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Real Analysis proof

  1. Nov 10, 2009 #1
    1. The problem statement, all variables and given/known data

    Let f be any function on the real line and suppose that: |f(x)-f(y)|<=|x-y|^2 for all x,y in R. Prove that f is a constant function. Note: "<=" reads "less than or equal to"

    2. Relevant equations



    3. The attempt at a solution

    I have tried proof by contradiction, it seems to be the most obvious route in proving this statement. I started by assuming that there exists x,y in the domain of the function f(x) such that f(x) is not equal to f(y). I wasn't really able to proceed much further from there. Any help towards finishing this proof or perhaps a different approach would be greatly appreciated.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 10, 2009 #2

    lanedance

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    can you use derivatves?

    what is the magnitude of the derivative as x->y?
     
  4. Nov 10, 2009 #3

    Mark44

    Staff: Mentor

    It's not given that the function is even continuous, let alone differentiable, so we can't assume that f' exists.
     
  5. Nov 11, 2009 #4

    Dick

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    You could try and prove f' exists. Or you could try and split the interval between x and y into 2^n parts and see what the inequality tells you as n->infinity.
     
  6. Nov 11, 2009 #5

    lanedance

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    fair bump, I thought uniform continuity & differentiablilty would follow reasonably easy from the definition, though i do like Dick's 2nd suggestion
     
    Last edited: Nov 11, 2009
  7. Nov 11, 2009 #6

    jgens

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    Spliting the interval into n equal parts should suffice.
     
  8. Nov 11, 2009 #7

    jgens

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    How's this?

    Let [itex]0<|x-a|<\mathrm{min}(1,\varepsilon)[/itex]. From this, we have that [itex]0\leq|f(x)-f(a)|\leq|x-a|^2<|x-a|<\varepsilon[/itex]. This proves that [itex]f[/itex] is continuous.

    From the defining property of [itex]f[/itex], we know that,

    [tex]\frac{|f(x)-f(a)|}{|x-a|} \leq |x-a|[/tex]

    Since [itex]f[/itex] is continuous, evaluating the limit as [itex]x \to a[/itex], we find that [itex]f'(a) = 0[/itex] which proves that [itex]f[/itex] is a constant function.

    I realize this is really rough, but could this approach be used to prove the initial problem? It's real late nowso I'm sure that it's riddled with errors.
     
  9. Nov 11, 2009 #8
    Well yes for this you can show by definition that |f'(a)| = 0 for every a due to mean value theorem. You can run into troubles if you tried applying this to say, finding isometries on R. Obviously you only need to show differentiability, but anytime you see |f(x)-f(y)| bounded by something involving |x-y|, it's going to be continuous.
     
  10. Nov 11, 2009 #9
    I think your proof is fine. I remember doing this problem last year and I think that's basically the solution I used; there may have been some subtlety I overlooked, but I don' think so.
     
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