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Old February 3rd, 2009, 03:42 AM
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Default Continuity Question

4. Let f : R -> R be a function satisfying
f(x + y) = f(x) + f(y) for any x, y in R.
Using elementary algebra prove that f(0) = 0 and f(−x) = −f(x) for every x;
by induction prove that f(nx) = nf(x) for every x and every natural number n;
and then show that f(rx) = rf(x) for every x and any rational number r = n/m.

These parts i have done. I'm not sure how to go about doing the next bit:

*Prove that if f is also continuous then, for some constant c,
f(x) = cx for every x in R.*

thanks.
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Old February 3rd, 2009, 08:53 AM
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Quote:
Originally Posted by DeFacto View Post
4. Let f : R -> R be a function satisfying
f(x + y) = f(x) + f(y) for any x, y in R.
Using elementary algebra prove that f(0) = 0 and f(−x) = −f(x) for every x;
by induction prove that f(nx) = nf(x) for every x and every natural number n;
and then show that f(rx) = rf(x) for every x and any rational number r = n/m.

These parts i have done. I'm not sure how to go about doing the next bit:

*Prove that if f is also continuous then, for some constant c,
f(x) = cx for every x in R.*

thanks.
You necessarily have c=f(1), so let c=f(1). Check that f(r)=c r for every r\in\mathbb{Q} (it should be almost obvious), and then for any real x use a sequence of rationals (r_n)_n that converges toward x to conclude.
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Old February 10th, 2009, 05:17 AM
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Note, by the way, that there exist non-continuous functions f(x) such that f(x+y)= f(x)+ f(y) but they are VERY complicated. Not only are they discontinuous at every point, they are unbounded, both above and below, on any interval, no matter how small! For such an f, given any point in the plane, there exist a point on the graph of y= f(x) arbitrarily close to it.
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Old February 10th, 2009, 09:56 AM
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See this.
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