TITLE: There exists a unique function $u\in C^0[-a,a]$ which satisfies this property
QUESTION [7 upvotes]: The problem:

Let $a>0$ and let $g\in C^0([-a,a])$. Prove that there exists a unique function $u\in C^0([-a,a])$ such that $$u(x)=\frac x2u\left(\frac x2\right)+g(x),$$ for all $x\in[-a,a]$. 

My attempt At first sight I thought to approach this problem as a fixed point problem from $C^0([-a,a])$ to $C^0([-2a,2a])$, which are both Banach spaces if equipped with the maximum norm. However i needed to define a contraction, because as it stands it is not clear wether my operator $$(Tu)(x)=\frac x2u\left(\frac x2\right)+g(x)$$ is a contraction or not. Therefore I tried to slightly modify the operator and I picked a $c>a>0$ and defined $$T_cu=\frac 1cTu.$$ $T_cu$ is in fact a contraction, hence by the contraction lemma i have for granted the existence and the uniqueness of a function $u_c\in C^0([-a,a])$, which is a fixed point for $T_cu.$ Clearly this is not what I wanted and it seems difficult to me to finish using this approach. Am I right, is all what I have done useless? And if this were the case, how to solve this problem?
Thanks in advance.

REPLY [2 votes]: ...how to solve this problem?

A direct road is to define $u$ as
$$
u(x)=\sum_{n=0}^{+\infty}2^{-n(n+1)/2}x^ng(2^{-n}x),
$$
and to prove that $u$ is indeed continuous on $[-a,a]$. 
Hint: For every sequence of continuous functions $(g_n)_{n\geqslant0}$ such that $\|g_n\|_\infty\leqslant cA^n$ for every $n$, for some finite $c$ and $A$, the series $\sum\limits_{n=0}^{+\infty}2^{-n(n+1)/2}g_n$ defines a continuous function.