# Thread: Prove that f is a constant.

1. ## Prove that f is a constant.

Suppose f : R ------> R, and suppose that $\displaystyle \vert f(x) - f(y) \vert \leq (x - y)^2$ for all x, y in R.
Prove that f is a constant.

I tried solving this problem using the Mean Value Theorem, and showing that f'(xo) is always zero.

$\displaystyle \vert f'(xo) \vert = \vert \frac{f(x) - f(y)}{x -y} \vert \leq \frac{(x-y)^2}{(x-y)} \leq (x-y) \leq \epsilon$

Now, is there an argument I can use to let (x-y) be less than epsilon? Is there something that I'm overlooking?

Any insight would be welcomed,

Thanks

2. ## Re: Prove that f is a constant.

isnt $\displaystyle lim_{x \to x_0} \frac{|f(x)-f(x_0)|}{|x-x_0|}}$ another way of saying f is differentiable at $\displaystyle x_0$ if this limit exists?
then$\displaystyle lim_{x \to x_0} \frac{|f(x)-f(x_0)|}{|x-x_0|}} \leq lim_{x \to x_0}|x-x_0|$. The right side is going to 0. Since $\displaystyle x_0$ was an arbitrary point.
It means the function has 0 derivative for all points. Since $\displaystyle \mathbb{R}$ is path connected, for any $\displaystyle x_0,y_0$, there exists a parametrization g(t), such that $\displaystyle g(0) = x_0$ and $\displaystyle g(1)=y_0$

$\displaystyle \frac{|x_0 - y_0|}{1}$ = 0, which implies $\displaystyle x_0=y_0$ thus f is constant

We need to show this path connectivity part because f might be a different constant value at different parts in the domain (f=3 in one neighborhood, f=5 in another so f'(x) = 0 for both but they are not the same right?)

3. ## Re: Prove that f is a constant.

Jill90

follow please the steps below .

|f(x)-f(y)|<= (x-y)^2 implies |f(x)-f(y)|<=|x-y|^2

then |(f(x)-f(y))/(x-y)|<=|x-y| and taking into account a well known property of the moduli this imples: -|x-y|=<|(f(x)-f(y))/(x-y)|<= +|x-y|

Taking the limits when x goes to y the middle part is the derivative of f(x) and both |x-y| become zero.

as per the squezze theorem this implies that the derivative of f"(x) = 0

This implies that the function f(x) is CONSTANT .

MINOAS

4. ## Re: Prove that f is a constant.

Originally Posted by MINOANMAN
Jill90

follow please the steps below .

|f(x)-f(y)|<= (x-y)^2 implies |f(x)-f(y)|<=|x-y|^2

then |(f(x)-f(y))/(x-y)|<=|x-y| and taking into account a well known property of the moduli this imples: -|x-y|=<|(f(x)-f(y))/(x-y)|<= +|x-y|

Taking the limits when x goes to y the middle part is the derivative of f(x) and both |x-y| become zero.

as per the squezze theorem this implies that the derivative of f"(x) = 0

This implies that the function f(x) is CONSTANT .

MINOAS
you still have to assume path connectivity of any open set in $\displaystyle \mathbb{R}$. You have to atleast mention this part because f could be different constants on different open subsets if the open subsets of R are not path connected.