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Math Help - Additive Functions!

  1. #1
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    Additive Functions!

    By definition, a function is additive if f(x+y) = f(x) + f(y) for all x, y \in \mathbb{R}.
    How would you show that if f is continuous at some point x_0, then it is continuous at every point of \mathbb{R}?
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  2. #2
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    Quote Originally Posted by frenchguy87 View Post
    By definition, a function is additive if f(x+y) = f(x) + f(y) for all x, y \in \mathbb{R}.
    How would you show that if f is continuous at some point x_0, then it is continuous at every point of \mathbb{R}?
    given a \in \mathbb{R}, take the limit of both sieds of the identity f(x)+f(x_0)=f(x-a+x_0)+f(a) as x \to a.
    Last edited by NonCommAlg; February 27th 2010 at 05:41 PM.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by frenchguy87 View Post
    By definition, a function is additive if f(x+y) = f(x) + f(y) for all x, y \in \mathbb{R}.
    How would you show that if f is continuous at some point x_0, then it is continuous at every point of \mathbb{R}?
    A little more formal.

    Notice two things first, namely f(0)=f(0+0)=f(0)+f(0)\implies f(0)=0 and so 0=f(0)=f(x-x)=f(x)+f(-x)\implies f(-x)=-f(x). Thus,

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

    A little less "limity". Suppose that there existed some x_0\in\mathbb{R} such that f is continuous at that point. Let x_1\in\mathbb{R} be arbitrary. Given any \varepsilon>0 we have that

    \left|f(x_1)-f(x)\right|=\left|f\left(x_1-x\right)\right|=\left|-f(x_1-x)\right|=\left|f\left(-(x_1-x)\right)\right| =\left|f\left(x_0-(x_1-x)\right)-f(x_0)\right|. Notice that since f is continuous at x_0 we know that for the previously mentioned \varepsilon there exists

    some \delta>0 such that \left|x-x_0\right|<\delta\implies |f(x)-f(x_0)|<\varepsilon. So, let z=x_0-(x_1-x), then notice that choosing \left|x_1-x\right|<\delta\implies \left|z-x_0\right|<\delta\implies \left|f(z)-f(x_0)\right|=\left|f(x_1)-f(x)\right|<\varepsilon.

    The conclusion follows. The point of all of these is to shift the point of continuity to an arbitrary point.

    Note that from this it readily follows that the only function f:\mathbb{R}\mapsto\mathbb{R} which is continuous at a single point and f(x+y)=f(x)+f(y) is the function f(1)x.
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    First you need to prove that f(0)= 0. That is true because f(a+0)= f(a)+ f(0)= f(a).

    If f is continuous at any x= b, then it is continuous at x= 0.

    That is true because \lim_{x\to b}f(x)= \lim_{y\to 0}f(b+ y) where y= x- b. Then \lim_{y\to 0}f(b+ y)= f(b)+ \lim_{y\to 0} f(y). Since f is continuous at x= b, \lim_{x\to b} f(x)= f(b) so we have \lim_{x\to b}f(x)= f(b)= f(b)+ \lim_{y\to 0}f(y) and so [tex]\lim_{y\to 0}f(y)= 0= f(0) and f is continuous at 0.

    If f is continuous at x= 0 then it is continuous at every point, a.

    Almost the same proof: [tex]lim_{x\to a} f(x)= \lim_{y\to 0}f(a+ y) where y= x- a. But f(a+y)= f(a)+ f(y) so \lim_{y\to 0}f(a+y)= f(a)+ \lim_{y\to 0}f(y) and since f is continous at 0, \lim_{y\to 0}f(x)= f(0)= 0. That is, if f is continous at x= 0, then \lim_{x\to a} f(x)= f(a) so f is continuous at x= a.
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    f(0)=f(0+0)=f(0)+f(0)\implies f(0)=0
    I don't get this implication, how do you conclude that f(0)=0?
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  6. #6
    Moo
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    You have f(0)=f(0)+f(0)=2f(0)

    So 2f(0)-f(0)=0 and finally f(0)=0...
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