wish to find a zero of $g'$.
Since $g$ is a continuous function on $[a,b]$, it attains a maximum on $[a,b]$. This maximum cannot be at $a$, since $g'_{+}(a)>0$ so $g$ is locally increasing at $a$. Similarly, $g'_{-}(b)<0$, so $g$ is locally decreasing at $b$ and cannot have a maximum at $b$. So the maximum is attained at some $c \in (a,b)$. But then $g'(c)=0$ by \PMlinkname{Fermat's theorem}{FermatsTheoremStationaryPoints}. |
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