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Maximum-minimum theorem: Difference between revisions

New page: 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]\ri...
 
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==Proof==
==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>
[[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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