Mean Value Theorem Proof

**Larrioto** · January 26th 2009, 03:39 PM

Use the Mean Value Theorem to prove that for all a,b that belongs to the reel, we have :

$|sin(b)-sin(a)|$ $\leq$ $| b-a |$

**NonCommAlg** · January 26th 2009, 04:03 PM

Originally Posted by Larrioto

Use the Mean Value Theorem to prove that for all a,b that belongs to the reel, we have :

$|sin(b)-sin(a)|$ $\leq$ $| b-a |$

well, if you let $f(x)=\sin x,$ then by mean value theorem there exists $c$ between $a,b$ such that $f(b)-f(a)=(b-a)f'(c).$ that means $\sin b - \sin a = (b-a)\cos c.$

so $|\sin b - \sin a|=|b-a||\cos c| \leq |b-a|,$ because $|\cos c| \leq 1.$ it wan't very difficult, was it?

**Larrioto** · January 26th 2009, 04:15 PM

**Mathstud28** · January 26th 2009, 04:25 PM

Originally Posted by Larrioto

Choose any value $c$ , we know there exists a corresponding interval $[a,b]$ such that $|f'(c)|=\left|\frac{f(b)-f(a)}{b-a}\right|\leqslant M$ and since $c$ was arbitrary this must work for every interval $[a,b]$ .

Another almost identical method would be to assume that there exists an interval $[a,b]$ such that $\left|\frac{f(b)-f(a)}{b-a}\right|>M$ . But since $f$ is differentiable we can find a $c\in(a,b)$ such that $|f'(c)|=\left|\frac{f(b)-f(a)}{b-a}\right|\implies |f'(c)|>M$ which is a contradiction.

This is called Lipschitz's continuity by the way.

If you have any questions at all feel free to ask

**Larrioto** · January 26th 2009, 04:38 PM

Originally Posted by Mathstud28

Choose any value $c$ , we know there exists a corresponding interval $[a,b]$ such that $|f'(c)|=\left|\frac{f(b)-f(a)}{b-a}\right|\leqslant M$ and since $c$ was arbitrary this must work for every interval $[a,b]$ .

Another almost identical method would be to assume that there exists an interval $[a,b]$ such that $\left|\frac{f(b)-f(a)}{b-a}\right|>M$ . But since $f$ is differentiable we can find a $c\in(a,b)$ such that $|f'(c)|=\left|\frac{f(b)-f(a)}{b-a}\right|\implies |f'(c)|>M$ which is a contradiction.

This is called Lipschitz's continuity by the way.

If you have any questions at all feel free to ask

Do you mean that whatever number we choose to be c , it's derivative will be 0 , therefore we can say that if $M\geq 0$ then it is $\geq \left|\frac{f(b)-f(a)}{b-a}\right|$ ?

**Mathstud28** · January 26th 2009, 04:59 PM

Originally Posted by Larrioto

Do you mean that whatever number we choose to be c , it's derivative will be 0 , therefore we can say that if $M\geq 0$ then it is $\geq \left|\frac{f(b)-f(a)}{b-a}\right|$ ?

Im not quite sure what you mean by that...If the first proof doesn't make sense consider the second one. It reads better anyways.

**Larrioto** · January 26th 2009, 05:09 PM

ok , i get it , it's just that f(c) can be replaced by f(x), that was confusing me ...

**NonCommAlg** · January 26th 2009, 05:46 PM

Originally Posted by Mathstud28

Choose any value $c$ , we know there exists a corresponding interval $[a,b]$ such that $|f'(c)|=\left|\frac{f(b)-f(a)}{b-a}\right|$ ...

this, although an interesting claim, but is not what mean value theorem says: you choose an interval [a,b] and then you find a < c < b that satisfies the condition. what you're doing here is

reversing things, i.e. you choose c and then you claim that there exists an interval [a,b] containing c with that condition! i'm not sure if this is always possible!

**Mathstud28** · January 26th 2009, 06:10 PM

Originally Posted by NonCommAlg

this, although an interesting claim, but is not what mean value theorem says: you choose an interval [a,b] and then you find a < c < b that satisfies the condition. what you're doing here is

reversing things, i.e. you choose c and then you claim that there exists an interval [a,b] containing c with that condition! i'm not sure if this is always possible!

Mathworld has an "iff" in it :S...that is why I supplied a second proof that did not require that just in case. Sorry if there has been any confusion.

**NonCommAlg** · January 26th 2009, 07:15 PM

Originally Posted by Mathstud28

Mathworld has an "iff" in it :S...that is why I supplied a second proof that did not require that just in case. Sorry if there has been any confusion.

here's a simple counter-example: let $f(x)=x^3$ and choose $c=0.$ geometrically it's clear that there are no $a \neq b$ such that $f'(0)=\frac{f(b)-f(a)}{b-a}.$ you can also prove it very easily:

if there exist such $a,b,$ then: $0=\frac{b^3-a^3}{b-a}=a^2+ab+b^2,$ which is possible iff $a=b=0,$ because: $0=2a^2 + 2ab+2b^2=(a+b)^2+a^2+b^2.$

regarding Larrioto's question:

by mean value theorem for any real numbers $a,b,$ there exists $c$ (between $a,b$ ) such that $f(b)-f(a)=(b-a)f'c).$ thus: $|f(b)-f(a)|=|b-a||f'(c)| \leq M|b-a|.$

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