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Difference between revisions of "Maximum-minimum theorem"
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==References==
 
==References==
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R.G. Bartle, D.R. Sherbert, <i>Introduction to Real Analysis</i>
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R.G. Bartle, D.R. Sherbert, <i>Introduction to Real Analysis</i>, John Wiley & Sons
  
 
==See Also==
 
==See Also==
 
<UL>
 
<UL>
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<LI>Calculus</LI>
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<LI>[[Calculus]]</LI>
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<LI>Bolzano's theorem</LI>
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<LI>[[Bolzano's theorem]]</LI>
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<LI>[[Closed set]]</LI>
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<LI>[[Topology]]</LI>
 
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</UL>
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[[Category:Theorems]]

Latest revision as of 11:14, 30 May 2019

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 $f:[a,b]\rightarrow\mathbb{R}$

Let $f$ be continous on $[a,b]$

Then, $f$ has an absolute maximum and an absolute minimum on $[a,b]$

Proof

We will first show that $f$ is bounded on $[a,b]$...(1)

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

As $[a,b]$ is bounded, $\left\langle x_n\right\rangle$ is bounded.

By the Bolzano-Weierstrass theorem, there exists a sunsequence $\left\langle x_{n_r}\right\rangle$ of $\left\langle x_n\right\rangle$ which converges to $x$.

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

Similarly we can show that $f$ is bounded below

Now, Let $M=\sup\{f([a,b])\}$

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

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

As $f$ is continous at $x$, $f(x)\in V_{\frac{1}{n}}(M)\forall n$

i.e. $f(x)=M$

References

R.G. Bartle, D.R. Sherbert, Introduction to Real Analysis, John Wiley & Sons

See Also

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