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Real analysis

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data
    f:R->R
    f satisfies that for all x,y f(x+y) = f(x) + f(y)

    show that f is continuous on R


    3. The attempt at a solution
    I assumed that limit of f at 0 existed. then i showed that that limit must be zero and that f(0)=0, so f is continuous at 0.
    From there, i broke it up into different cases (integer, inverse of an integer, rational number, irrational number) and showed that f must be continuous at each of them.

    My question is whether or not my assumption that limit of f at 0 existed is a valid assumption to make. Is it even necessarily true? if so how would i go about proving it?
     
  2. jcsd
  3. Nov 17, 2008 #2
    I think you are coming about this problem in the wrong way.

    while it is true for a continuous function, that the limit at zero will be defined, you cant assume this function is continuous in your proof.

    you should know 2 definitions of continuity. and epsilon delta one and a limits one. Ill right them out for you.


    a function f is continous at a point if [tex]\forall \epsilon <0 \exists \delta >0[/tex] such that |x-x0|< [tex]\delta\Rightarrow[/tex]|f(x)-f(x0)|<[tex]\epsilon[/tex]

    Also

    for any sequence Xn with [tex]Lim_{n \rightarrow infinity} X_{n}[/tex]= L
    A function f is continuous if f(Xn)=f(L) as n[tex]\rightarrow[/tex] infinity
     
  4. Nov 17, 2008 #3

    Dick

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    There's a problem with this question. There are functions that satisfy f(x+y)=f(x)+f(y) which are discontinuous. You DO need to assume that the function is continuous someplace. Can you state the original problem in full?
     
  5. Nov 17, 2008 #4
    can you give me an example of such a function?
    i think that is my problem
     
  6. Nov 17, 2008 #5

    Dick

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  7. Nov 17, 2008 #6

    HallsofIvy

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    The set of all real numbers is a vector space over the field of rational numbers. Choose a basis for that vector space. That's the "hard" part. Since the real numbers are uncountable and the rational numbers are uncountable, that basis is not only infinite, it is uncountable. But, since the rational numbers themselves are a one-dimensional subspace of that vector space, we can certainly take 1 to be a basis "vector". Since [itex]\sqrt{2}[/itex] is not rational, we can take [itex]\sqrt{2}[/itex] as another basis "vector". Define f(1)= 1, f([itex]\sqrt{2}[/itex])= 2,and f(x)= 0 for any other basis "vector". Finally define f(x) for any real number x by "linearity". Since f is a linear transformation on the real numbers, f(x+y)= f(x)+ f(y) and f(ax)= af(x) for x any rational number. It is (relatively) easy to prove that if f(x+y)= f(x)+ f(y) and f is continuous, then f(x)= Cx for some constant f. Since it is clear that the f(x) defined here satisfies f(x+y)= f(x) but is NOT of the form "Cx", it is not continuous.
     
  8. Nov 17, 2008 #7

    Dick

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    You can also explicitly show that the function Halls gave you is discontinuous. If q_i is a sequence of rational numbers approaching sqrt(2) then f(q_i) approaches sqrt(2), but f(sqrt(2))=2. Not continuous. What you can't determine is what is f(sqrt(3)) or f(pi) or f(e) are. As he said, you have to make an uncountably infinite number of choices to define the function on all other irrationals. You need the axiom of choice via Zorn's Lemma to show there is a basis. If you don't get this, I think the question may not be what you think it is, or it might just be a mistake.
     
  9. Dec 3, 2008 #8
    I have edited in an adjustment to the problem statement.

    Does this work now?

    So far I cant quite prove this either. I'll edit this post a little later with my attempt ( i have class in 5 minutes)
     
  10. Dec 3, 2008 #9

    Dick

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    If "and 0 is in C(f)" means you are assuming f is continuous at 0, then yes, now you can prove it.
     
  11. Dec 3, 2008 #10
    My professor dropped a hint (I think) and asked how I would do it if I used sequences.

    So I begin like this:

    Let a be in the domain and the sequence Xn ->0. ( since we know that the function is continuous at 0)

    Now I can write something that looks at least similar to the problem posed :

    a + Xn -> a and f(a + Xn) -> f(a) + f(Xn)

    but how can I relate that to continuity?
     
  12. Dec 3, 2008 #11

    Dick

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    Very easily. Write down the definition of continuity in terms of sequences and think about it.
     
  13. Dec 3, 2008 #12

    Office_Shredder

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    You want to show if yn is a sequence converging to a, then f(yn) converges to f(a). So start by defining xn = yn - a

    xn converges to 0 like you have.... and f(yn) = f(a) + f(xn)

    But what does f(xn) converge to?
     
  14. Dec 3, 2008 #13
    the sequential definition of continuity says that f(Xn)->f(0) ( in this case )

    so f(Yn) = f(a) + f(Xn) -> f(a) + f(0)

    Since Xn = Yn - a and we have that

    f(Xn)= f(Yn) - f(a) -> f(a) - f(a) = 0 (this seems reasonable given the way Xn is defined)

    Then f(Yn) -> f(a)

    which fits the sequential definition of continuity given above.

    This seems reasonable but I dont feel..' complete' about it...
     
  15. Dec 3, 2008 #14

    Dick

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    yn->a. You want to show f(yn)->f(a). Define xn=yn-a. Then xn->0. If xn->0 then f(xn)->f(0) since f is continuous at 0. What is f(0)? f(yn)=f(a+xn)=f(a)+f(xn). Can you finish this up and feel 'complete'?
     
  16. Dec 4, 2008 #15

    HallsofIvy

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    I would not use sequences. Look at [itex]\lim_{x\rightarrow a} f(x)[/itex] where a is any real number. Let h= x- a. Then x= a+ h so f(x)= f(a+ h)= f(a)+ f(h). The limit becomes [itex]\lim_{h\rightarrow 0} f(a)+ f(h)= f(a)+ \lim_{h\rightarrow 0} f(h)[/itex].
    Knowing that f is continuous at 0, what is that limit?
     
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