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Replies: 10   Last Post: Apr 25, 2013 1:01 PM

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 dan.ms.chaos@gmail.com Posts: 409 Registered: 3/1/08
Posted: Apr 24, 2013 12:54 PM

On Apr 24, 5:38 pm, dullr...@sprynet.com wrote:
> On Wed, 24 Apr 2013 02:19:43 -0700, William Elliot <ma...@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...

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

Date Subject Author
4/24/13 William Elliot
4/24/13 David C. Ullrich
4/24/13 dan.ms.chaos@gmail.com
4/24/13 Herman Rubin
4/24/13 David C. Ullrich
4/25/13 David Petry
4/25/13 Herman Rubin
4/25/13 dan.ms.chaos@gmail.com
4/24/13 David C. Ullrich
4/24/13 William Elliot
4/25/13 David C. Ullrich