There exists a unique function $u\in C^0[-a,a]$ which satisfies this property 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.

Your approach will work if $a<2$, in the general case, write 
$$u(x)=\frac x2u\left(\frac x2\right)+g(x)$$ as 
$$u(x)=x^k\alpha_ku\left(\frac x{2^k}\right)+F_k(g)(x)$$
where $F_k$ is a functional of $g$ and $\alpha_k$ satisfy the recurrence relation 
$\alpha_{k+1}=\frac{\alpha_k}{2^{k+1}}$. Therefore, $a^{k+1}\alpha_k$ converges to $0$ and you apply Banach fixed point for $T_k(u)(x)=x^k\alpha_ku\left(\frac x{2^k}\right)+F_k(g)(x)$ for a $k$ large enough.
