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Topic: Ask-an-Analysis problem
Replies: 10   Last Post: Apr 25, 2013 1:01 PM

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David C. Ullrich

Posts: 3,033
Registered: 12/13/04
Re: Ask-an-Analysis problem
Posted: Apr 24, 2013 10:38 AM
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On Wed, 24 Apr 2013 02:19:43 -0700, William Elliot <marsh@panix.com>
wrote:

>Assume for f:[0,1] -> R that there's some c /= 0,1 with
>for all x in [0,1/2], f(x) = c.f(2x).
>
>Show there's some k with for all x in [0,1], f(x) = kx.



Not true. It's well known that there exists a nowhere-
continuous function f : R -> R such that
f(x+y) = f(x) + f(y) for all x, y.

Maybe you omitted a hypothesis? Seems like it may be
true for continuous f...







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