View source for Maximum-minimum theorem

The '''Maximum-minimum theorem''' is a result about continous functions that deals with a property of intervals rather than that of the function itself.

==Statement==
Let <math>f:[a,b]\rightarrow\mathbb{R}</math>

Let <math>f</math> be continous on <math>[a,b]</math>

Then, <math>f</math> has an [[Extrema|absolute maximum]] and an [[Extrema|absolute minimum]] on <math>[a,b]</math>

==Proof==
We will first show that <math>f</math> is bounded on <math>[a,b]</math>...(1)

Assume if possible <math>\forall n\in\mathbb{N}\exists x_n\in [a,b]</math> such that <math>f(x_n)>n</math>

As <math>[a,b]</math> is bounded, <math>\left\langle  x_n\right\rangle</math> is bounded.

By the [[Bolzano-Weierstrass theorem]], there exists a sunsequence <math>\left\langle x_{n_r}\right\rangle</math> of <math>\left\langle x_n\right\rangle</math> which converges to <math>x</math>. 

As <math>[a,b]</math> is closed, <math>x\in [a,b]</math>. Hence, <math>f</math> is continous at <math>x</math>, and by the [[Limit|sequential criterion for limits]] <math>f(x_n)</math> is convergent, contradicting the assumption. 

Similarly we can show that <math>f</math> is bounded below

Now, Let <math>M=\sup\{f([a,b])\}</math>

By the [[Gap lemma]], <math>\forall n\in\mathbb{N}</math>, <math>\exists x_n</math> such that <math>M-f(x_n)<\frac{1}{n}</math>

As <math>\left\langle x_n\right\rangle</math> is bounded, by [[Bolzano-Weierstrass theorem]], <math>\left\langle x_n\right\rangle</math> has a subsequence <math>\left\langle x_{n_r}\right\rangle</math> that converges to <math>x\in [a,b]</math>

As <math>f</math> is continous at <math>x</math>, <math>f(x)\in V_{\frac{1}{n}}(M)\forall n</math>

i.e. <math>f(x)=M</math>

==References==
R.G. Bartle, D.R. Sherbert, <i>Introduction to Real Analysis</i>, John Wiley & Sons

==See Also==
<UL>
<LI>[[Calculus]]</LI>
<LI>[[Bolzano's theorem]]</LI>
<LI>[[Closed set]]</LI>
<LI>[[Topology]]</LI>
</UL>