Let $f:(0,\infty)\to\mathbb{R}$ be a continuous function such that for all $x\in(0,\infty)$, it holds that $f(2x)=f(x)$. Show that the function defined by the equation $$g(x)=\int_x^{2x}\frac{f(t)}{t}dt$$ for $x>0$ is a constant function.
My approach: It is given that $f$ is continuous $\forall x>0$. Thus from the definition of $g$ we can conclude by the Leibniz Rule, that $g$ is differentiable for all $x>0$ and $$g'(x)=\frac{f(2x)}{x}-\frac{f(x)}{x}.$$ Now since $f(2x)=f(x)$ holds true $\forall x>0,$ we can conclude that $g'(x)=0,$ $\forall x>0$.
Now let us choose an infinitesimal real number $\epsilon>0$ and assume that $f(0+\epsilon)=f(\epsilon)=L$. Also let $x_1>\epsilon$ be arbitrary. Now since $g$ is differentiable for all $x>0$, implies that $g$ is differentiable on the closed interval $[\epsilon,x_1]$, which in turn implies that $g$ is continuous on the closed interval $[\epsilon,x_1]$ and differentiable on the open interval $(\epsilon,x_1)$ with $g'(x)=0$ for all $x\in(\epsilon,x_1)$. Thus we can conclude that $g$ is constant, that is, $g(x)=c,$ for all $x\in[\epsilon,x_1]$, where $c$ is some constant. Actually by MVT, we can show that $c=L$.
Now since $x_1$ was arbitrary, we can conclude that $g$ is a constant function in its domain.
Is the approach correct and rigorous enough?