I need to find all continuous functions satisfying:

$$3f(2x+1)=f(x) + 5x$$

The functional equation looks simple but I am unable to solve it. I tried to convert it into a Cauchy type equation but I wasn't able to do so.

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    $\begingroup$ Is there any restriction given for $x$? It is rather simple to determine the value of $f(-1)$ but I am not sure whether this is helpful or not. $\endgroup$ – mrtaurho Dec 15 '18 at 10:53
  • $\begingroup$ If the domain of $f$ is all real numbers then Song's answer is correct. If the domain is allowed to be $x>-1$ then Cesareo's answer is correct and includes Song's answer (where $C_0=-1/2$). So the domain of your function $f$ is relevant. What is the domain? $\endgroup$ – Rory Daulton Dec 15 '18 at 12:52

Let $g(x) = f(x-1)$. Then we have $$ 3g(2x+2) = g(x+1) + 5x, $$ or equivalently $$ g(x)-\frac{1}{3}g(\frac{x}{2}) = \frac{5x-10}{6}=: \phi(x).$$ Note that $g(0) = -2.5$. Hence we have $$\begin{eqnarray} g(x) = g(x) -\lim_{j\to\infty}3^{-j}g(2^{-j}x) &=& \sum_{j=0}^\infty \left(3^{-j} g(2^{-j}x) -3^{-j-1}g(2^{-j-1}x)\right)\\ &=& \sum_{j=0}^\infty 3^{-j}\phi(2^{-j}x)\\ &=&\frac{1}{6}\sum_{j=0}^\infty 3^{-j}(5\cdot2^{-j}x-10)\\ &=&\frac{6x-15}{6} = x -\frac{5}{2}. \end{eqnarray}$$ This establishes $f(x) = x -\frac{3}{2}$.

$\textbf{EDIT:}$ I implicitly assumed that the domain of definition of $g$ is $\mathbb{R}$. If the domain of $g$ contains $0$, then the unique continuous solution is given by $g(x) = x-\frac{5}{2} $ as we can see from the above argument. Otherwise, the argument collapses, and one can see that $$g(x) = ( x-\frac{5}{2}) + h(x)$$ is a solution of $g(x)-\frac{1}{3}g(\frac{x}{2}) = \phi(x)$ whenever it holds that $$ h(x) = \frac{1}{3}h(\frac{x}{2})\quad\cdots(*). $$ Note that any continuous function $k : [1,2]\to\mathbb{R}$ with $k(2) = \frac{1}{3}k(1)$ can be extended uniquely to continuous $\overline{k} :(0,\infty)\to\mathbb{R}$ satisfying $(*)$. This shows that there are as many solutions $$g:x\in(0,\infty)\mapsto ( x-\frac{5}{2}) + \overline{k}(x)$$ as there are $k:[1,2] \to\mathbb{R}$ with $k(2) = \frac{1}{3}k(1)$. And the same is also true for $g(x)$ on $(-\infty,0)$.

  • $\begingroup$ Is this one doable with Banach fixed point theorem? $\endgroup$ – Aqua Apr 12 '19 at 13:20
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    $\begingroup$ Sorry for being late, but now that I think about it, we can apply BFPT to $h(x) = f(x-1) - x +\frac 5 2$, $h(x) = \frac 1 3 h\left(\frac x 2\right)$, because $h$ must be bounded on $\Bbb R$ and $\Phi : f(\cdot) \mapsto \frac 1 3 f(\cdot/2)$ is a well-defined contraction on $C_b(\Bbb R)$. $\endgroup$ – Song Jul 25 '19 at 1:56

This is a linear difference equation that can be easily solved as

$$ f(x) = f_h(x)+f_p(x) $$

the homogeneous solution gives

$$ f_h(x) = C_0 3^{1-\log_2(x+1)} $$

The complete solution is

$$ f(x) = \frac{1}{2} \left(\left(2 C_0+1\right) 3^{1-\log_2 (x+1)}+2 x-3\right) $$






$f(2^x-1)=3^{-x}\Theta(x)+2^x+\dfrac{10x-5}{6}$ , where $\Theta(x)$ is an arbitrary periodic function with unit period (according to http://eqworld.ipmnet.ru/en/solutions/fe/fe1102.pdf)

$f(x)=3^{-\log_2(x+1)}\Theta(\log_2(x+1))+x+\dfrac{10\log_2(x+1)+1}{6}$ , where $\Theta(x)$ is an arbitrary periodic function with unit period


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