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extreme value theorem (Theorem)

Extreme Value Theorem. Let $ a$ and $ b$ be real numbers with $ a<b$, and let $ f$ be a continuous, real valued function on $ [a,b]$. Then there exists $ c,d\in[a,b]$ such that $ f(c)\leq f(x) \leq f(d)$ for all $ x\in[a,b]$.

Proof. We show only the existence of $ d$. By the boundedness theorem $ f([a,b])$ is bounded above; let $ l$ be the least upper bound of $ f([a,b])$. Suppose, for a contradiction, that there is no $ d\in[a,b]$ such that $ f(d)=l$. Then the function

$\displaystyle g(x) = \frac{1}{l-f(x)} $
is well defined and continuous on $ [a,b]$. Since $ l$ is the least upper bound of $ f([a,b])$, for any positive real number $ M$ we can find $ \alpha\in[a,b]$ such that $ f(\alpha)>l-\frac{1}{M}$, then
$\displaystyle M < \frac{1}{l-f(\alpha)} \textrm{.}$
So $ g$ is unbounded on $ [a,b]$. But by the boundedness theorem $ g$ is bounded on $ [a,b]$. This contradiction finishes the proof.



"extreme value theorem" is owned by classicleft.
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Other names:  Weierstrass extreme value theorem
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Cross-references: unbounded, positive, well defined, contradiction, least upper bound, bounded, boundedness theorem, proof, function, continuous, real numbers

This is version 4 of extreme value theorem, born on 2004-07-25, modified 2004-07-28.
Object id is 6023, canonical name is ExtremeValueTheorem.
Accessed 6794 times total.

Classification:
AMS MSC: 26A06 (Real functions :: Functions of one variable :: One-variable calculus)
Pending Errata and Addenda
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