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

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Posts: 409
Registered: 3/1/08
Re: Ask-an-Analysis problem
Posted: Apr 24, 2013 12:54 PM
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On Apr 24, 5:38 pm, wrote:
> On Wed, 24 Apr 2013 02:19:43 -0700, William Elliot <>
> 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...

Maybe if you restrict it such that f:[0,1] -> [0,1]?
Just a hunch,not sure.

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