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Math Help - Prove cts function with midpoint property is linear

  1. #1
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    Prove cts function with midpoint property is linear

    Hi,
    I would like to show that a continuous map a,b)\to \mathbb{R}" alt="fa,b)\to \mathbb{R}" /> which has the property that f\left(\frac{x+y}{2}\right) = \frac{f(x)+f(y)}{2} for x,y\in (a,b) satisfies
    f(x) = Ax + B for some constants A, B.

    I can prove it on the closed interval [0,1]: in this case it suffices to show f(x)=Ax with f(0)=0. I prove it holds on the dyadic rationals of the form \frac{m}{2^k} by an induction argument, and extending to the closed interval by continuity. BUT my argument needs the endpoints of the interval: I am progressively subdividing; f(\frac{1}{2}) = \frac{f(0)+f(1)}{2} = \frac{f(1)}{2}, etc, and I cannot see how to prove it on an open interval.

    Many thanks
    Last edited by jacobi; October 10th 2010 at 04:37 AM.
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  2. #2
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    Quote Originally Posted by jacobi View Post
    Hi,
    I would like to show that a continuous map a,b)\to \mathbb{R}" alt="fa,b)\to \mathbb{R}" /> which has the property that f\left(\frac{x+y}{2}\right) = \frac{f(x)+f(y)}{2} for x,y\in (a,b) satisfies
    f(x) = Ax + B for some constants A, B.

    I can prove it on the closed interval [0,1]: in this case it suffices to show f(x)=Ax with f(0)=0. I prove it holds on the dyadic rationals of the form \frac{m}{2^k} by an induction argument, and extending to the closed interval by continuity. BUT my argument needs the endpoints of the interval: I am progressively subdividing; f(\frac{1}{2}) = \frac{f(0)+f(1)}{2} = \frac{f(1)}{2}, etc, and I cannot see how to prove it on an open interval.

    Many thanks

    So, in fact, you can prove it in any interval of the form [\alpha,\,\beta]\subset (0,1) , right? Well, you can choose say \alpha=\frac{1}{n}\,,\,\beta=1-\frac{1}{n} and let n\to\infty ...

    Tonio

    Pd. Where does continuity come into play in the above?
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  3. #3
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    Oh, of course, I see now: given x, y in (0,1) they'll lie in some closed interval. Duh.

    Thanks, Tonio!
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  4. #4
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    Oh, of course, I see now: given x, y in (0,1) they'll lie in the closed subinterval [x,y]. Duh.

    Thanks, Tonio!

    Pd. Where does continuity come into play in the above?
    If you're referring to my work, I used continuity to extend the dyadic rationals to the interval. If you're referring to your post, you're using that a_n\to a implies f(a_n)\to f(a).
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  5. #5
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    Actually I am mistaken. For each \alpha_n = \frac{1}{n}, \beta_n = 1-\frac{1}{n} we get different constants A_n, B_n. How can we guarantee that these converge?
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  6. #6
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    Quote Originally Posted by jacobi View Post
    Actually I am mistaken. For each \alpha_n = \frac{1}{n}, \beta_n = 1-\frac{1}{n} we get different constants A_n, B_n. How can we guarantee that these converge?
    hehe...it took you a while . Never mind, we all passed through these messes. So you have different constants...or do you?

    Take n, n+1 and their respective constants...and NOW use continuity!

    Tonio
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