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Math Help - Solving a functional equation

  1. #1
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    Solving a functional equation

    let \ f:R \to R  \ be \ continuous \ with \ f(x+y)= f(x)+f(y) \ , \ \forall x,y \in R.
    let \ a=f(1) . prove \  that  \ f (x)= ax , \forall x \in R
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by flower3 View Post
    let \ f:R \to R  \ be \ continuous \ with \ f(x+y)= f(x)+f(y) \ , \ \forall x,y \in R.
    let \ a=f(1) . prove \  that  \ f (x)= ax , \forall x \in R
    What have you tried? What do you know? These are all important questions. There are ways to do this quicker than others, if you have the right knowledge.
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  3. #3
    Senior Member Dinkydoe's Avatar
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    Ehm, I looked at some of his earlier posts and it seems he usually only
    drops a question and doesn't really seem to care about the answer. He doesn't really want to be taken serious I guess.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Dinkydoe View Post
    Ehm, I looked at some of his earlier posts and it seems he usually only
    drops a question and doesn't really seem to care about the answer. He doesn't really want to be taken seriously I guess.
    Yeah, yeah I know. You'll notice that I answered a lot of them.
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    Quote Originally Posted by Drexel28 View Post
    Yeah, yeah I know. You'll notice that I answered a lot of them.
    Hey. So should I answer this or not? What do you say?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by bandedkrait View Post
    Hey. So should I answer this or not? What do you say?
    Goo ahead
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     f(x)=f(x. 1)=xf(1)=xa=ax for all x\in\mathbb R.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by flower3 View Post


     f(x)=f(x. 1)=xf(1)=xa=ax for all x\in\mathbb R.
    Try for every x\in\mathbb{N} or something similar. It follows by induction that f(n)=f\left(\underbrace{1+\cdots+1}_{n\text{ times}}\right)=nf(1) but to say it for \mathbb{R} is not so simple.
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  9. #9
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    Okay, I'm solving this purely for personal gratification :-/


    f(x+y) = f(x) + f(y)

    Let y=0
    so f(x)=f(x)+f(0)

    i.e. f(0)=0

    Now, differentiate both sides partially w.r.t x,

    f'(x+y) = f'(x)................(1)
    Again differentiate partially w.r.t y,
    f'(x+y) = f'(y)................(2)

    from (1) and (2)

    f'(x)=f'(y)

    since, x and y are arbitrary, f'(x)=k for all x, where k is a constant (assuming total continuity and differentiability for both f(x) and f'(x)

    since f'(x) is a constant,

    f(x) = \int(kdx)

    =kx+c

    since f(0)=0 c=0 , and since f(1)=a, k=a

    Thus f(x)=ax for all x
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  10. #10
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by bandedkrait View Post
    Okay, I'm solving this purely for personal gratification :-/


    f(x+y) = f(x) + f(y)

    Let y=0
    so f(x)=f(x)+f(0)

    i.e. f(0)=0

    Now, differentiate both sides partially w.r.t x,

    f'(x+y) = f'(x)................(1)
    Again differentiate partially w.r.t y,
    f'(x+y) = f'(y)................(2)

    from (1) and (2)

    f'(x)=f'(y)

    since, x and y are arbitrary, f'(x)=k for all x, where k is a constant (assuming total continuity and differentiability for both f(x) and f'(x)

    since f'(x) is a constant,

    f(x) = \int(kdx)

    =kx+c

    since f(0)=0 c=0 , and since f(1)=a, k=a

    Thus f(x)=ax for all x
    You of course know this doesn't work because you assume differentibility. In fact, we only need that it is differentiable at zero to proceed like you did since \lim_{\Delta x\to 0}\frac{f\left(x+\Delta x\right)-f\left(x\right)}{\Delta x}=\lim_{\Delta x\to 0}\frac{f\left(\Delta x\right)}{\Delta x}=f'(0) since f(0)=f(0+0)=2f(0)\implies f(0)=0. But, this isn't now you do the problem.
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  11. #11
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    Quote Originally Posted by bandedkrait View Post
    Okay, I'm solving this purely for personal gratification :-/


    f(x+y) = f(x) + f(y)

    Let y=0
    so f(x)=f(x)+f(0)

    i.e. f(0)=0

    Now, differentiate both sides partially w.r.t x,

    f'(x+y) = f'(x)................(1)
    Again differentiate partially w.r.t y,
    f'(x+y) = f'(y)................(2)

    from (1) and (2)

    f'(x)=f'(y)

    since, x and y are arbitrary, f'(x)=k for all x, where k is a constant (assuming total continuity and differentiability for both f(x) and f'(x)

    since f'(x) is a constant,

    f(x) = \int(kdx)

    =kx+c

    since f(0)=0 c=0 , and since f(1)=a, k=a

    Thus f(x)=ax for all x
    A question: Aren't you assuming function is differentiable?
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  12. #12
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    Quote Originally Posted by aman_cc View Post
    A question: Aren't you assuming function is differentiable?

    Yes. But I don't see how the relation can be proved otherwise. However, it's more appropriate to use the definition of derivative itself, like drexel28 did.
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  13. #13
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    Quote Originally Posted by bandedkrait View Post
    Yes. But I don't see how the relation can be proved otherwise. However, it's more appropriate to use the definition of derivative itself, like drexel28 did.
    Yes - I saw Drexel28's post. I trust the proof will involve some form of mean value theorem for continuous function. Will be happy if shown that it can be done in a different way as well.
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  14. #14
    Senior Member Dinkydoe's Avatar
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    I think the best approach is to show for all rational numbers:

    Since f(1) =f(\frac{p}{p}) = p\cdot f(\frac{1}{p}) = a \Rightarrow f(\frac{1}{p}) = \frac{a}{p}.

    It simply follows now for all rational numbers f(\frac{p}{q}) = p\cdot f(\frac{1}{q}) = a\cdot \frac{p}{q}

    Now with an argument using continuity and the fact that every real number can be written as a infinite sum of rational numbers, it can be shown for all x\in \mathbb{R}
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  15. #15
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Dinkydoe View Post
    I think the best approach is to show for all rational numbers:

    Since f(1) =f(\frac{p}{p}) = p\cdot f(\frac{1}{p}) = a \Rightarrow f(\frac{1}{p}) = \frac{a}{p}.

    It simply follows now for all rational numbers f(\frac{p}{q}) = p\cdot f(\frac{1}{q}) = a\cdot \frac{p}{q}

    Now with an argument using continuity and the fact that every real number can be written as a infinite sum of rational numbers, it can be shown for all x\in \mathbb{R}
    Exactly. This is how you do it. Let x\in\mathbb{R} be arbitrary. Since \mathbb{Q} is dense in \mathbb{R} there exists some sequence \left\{q_n\right\}_{n\in\mathbb{N}} such that q_n\to x. And since f is continuous we know that x_n\to x\implies f(x_n)\to f(x). So, we have that f(x)=\lim\text{ }f(q_n)=\lim\text{ }\left\{a\cdot q_n\right\}=a\lim\text{ }q_n=ax.
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