<s>[INST] Uniqueness of solution for the functional equation $f(x)=\frac{x}{2}f\left(\frac{x}{2}\right)+h(x)$ Let $h \in \mathcal{C}:=C([-a,a])$, where $a>0$. Prove that there exists a unique function $f \in \mathcal{C}$ such that
$$
f(x)=\frac{x}{2}f\Big(\frac{x}{2}\Big)+h(x)\quad \forall x \in [-a,a].
$$
 [/INST] If $f$ is a solution then for all $x \in [-a,a]$ we get by applying the functional equation $n$ times
\begin{align*}
f(x) &= \frac{x}{2} f\left(\frac{x}{2}\right) + h(x)\\
&= \frac{x^2}{2^{1+2}} f\left(\frac{x}{4}\right) + \frac{x}{2} h\left(\frac{x}{2}\right) \\
&= \frac{x^2}{2^{1+2+3}} f\left(\frac{x}{8}\right) + \frac{x^2}{2^{1+2}} h\left(\frac{x}{4}\right) + \frac{x}{2} h\left(\frac{x}{2}\right) + h(x)\\
&= \ldots\\
&= \frac{x^n}{2^{n(n-1)/2}} f\left(\frac{x}{2^n}\right) + \sum_{k=0}^{n-1} \frac{x^k}{2^{k(k-1)/2}} h\left(\frac{x}{2^k}\right)
\end{align*}
The first summand tends to zero for $n \to \infty$, as do the summands inside the second sum for $k \to \infty$. So we would like to define
$$f(x) = \sum_{k=0}^{\infty} \frac{x^k}{2^{k(k-1)/2}} h\left(\frac{x}{2^k}\right).$$
This $f$ satisfies the functional equation, we just have to show that the series converges and defines a continuous function on $[-a,a]$. If we can show that the series converges uniformly on $[-a,a]$ then the our limit function $f$ will be continuous as a uniform limit of continuous functions.
For large $k$, say $k \geq k_0$, we have $|h(x/2^k)| \leq |h(0)| + \varepsilon$, by continuity of $h$, so
$$\sum_{k \geq k_0} \left| \frac{x^k}{2^{k(k-1)/2}} h\left(\frac{x}{2^k}\right) \right|
\leq \left(|h(0)| + \varepsilon\right) \sum_{k \geq k_0} \frac{a^k}{2^{k(k-1)/2}}$$
and the right hand side converges independently of $x$.
For the uniqueness, let $f$ and $g$ be two solutions to the functional equation. Then $f-g$ satisfies the functional equation for $h \equiv 0$. Therefore we can assume $h \equiv 0$ and show that $f \equiv 0$ is the only solution. Applying the functional equation $n$ times as above shows
$$f(x) = \frac{x^n}{2^{n(n-1)/2}} f\left(\frac{x}{2^n}\right)$$
for all $n$. But $f(x/2^n)$ is bounded by continuity of $f$, so this converges to $0$ for $n \to \infty$. Thus $f \equiv 0$.
</s>