# Derivative proof

1. Jul 4, 2008

### tvguide123

1. The problem statement, all variables and given/known data
Suppose that |f(x) - f(y)| $$\leq$$ |x - y|n
for n > 1

Prove that f is constant by considering f '

2. Relevant equations
Well
f'(a) = limit as x->a [f(x) - f(a)]/[x-a]

3. The attempt at a solution

I'm really not sure how the derivative of "f" is going to show that "f" is constant since I cannot actually calculate the derivative.

Any help or hints would be greatly appreciated, this ones got me stumped.

Thanks :)

2. Jul 4, 2008

### Mute

Divide both sides of the inequality by |x-y|, then take the limit as x -> y. Use your definition of the derivative on the LHS and evaluate the RHS in this limit. What does the result tell you?

3. Jul 4, 2008

### tvguide123

Ok I tried doing that but I'm having trouble with the absolute value signs.
This is what I got but I'm not sure if the absolute value signs are in the right place for the LHS.

|f[x] - f[y] / x-y | is less than or equal to |x-y|^n-1

then i take limit as x->y for both sides.

Paying attention to the RHS. lim x->y |x-y|^n-1.
It seems like the limit of the RHS is approaching 0 which means that |f[x] - f[y] / x-y |is less than or equal to 0.

Hence f'(x) = 0 and thus f(x) is constant.

Would these steps be allowable and are my absolute value signs in the right place for the LHS.

Thanks for the quick reply Mute.

4. Jul 5, 2008

### dirk_mec1

Correct! And your absolute signs are in the right place.

5. Jul 6, 2008

### tvguide123

Hmm wait a minute.

f'(x) = limit as x->y [f(x) - f(y)]/[x-y]

is the formula for the derivative.

Does this formula still hold if [f(x) - f(y)]/[x-y] is in absolute value signs?

Last edited: Jul 6, 2008
6. Oct 18, 2008

### sirchamb2

For any real-valued function g(x):

|g(x)| is always real-valued and positive so |g(x)| <= 0 then |g(x)| = 0

Also |g(x)| = 0 if and only if g(x) = 0

So if |f'(x)| = 0 then f'(x) = 0 so f(x) = C