Results 1 to 4 of 4

Thread: Prove that f is a constant.

  1. #1
    Newbie
    Joined
    Dec 2012
    From
    United States
    Posts
    8

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member jakncoke's Avatar
    Joined
    May 2010
    Posts
    387
    Thanks
    80

    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?)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Feb 2013
    From
    Saudi Arabia
    Posts
    445
    Thanks
    86

    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member jakncoke's Avatar
    Joined
    May 2010
    Posts
    387
    Thanks
    80

    Re: Prove that f is a constant.

    Quote Originally Posted by MINOANMAN View Post
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Dec 8th 2011, 08:16 AM
  2. to prove f is constant
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: May 14th 2011, 03:59 AM
  3. Prove that f(x) is constant...?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 27th 2009, 02:51 PM
  4. prove f(x) is constant
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jan 22nd 2009, 09:13 AM
  5. Prove a function is constant.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 17th 2007, 12:23 AM

Search Tags


/mathhelpforum @mathhelpforum