2

Find all functions $f : \mathbb{R}^+ \to \mathbb{R}^+$ satisfying $$f(x)+f(\frac{x}{2})= \frac{x}{2}$$ $\forall x \in \mathbb{R}^+$.

My Attempt :

$-\frac{x}{3} + f(x) = \frac{x}{6} - f(\frac{x}{2})$

Let $g(x) = f(x) - \frac{x}{3}$

so $g(x)=-g(\frac{x}{2})\;$ $\forall x \in \mathbb{R}^+$

then $g(x)=g(\frac{x}{4})$

Please suggest how to proceed.

user403160
  • 3,286
  • 2
    If we assume that $f$ is continuous at $0$, by plugging in $x = 0$ we obtain $f(0) = 0$. Then also $g(0) = 0$. For any $x > 0$ we have:

    $$g(x) = g\left(\frac{x}{4}\right) = g\left(\frac{x}{16}\right) = \ldots = g\left(\frac{x}{4^n}\right) \xrightarrow{n\to\infty} g(0) = 0$$

    So, $g \equiv 0$, meaning $f(x) = \frac{x}3, \forall x > 0$. But, $f$ is not even defined at $0$, let alone continuous.

     – mechanodroid Nov 12 '17 at 15:11
  • Why $f$ is not defined at $0$ ? – user403160 Nov 12 '17 at 15:35
  • Well usually $R^+$ means positive numbers, thus 0 is excluded. – nonuser Nov 12 '17 at 15:50
  • @mechanodroid, "If we assume that $f$ is continuous at $0$, by plugging in $x=0$ we obtain $f(0)=0$. Then also $g(0)=0$." Since $0$ is not in the domain, why can we get $f(0)=0$ and $g(0)=0$ ? – user403160 Nov 13 '17 at 15:30
  • We can't, not in this case. I was just pointing out the solution of the same functional equation where $f : [0, +\infty\rangle \to [0, +\infty\rangle$ and $f$ is assumed to be continuous at $0$. – mechanodroid Nov 13 '17 at 15:33
  • @mechanodroid, So we can't conclude that $g(0)=0$ and $f(x) = \frac{x}3, \forall x > 0$, right ? – user403160 Nov 13 '17 at 15:47
  • @carat No. I was talking about a different problem, basically. – mechanodroid Nov 13 '17 at 15:48
  • @mechanodroid, could you please solve my problem using knowledge at high school level ? – user403160 Nov 13 '17 at 15:56
  • You have a solution already. It's not a unique function. I have no idea how to solve it more elementary. – mechanodroid Nov 13 '17 at 15:58
  • @mechanodroid. Thank you. – user403160 Nov 13 '17 at 16:17

1 Answers1

4

Let $x=2^z$. The equation is

$$f(2^z)+f(2^{z-1})=2^{z-1},$$ or

$$g(z)+g(z-1)=2^{z-1},$$ which is an ordinary linear recurrence.

The solution of the homogeneous equation $g(z)+g(z-1)=0$ is, by induction,

$$g(z)=(-1)^{\lfloor z\rfloor} g(\{z\})$$ where $\{z\}$ denotes the fractional part.

A particular solution of the non-homogeneous equation is

$$\frac23 e^{z-1}$$ (by undeterminate coefficients on $g(z)=ce^{z-1}$).

Finally,

$$f(x)=(-1)^{\lfloor\log_2x\rfloor}g(\{\log_2x\})+\frac x3$$ for some function $g$ defined on $[0,1)$.

If you want $f$ continuous, then $g$ must be continuous and $f(1^-)=f(1^+)$, or

$$(-1)^{-1}g(1^-)=(-1)^0g(0^+).$$