Help in solving a simple functional equation: $3f(2x+1)=f(x) + 5x$

functional-equations

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 $(-\infty,0)$.