1. Not finding help here? Sign up for a free 30min 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!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

  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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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: Nov 15, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?