Proving Constant Function f from Derivative: f'(a)

[tvguide123]

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]

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?

 

[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.

 

[dirk_mec1]

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

 

[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?

 

[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