1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Baby Rudin problem

  1. Jan 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Let f be defined for all real x and suppose that,
    [itex]\left|f(x) - f(y)\right|\leq(x-y)^2[/itex]

    for all real x and y. Prove that f is constant.

    3. The attempt at a solution
    First of all, is following allowed. Since f is constant then [itex]\left|f(x) - f(y)\right|=0[/itex], and form here to build my proof?

    Here is what i dont understand. If i suppose that [itex]\left|f(x) - f(y)\right|\leq(x-y)^2[/itex] then does that imply that [itex]f(x)[/itex] is always greater then [itex]f(y)[/itex]?
    Since [itex](x-y)^2>0[/itex] when [itex]x≠y[/itex] then that means that [itex]\left|f(x) - f(y)\right|> 0[/itex] because if it were to be [itex]\left|f(x) - f(y)\right|= 0[/itex] we would violate [itex]x≠y[/itex] . Onliy way that [itex]\left|f(x) - f(y)\right|> 0[/itex] is that [itex]f(x)[/itex] is always greater then [itex]f(y)[/itex]

    But this leads me to contradiction since [itex]f[/itex] should be constant i.e. [itex]\left|f(x) - f(y)\right|=0[/itex].

    On the other hand [itex](x-y)^2=0[/itex] iff [itex]x=y[/itex] then that means that [itex]\left|f(x) - f(y)\right|= 0[/itex].

    So perhaps my question boils down to is can i take that [itex]x=y[/itex] or am i only allowed to that [itex]x≠y[/itex]???

    Also i am trying to think of all things that will prove that some function is constant. First thing that comes to mind is that first derivative is 0. Can that be a starting point?
  2. jcsd
  3. Jan 24, 2013 #2


    Staff: Mentor

    No, this isn't allowed. You can't assume that f is constant - that's what you need to show. Starting with the given inequality, you need to arrive at the conclusion that f is constant.
    No it doesn't imply that. |f(x) - f(y)| is always >= 0, regardless of which function is larger.
  4. Jan 24, 2013 #3


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It sounds promising, but you aren't given that the function is differentiable. Can you show it is differentiable based on the given inequality?

    One observation is that the function is uniformly continuous, because if [itex]|x - y| \leq 1[/itex] we have
    $$|f(x) - f(y)| \leq (x-y)^2 \leq |x - y|$$
    In fact, this shows that [itex]f[/itex] is Lipschitz, a stronger condition than uniform continuity, but weaker than differentiability. Perhaps with a bit more work you can show differentiability?
  5. Jan 25, 2013 #4
    Try dividing both sides by [itex] \mid x-y\mid [/itex] and then find a way to look at it as [itex] f'(y)[/itex].
  6. Jan 25, 2013 #5


    User Avatar
    Homework Helper

    This is a dumb question (like so many in baby Rudin, what a terible book). Clearly f is differentiable, but I prefer to note

    $$|\mathrm{f}(x)-\mathrm{f}(y)|=\left| \sum^n_{k=1}(\mathrm{f}(z_k)-\mathrm{f}(z_{k-1}))\right|\le \sum^n_{k=1}|\mathrm{f}(z_k)-\mathrm{f}(z_{k-1})|\le \sum^n_{k=1}(z_k-z_{k-1})^2=(x-y)^2/n\le \epsilon$$
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook