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Continuous function proof

  1. May 13, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose f is continuous, f(1) = 5, and f(nx) = [f(x)]^n where n is any integer and x is any real number. Prove that f(x) = 5^x for all real x.


    2. Relevant equations



    3. The attempt at a solution

    I've proved that f(x) = 5^x for rational x. Now I have to extend this to irrational x as well. Let x_n be a sequence of rational numbers converging to an irrational number z. By the continuity of f we have [tex] f(z) = \lim_{x_n \rightarrow z} f(x_n) = f(\lim_{x_n \rightarrow z} x_n) \Rightarrow f(z) = \lim_{x_n \rightarrow z} 5^{x_n} = 5^{\lim_{x_n \rightarrow z}} = 5^z [/tex].

    My question is, how can I rewrite this proof to make use of the epsilon-delta definition of limit. Here is my attempt for that:

    Since x_n converges to z and f is continuous, then [tex] |f(x_n) - f(z)| = |5^{x_n} - f(z)| < \epsilon [/tex] whenever [tex] 0 < |x_n - z| < \delta [/tex].

    If you take f(z) = 5^z then |5^z||5^(x_n - z) - 1| < epsilon. 5^(x_n - z) has a limit of 1 and so the |5^(x_n - z) - 1| goes to 0, and so the quantity on the left hand side of the inequality can be made as small as we please, and so f(x) = 5^x for irrational x too.
     
  2. jcsd
  3. May 13, 2009 #2

    jbunniii

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    You know that

    [tex]g(x) = 5^x[/tex] for all real x

    is a continuous function that agrees with f for every rational x.

    Suppose you have another continuous function h(x) that also satisfies

    [tex]h(x) = 5^x[/tex] for all rational x.

    Then

    [tex]d(x) = f(x) - g(x)[/tex] is a continuous function that is equal to zero for every rational x. If you can show that this implies that d(x) has to be zero for all real x, then you're done. Try assuming that [tex]d(x) \neq 0[/tex] for some real x, and look for a contradiction.
     
  4. May 13, 2009 #3
    Suppose there exists an irrational number, z, such that d(z) > 0. Since d is continuous, the following inequality should hold for rational x: |d(x) - d(z)| = |d(z)| < epsilon. Since d(z) > 0, then |d(z)| > 0. d is continuous so the inequality should hold for all positive epsilon, including epsilon < |d(z)| since d(z) is a fixed quantity. This results in a contradiction. The proof for d(z) < 0 is the same.

    And so d(x) = f(x) - g(x) = 0 for all real x.
     
  5. May 13, 2009 #4

    jbunniii

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    I didn't quite follow your proof; I don't think it's quite right. See if you can fill in the details of the following sketch:

    Suppose [tex]d(z) > 0[/tex] for some irrational z. Since d is continuous, this implies that [tex]d(x) > 0[/tex] for all x in some interval containing z. But any interval contains rational numbers, and [tex]d(x) = 0[/tex] for all rational x; thus, we have a contradiction.
     
  6. May 13, 2009 #5
    I don't see what I can add to your sketch. It seems like a perfectly good proof.

    I will try to re-word my proof.

    Suppose there is a z that is irrational such that d(z) > 0. d is continuous, so we have
    |d(x) - d(z)| < epsilon for all x sufficiently close to z. Now, we can always find rational x values close enough to z so that the inequality |d(x) - d(z)| < epsilon holds for all positive epsilon, and for rational x values in a neighborhood of z. Since d(x) = 0 for rational x, the inequality is now 0 < |d(z)| < epsilon, but this inequality has to hold for all epsilon, including epsilon < |d(z)| and so this is a contradiction.

    The same proof can be used for d(z) < 0. So d(x) = 0 for all x values.
     
  7. May 13, 2009 #6

    jbunniii

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    The order in which you are saying things still seems a bit confusing. For example, you claim that

    [tex]|d(x) - d(z)| < \epsilon[/tex]

    without first defining what [tex]\epsilon[/tex] is.

    The main assertion from my sketch that needs to be justified is

    "Suppose [tex]d(z) > 0[/tex] for some irrational z. Since d is continuous, this implies that [tex]d(x) > 0[/tex] for all x in some interval containing z."

    So how about if I suggest how to start that justification:

    "Continuity of d at the point z means that, given [tex]\epsilon > 0[/tex], there exists [tex]\delta > 0[/tex] such that for all [tex]z - \delta < x < z + \delta[/tex], we have [tex]|d(x) - d(z)| < \epsilon[/tex]..."

    Now, what happens if you choose [tex]\epsilon = d(z)/2[/tex]?
     
  8. May 14, 2009 #7
    Well, assuming d(z) > 0, we know from the continuity of d that for every positive epsilon there exists a positive delta such that d(z) - epsilon < d(x) < d(z) + epsilon whenever 0 < |x-z| < delta. Taking epsilon = d(z)/2, we have d(z) - d(z)/2 = d(z)/2 < d(x).

    d(z)/2 is positive and therefore d(x) is positive in some neighborhood of z.


    Going back to the proof I was working on, I was just being lazy by not writing "for every epsilon..."

    Here is another version of it:

    Suppose x_n is a sequence of rational numbers converging to an irrational number z. Then by the continuity of d, for every positive epsilon there exists a positive delta such that |f(x_n) - f(z)| < epsilon whenever 0 < |x_n - z| < delta. Now, f(x_n) = 0 for all x_n since x_n is a sequence of rational numbers, and so we have |f(z)| < epsilon.

    Since d is continuous, we can choose epsilon of any size (provided x_n is close enough to z), say we choose epsilon < |f(z)|, then this results in a contradiction.
     
  9. May 14, 2009 #8

    jbunniii

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    I like the first paragraph. It allows you to conclude that |f(z)| < epsilon, for every epsilon > 0.

    From there, you don't even need to argue about a contradiction, nor do you need to mention anything about the rational sequence again.

    Just the fact that |f(z)| < epsilon for every epsilon > 0 implies immediately that f(z) = 0.
     
  10. May 14, 2009 #9
    :rofl: Very true, don't know how I didn't see that.

    Thanks for helping!
     
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