Prove that this function is constant on I (absolute values, derivatives)

[llama_guy]

Stuck on the following problem

f is a function on the interval I. We know that there is a constant K>0 such that
|f(a)-f(b)| <= K*|a-b|^2 for all a,b in I

Prove that the function f is constant on I.
(hint: start by calculating the derivative using the definition of the derivative)

So I do that.
f'(x)=f(x+h)-f(x)/h

I insert arbitrary values of x on I:
x=b, x+h=a, so that h=a-b

The derivative then becomes:
f'(b)=(f(a)-f(b))/(a-b)

Remove absolute value from the above inequality:

-K*|a-b|^2 <=f(a)-f(b) <= K*|a-b|^2

Insert the derivative by dividing with (a-b):
-K*|a-b|^2/(a-b) <= f'(b) <= K*|a-b|^2/(a-b)

I know I have to somehow prove that f'(x) = 0, which means K*|a-b|^2/(a-b) = -K*|a-b|^2/(a-b) = 0 has to be true.
Assuming h is positive, then a>b and |a-b|=(a-b)
-K*|a-b|^2/(a-b) <= f'(b) <= K*|a-b|^2/(a-b)
-K*(a-b)^2/(a-b) <= f'(b) <= K*(a-b)^2/(a-b)
Simplify:
-K*(a-b) <= f'(b) <= K*(a-b)

I'm stuck here. How does -K*(a-b) = K*(a-b) = 0? Am I even approaching this problem correctly?

EDIT: If it's the derivative then h->0 so (a-b)->0 so -K*(a-b) is the left limit and K*(a-b) is the left limit; both going towards 0, hence f'(b) = 0 - am I correct? And this goes for all values of b on the interval, hence the function is a constant on the interval.

 

[HallsofIvy]

Stuck on the following problem
So I do that.
f'(x)=f(x+h)-f(x)/h

This is NOT the definition of the derivative!
check that.

I insert arbitrary values of x on I:
x=b, x+h=a, so that h=a-b

The derivative then becomes:
f'(x)=f(a)-f(b)/(a-b)

Remove absolute value from the above inequality:

-K*|a-b|^2 <=f(a)-f(b) <= K*|a-b|^2

Insert the derivative by dividing with (a-b):
-K*|a-b|^2/(a-b) <= f'(x) <= K*|a-b|^2/(a-b)

I know I have to somehow prove that f'(x) = 0, which means K*|a-b|^2/(a-b) = -K*|a-b|^2/(a-b) = 0 has to be true.
Assuming h is positive, then a>b and |a-b|=(a-b)
-K*|a-b|^2/(a-b) <= f'(x) <= K*|a-b|^2/(a-b)
-K*(a-b)^2/(a-b) <= f'(x) <= K*(a-b)^2/(a-b)
Simplify:
-K*(a-b) <= f'(x) <= K*(a-b)

I'm stuck here. How does -K*(a-b) = K*(a-b) = 0? Am I even approaching this problem correctly?

 

[llama_guy]

Sorry, I made that a little unclear.
f'(x)=f(x+h)-f(x)/hI did mean the definition of the derivative is f'(x)=(f(x+h)-f(x))/h, h->0 (I tend to just scribble things a little simplified, sorry).

You just quoted the rest? Were you meaning to say something or was it correct? S:

(also, see my edit, am I on the right track here)

 

[DrPhil]

Stuck on the following problem
f is a function on the interval I. We know that there is a constant K>0 such that
|f(a)-f(b)| <= K*|a-b|^2 for all a,b in I

Prove that the function f is constant on I.
(hint: start by calculating the derivative using the definition of the derivative)

So I do that.
f'(x) = lim(h->0)[f(x+h)-f(x)]/h

I insert arbitrary values of x on I:
x=b, x+h=a, so that h=a-b

The derivative then becomes:
f'(b) = lim(a->b)[f(a)-f(b)]/(a-b)

Remove absolute value from the above inequality:

-K*|a-b|^2 <=f(a)-f(b) <= K*|a-b|^2

Insert the derivative by dividing with (a-b)-K*|a-b|^2/(a-b) <= f'(x) <= K*|a-b|^2/(a-b)

I know I have to somehow prove that f'(x) = 0, which means K*|a-b|^2/(a-b) = -K*|a-b|^2/(a-b) = 0 has to be true.
Assuming h is positive, then a>b and |a-b|=(a-b)
-K*|a-b|^2/(a-b) <= f'(x) <= K*|a-b|^2/(a-b)
-K*(a-b)^2/(a-b) <= f'(x) <= K*(a-b)^2/(a-b)
Simplify:
-K*(a-b) <= f'(x) <= K*(a-b)

I'm stuck here. How does -K*(a-b) = K*(a-b) = 0? Am I even approaching this problem correctly?

\(\displaystyle \displaystyle |f'(x)| = \lim_{h \to 0+}\left[\dfrac{|f(x+h) - f(x)|}{h}\right] \le \lim_{h \to 0+}\left[\dfrac{K\ h^2}{h}\right] = \lim_{h \to 0+} [K\ h] = 0 \)

 

[llama_guy]

I managed to find the solution, though not as elegantly as you, Phil. Thanks! : D