# Derivative proof

## Homework Statement

Suppose that |f(x) - f(y)| $$\leq$$ |x - y|n
for n > 1

Prove that f is constant by considering f '

## Homework Equations

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

## 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 :)

Mute
Homework Helper
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?

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.

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

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:
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