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Homework Help: Real analysis show f is continuous

  1. Nov 11, 2007 #1
    1. The problem statement, all variables and given/known data
    suppose f is a function with the property f(x+y)=f(x)+f(y) for x,y in th reals. suppose f is continuous at 0. show f is continuous everywhere.


    2. Relevant equations



    3. The attempt at a solution

    I do not understand how to show that f is continuous everywhere.
     
  2. jcsd
  3. Nov 11, 2007 #2

    morphism

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    Try to see what properties f has. For example, try to prove that f(x-y) = f(x)-f(y). (Hint: Prove first that f(0)=0 and f(-x)=-f(x).)

    Then use an epsilon-delta argument to prove that f is continuous everywhere.
     
  4. Nov 11, 2007 #3

    HallsofIvy

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    Use the definition of continuous: f is continuous if and only if [itex]\lim_{x\rightarrow a}f(x)= f(a)[/itex].

    Now, let h= x- a and write that in terms of [itex]\lim_{h\rightarrow 0}[/itex].

    Exactly the same proof can be used to show that if such an f is continuous at any specific x, then it is continuous at x=0. Putting the two together, if a function satisfying f(x+y)= f(x)+ f(y) is continuous for any x, then it is contnuous for all x.

    And, by the way, f must be of the form f(x)= Cx for some number C. The really remarkable thing is that there exist function satisfying f(x+y)= f(x)+ f(y) that are not continuous!
     
  5. Nov 11, 2007 #4
    ok I have deterimed that i can not do this problem. I do not not even understand how two deduce that f(0)=0
     
  6. Nov 11, 2007 #5

    morphism

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    You know that f(x+y)=f(x)+f(y). What happens if you plug in x=y=0?
     
  7. Nov 11, 2007 #6
    you have f(0)=f(0)+f(0)
     
  8. Nov 11, 2007 #7

    morphism

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  9. Nov 11, 2007 #8
    how is this 0
     
  10. Nov 11, 2007 #9
    f(0)-f(0)=f(0)+f(0)-f(0)
    0=f(0)
     
  11. Nov 11, 2007 #10

    morphism

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    That's good. Now can you show that f(-x)=-f(x)? (Try to plug in "clever" values for x and y in f(x+y)=f(x)+f(y).)
     
  12. Nov 12, 2007 #11

    CompuChip

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    If you prefer, I also think an [itex]\epsilon-\delta[/itex] proof is not very hard in this case.
     
  13. Nov 12, 2007 #12
    In fact we can solve this functional equation. Suppose that f is continous at 0. Then f(0+0)=f(0)+f(0) so f(0)=0. Now [tex]\lim_{y\to 0}f(x+y) = \lim_{y\to 0}f(x)+f(y) = f(x)+f(0)=f(y)[/tex] so [tex]f[/tex] is continous everywhere. Let [tex]f(1)=c[/tex]. But using the argument f(1+...+1) = nf(1) you can show that f(n)=nc. Now consider a rational number p/q with p,q>0 so if x=p/q then qx=p and so f(qx) = p so f(x+...+x)=p thus pf(x)=ap; that means f(p/q) = a*(p/q). We can extend this argument to negative also easily. And so if x is any real number there is a sequence of rationals converging to it so f(x) = a*x because this is true for rationals. Thus, the only possible functions are the lines through the origin.
     
  14. Nov 13, 2007 #13

    HallsofIvy

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    There do, however, exist non-continuous solutions to that functional equation. In distinction to the continuous solutions, all of the form f(x)= Cx, non-continuous solutions are remarkably messy- there are "dense in the plane".
     
  15. Nov 14, 2007 #14
    Trivial? Use axiom of choice and contruct the Hamel Basis.
     
  16. Nov 15, 2007 #15

    HallsofIvy

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    I didn't say it wasn't easy! I'm not sure I would consider contructing a Hamel basis trivial, though. This is one of those situations where it is easy to tell HOW to do something, just very difficult to do it. "First construct a basis for the real numbers as a vector space over the rational numbers"- It's very easy to see that we can use "1" as a basis element and then [itex]\sqrt{2}[/itex] as another, ... but not clear to me how I could give the entire basis. Now define f(1)= 1, f([itex]\sqrt{2}[/itex])= 2, f(v)= 0 for v any other member of the basis, and define f(x) for any real number x "by linearity". Then we have a function satisfying f(x+ y)= f(x)+ f(y) (and f(rx)= rf(x) for any rational r) but which is not continuous.

    (I think we have gotten well off the original question now!)
     
    Last edited by a moderator: Nov 15, 2007
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