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Difference between revisions of "Maximum-minimum theorem"
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Assume if possible <math>\forall n\in\mathbb{N}\exists x_n\in [a,b]</math> such that <math>f(x_n)>n</math>
 
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.
+
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\langlex_{n_r}\right\rangle</math> of <math>\left\langle x_n\right\rangle</math> which converges to <math>x</math>.  
+
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.  
 
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.  
Line 30: Line 30:
  
 
i.e. <math>f(x)=M</math>
 
i.e. <math>f(x)=M</math>
 +
 +
==References==
 +
R.G. Bartle, D.R. Sherbert, <i>Introduction to Real Analysis</i>
 +
 +
==See Also==
 +
<UL>
 +
<LI>Calculus</LI>
 +
<LI>Bolzano's theorem</LI>
 +
</UL>

Revision as of 23:41, 14 February 2008

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

See Also

  • Calculus
  • Bolzano's theorem
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