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 February 29th, 2012, 09:53 AM #1 Newbie   Joined: Nov 2009 Posts: 16 Thanks: 0 Functional equation Here is my problem: There is given the functional equation: $f(2x)+f(1/2)=f(x)+f(x+1/2)$ for $x \in [0,1/2]$. We also know that $f(0)=-1$ and $f(1)=1$. Additionally, we assume the continuity and strict monotonicity of $f$. Is it possible to get any information on $f(1/2)$? In particular, can we prove that $f(1/2)=0$? Thank you for hints and help.
 February 29th, 2012, 10:38 AM #2 Senior Member   Joined: Feb 2012 Posts: 628 Thanks: 1 Re: Functional equation Show that the only function satisfying that functional equation and the given information about the function is $f(x)= 2x - 1$, thus proving that $f(1/2)= 0$.
 February 29th, 2012, 11:50 AM #3 Newbie   Joined: Nov 2009 Posts: 16 Thanks: 0 Re: Functional equation OK. However, I would be grateful for any tips how to prove it. The equation which I wrote is a derivative of the Hosszu functional equation (solution of this equation under some assumption is a linear function). But Hosszu equation involves two variables, not one. How to handle with this?
 February 29th, 2012, 12:04 PM #4 Senior Member   Joined: Feb 2012 Posts: 628 Thanks: 1 Re: Functional equation I am not familiar with the Hosszu functional equation. What is the equation, and under what assumption is the equation a linear function? Because essentially what you have to prove is that the functional equation you gave ($f(2x) + f(1/2)= f(x) + f(x + 1/2)$) must represent a linear function.
 February 29th, 2012, 12:57 PM #5 Newbie   Joined: Nov 2009 Posts: 16 Thanks: 0 Re: Functional equation The Hosszu functional eq. is $f(x)+f(y)=f(x+y-xy)+f(xy)$. If $f$ is continuous and satisfies this eq., then f is linear. Do you maybe have any idea how to solve an original problem? Best regards.
 February 29th, 2012, 01:20 PM #6 Senior Member   Joined: Feb 2012 Posts: 628 Thanks: 1 Re: Functional equation One more question about the Hosszu functional equation: does the function have to satisfy that equation for all x and all y in order to be linear?
 February 29th, 2012, 11:37 PM #7 Newbie   Joined: Nov 2009 Posts: 16 Thanks: 0 Re: Functional equation If I knew this, I wouldn't ask. So far, I haven't found any results on Hosszu equation on restricted domain, so in my opinion the equation must be satisfied for all x and y and then the function is linear.
 March 1st, 2012, 10:01 AM #8 Senior Member   Joined: Feb 2012 Posts: 628 Thanks: 1 Re: Functional equation You can use the functional equation $f(2x) + f(1/2)= f(x) + f(x + 1/2)$ recursively as follows: Set x = 1/4. Then you get $f(1/2) + f(1/2)= f(1/4) + f(3/4)$, or (1) $2f(1/2)= f(1/4) + f(3/4)$. Then set x = 1/8 to yield $f(1/4) + f(1/2)= f(1/ + f(5/" /> and set x = 3/8 to yield $f(3/4) + f(1/2)= f(3/ + f(7/" />. Adding these two equations yields $f(1/4) + f(3/4) + 2f(1/2)= f(1/ + f(3/ + f(5/ + f(7/" />, and substituting from (1) yields $4f(1/2)= f(1/ + f(3/ + f(5/ + f(7/" />. By using this process recursively, you can show that $f(1/2)= \frac{1}{2^{n - 1}} \sum_{k = 1}^{2^{n - 1}} f(\frac{2k - 1}{2^n})$ and thus $f(1/2)= \lim_{n \rightarrow \infty} \frac{1}{2^{n - 1}} \sum_{k = 1}^{2^{n - 1}} f(\frac{2k - 1}{2^n})$ = $\int_0^1 f(x) dx$. Now, using the functional equation again, we have $f(2x) - f(x + 1/2)= f(x) - f(1/2)$ and thus $\int_0^1 f(2x) - f(x + 1/2) dx= \int_0^1 f(x) - f(1/2) dx = \int_0^1 f(x) dx - \int_0^1 f(1/2) dx = f(1/2) - f(1/2) = 0$. Hence $\int_0^1 f(2x) dx= \int_0^1 f(x + 1/2) dx$. I'm not sure if you can do anything with that, but that's my attempt at doing something useful with this problem.

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