Prove that $f(x)$ is continuous over $\mathbb{R}.$ Problem
Let $f(x)$ be continuous at $x=0$ and satisfy that $f(0)=0$,
$$3f(x)-4f(4x)+f(16x)=3x ,\forall x \in \mathbb{R}.$$
Prove that $f(x)$ is continuous over $\mathbb{R}.$
Comment
Maybe, we can obtain
$$3[f(x)-f(4x)]-[f(4x)-f(16x)]=3x.$$
Denote
$$g(x):=f(x)-f(4x).$$
Then
$$3g(x)-g(4x)=3x.$$
Can we apply the induction from here? I'm not sure.
 A: Note that $f_0(x)=x$ is a solution. Set $g(x)=f(x)-x$ the difference to the particular solution, then $g$ solves the homogeneous equation
$$
3g(x)−4g(4x)+g(16x)=0
$$
and it is continuous in $x=0$. 

Set $h(x)=3g(x)-g(4x)$, then $h(x)-h(4x)=0$, $h(0)=0$ and $h$ is continuous at $x=0$. By induction 
$$
h(x)=h(4^{-n}x)
$$
for all $n$  and because of continuity at $0$ one gets $$h(x)=h(0)=0.$$ 

So $3g(x)=g(4x)$, again by induction
$$
g(x)=3^ng(4^{-n}x)
$$
for all $n$. Now define $g$ any way you want for $|x|\in [1,4)$, but so that it is overall bounded, $|g(x)|\le M$, on this set. For any $x\ne 0$ there is exactly one $n$ so that $4^{-n}|x|\in[1,4)$. Then with that $n$ define 
$$
g(x)=3^ng(4^{-n}x)\implies f(x)=x+3^ng(4^{-n}x)
$$

Now this so constructed $f$ satisfies the assumption that is continuous at $x=0$, as $$|f(x)|\le4^{-m}+3^{-m-1}M$$ for $|x|\le 4^{-m}$, or $$f(x)=x+O(x^{\log_4(3)})$$ for $x\approx 0$. I see no further conditions that could constrain $g$ on $(-4,-1]\cup[1,4)$. Thus that $f$ is continuous on the whole of $\Bbb R$ can not be proven.
