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Continuous Function

  1. Oct 29, 2005 #1
    [SOLVED] Continuous Function

    "Let [tex]f:\mathbb{R}\rightarrow\mathbb{R}[/tex] such that [tex]\forall x,y\in\mathbb{R}[/tex],[tex]f(x+y)=f(x)+f(y)[/tex].

    Prove that if [tex]f[/tex] is continuous at 0, then it is continuous at all points."

    Assuming [tex]f[/tex] to be continuous at 0, then [tex]\forall\epsilon >0, \exists\delta >0[/tex] such that [tex] |f(x+y)-f(0)|<\epsilon[/tex] whenever [tex]|x+y|<\delta[/tex]. [tex]f(0)[/tex] is simply [tex]f(x-y)[/tex]. We can rewrite this as,

    [tex] |f(x)+f(y)-f(x-y)|<\epsilon[/tex] whenever [tex]|x+y|<\delta[/tex].

    However I can't conclude that it is continuous on all points. In fact, this seems to be a 2 variable function, i.e., RXR->R, and it is not something we have covered yet.
     
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  3. Oct 29, 2005 #2

    Hurkyl

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    Why do you think f(0) = f(x-y)?
    What does it mean for f to be continuous at a point a?
     
  4. Oct 29, 2005 #3
    f is continuous at a if the limit exists and is equal to f(a). f(0) = f(x-y) if x = y. How else would x+y=0?
     
  5. Oct 29, 2005 #4
    f is not a function of two variables. It satisfies a condition which involves two variables. (If x and y are real numbers, then x + y is just /one/ real number right?).
     
  6. Oct 29, 2005 #5

    Hurkyl

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    Well, you've written everything else out in epsilon-deltas, why not this?

    Why should x+y=0?
     
  7. Oct 29, 2005 #6
    I thought I had written it out already.

    What is f(0) then?
     
  8. Oct 29, 2005 #7

    Hurkyl

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    You can figure out f(0), but I just realized an important mistake: you have the continuity of f at 0 wrong!

    It said right at the beginning that f is a function of one variable, and everywhere in your post that you use f, you write it as a function of one variable, so I'm not sure why you think it's a function of two variables.

    (And, incidentally, what you've written doesn't resemble the definition of the continuity of a two-variable function)

    So let's work on getting that bit right first... would you care to try it again? what does it mean for any function f of one real variable to be continuous at zero?

    (If you want, it might also help for you to try to elaborate why you think it's a function of two variables, so we can quickly dispel that confusion... at least if you don't work it out yourself before you hit "submit"!)
     
  9. Oct 29, 2005 #8
    f:I->R is continuous at 0 if for every e>0, there exists d>0 such that |f(x)-f(0)|<e for all x in I satisfying |x|<d.

    So if f(x+y) = f(x)+f(y), then f is continuous if |f(x)+f(y)-f(0)|<e whenever |x+y|<d, or so I think. This is, of course, just me plugging things blindly into the definition.

    So, maybe I could arrange the equation in such a way that f(x+y)-f(y)=f(x). In this case, f(x) is continuous at 0 if |f(x+y)-f(y)-f(0)|<e whenever |x|<d. This seems to make a little more sense...
     
  10. Oct 29, 2005 #9

    Hurkyl

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    This is, indeed, almost exactly what it means for f to be continuous at 0. (You meant to say for all x satisfying 0 < |x| < d)

    So the next question is what are you trying to plug into what, and why? My best guess is that you are mechanically trying to plug in an expression that defines f evaluated at a point, and that you have become so fixated on performing this step that you are randomly trying to accomplish it, despite the fact you have not been given (and cannot possibly determine) an expression defining f.

    My advice is to step back and think about what you're doing. Stop trying to blindly follow an algorithm (which won't work here), and try and figure out what needs to be done. I notice that you still haven't written out the goal you're trying to reach... (at least you haven't in the forum)
     
  11. Oct 29, 2005 #10
    I must prove that continuity at 0 implies continuity on all of R, which is a consequence of how the function is defined. However, I don't fully understand the function, other than it looks like what an isomorphism between rings must satisfy... which is odd because this is an analysis assignment.
     
  12. Oct 29, 2005 #11

    Hurkyl

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    You also seem to have thought that writing things in terms of epsilon-deltas will help with this problem... so why won't you write the thing you're trying to prove that way?

    You can do analysis on rings. In fact, you're doing it right now, since [itex]\mathbb{R}[/itex] is a ring! You've observed that the given condition resembles the idea of a ring isomorphism... maybe you can draw upon what you know about them to help you with this problem. (Then again, maybe it will lead you astray!)

    I'm not yet sure if this has sunk in yet, but you haven't been told exactly what f is. You've only been told it satisfies some conditions: there could very well be lots of functions that satisfy those conditions. You're going to have to be happy with those conditions and figure out what they imply... you will not be able to determine exactly what f is.

    (Maybe you can find one of these functions, and it might help you figure out how to prove this theorem for all of them!)
     
    Last edited: Oct 29, 2005
  13. Oct 29, 2005 #12

    Physics Monkey

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    Icebreaker, I don't see what is so mysterious about this function. You could guess what it was right now, couldn't you? (More properly, you could a guess a one parameter family of functions.) After you prove the global continuity from the assumed continuity at zero, you can even prove that the function is exactly what you would expect. Why don't you follow Hurkyl's advice and write out the continuity condition in terms of epsilon/delta or maybe try to find an example to give yourself some intuition.
     
  14. Oct 30, 2005 #13
    If f is continuous at 0, then for every e>0, there exists d>0 such that |f(x)-f(0)|<e whenever |x|<d. However, f(x)=f(x+0)=f(x)+f(0) by hypothesis, therefore continuity at 0 can be restated as |f(x)|<e whenever |x|<d. This implies that f(0) = 0.

    I'm missing the punchline. But up to this point, is it correct?
     
  15. Oct 30, 2005 #14

    HallsofIvy

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    Are you really required to write the proof in [itex]\epsilon, \delta[/itex] form?

    You should be able to see easily what f(0) is- that's important (use the fact that f(x+ 0)= f(x)+ f(0) for all x).

    Let a be any real number. If you let h= x- a, then
    [tex]lim_{x\rightarrow a} f(x)= lim_{h\rightarrow 0}f(a+ h)[/tex]
    Now how can you rewrite the right side so you can use the fact that f is continuous at 0?
     
  16. Oct 30, 2005 #15

    Hurkyl

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    So far, you are correct.

    Incidentally, f(x)=f(x+0)=f(x)+f(0) is enough to imply f(0) = 0.



    HoI: the key step either way is the same... I was hoping we could do it his way, since he thought of it! Also, I was trying not to tell him the key step, and it might have leapt out at him once he wrote out what it means for f to be continuous at all points in the epsilon-delta form.

    Sometimes, it's just better to do something the way you first see it. For instance, for most problems where you're "supposed" to use Zorn's lemma, I instead use the well-ordering principle, because that approach comes to me quicker. It would generally take me longer to work out a Zorn's lemma approach, except for the problems where the setup is practically given to you.
     
    Last edited: Oct 30, 2005
  17. Oct 30, 2005 #16
    Yes, unfortunately I must write it out in d,e form. Thanks for the help!
     
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