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Let $h \in \mathcal{C}:=C([-a,a])$, where $a>0$. Prove that there exists a unique function $f \in \mathcal{C}$ such that $$ f(x)=\frac{x}{2}f\Big(\frac{x}{2}\Big)+h(x)\quad \forall x \in [-a,a]. $$

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Uniqueness Let $f_1, f_2 \in \mathcal{C}$ be solutions of $$ f(x)=\frac{x}{2}f\big(\frac{x}{2}\Big)+h(x) \quad \forall x \in [-a,a]. $$ Then for every $x \in [-a,a]$ we have $$ |f_1(x)-f_2(x)|=\frac{|x|}{2}\Big|f_1\big(\frac{x}{2}\Big)-f_2\big(\frac{x}{2}\Big)\Big|. $$ Therefore, for every $x \in [-a,a]$ and every $n \in \mathbb{N}$ we have $$ |f_1(x)-f_2(x)|=\frac{|x|^n}{2^{1+2\ldots+n}}\Big|f_1\big(\frac{x}{2^n}\Big)-f_2\big(\frac{x}{2^n}\Big)\Big|=\frac{|x|^n}{2^{n(n+1)/2}}\Big|f_1\big(\frac{x}{2^n}\Big)-f_2\big(\frac{x}{2^n}\Big)\Big|. $$ Hence $$ \|f_1-f_2\|\le\frac{a^n}{2^{n(n+1)/2}}\|f_1-f_2\| \quad \forall n \in \mathbb{N}. $$ Since $$ \lim_{n\to \infty}\frac{a^n}{2^{n(n+1)/2}}=0, $$ it follows that $f_1\equiv f_2$.

Existence Let $g \in \mathcal{C}$ be arbitrary. Consider the sequence $(f_n)_n\subset \mathcal{C}$ given by $$ f_1(x)=g(x),\ f_n(x)=\frac{x}{2}f_{n-1}\Big(\frac{x}{2}\Big)+h(x) \quad \forall x \in [-a,a] $$ For every $n$ and every $x \in [-a,a]$ we have \begin{eqnarray} |f_n(x)-f_{n+1}(x)|&=&\frac{|x|}{2}\Big|f_{n-1}\Big(\frac{x}{2}\Big)-f_n\Big(\frac{x}{2}\Big)\Big|\\ &\vdots&\\ &=&\frac{|x|^{n-1}}{2^{1+2\ldots+(n-1)}}\Big|f_1\Big(\frac{x}{2^{n-1}}\Big)-f_2\Big(\frac{x}{2^{n-1}}\Big)\Big|\\ &\le&\frac{a^{n-1}}{2^{n(n-1)/2}}\|f_1-f_2\|, \end{eqnarray} i.e. $$ \|f_n-f_{n+1}\|\le \frac{a^{n-1}}{2^{n(n-1)/2}}\|f_1-f_2\| \quad \forall n \ge 1. $$ Hence, for every $m>n \ge 1$ we have \begin{eqnarray} \|f_n-f_m\|&\le&\sum_{k=n}^{m-1}\|f_k-f_{k+1}\|\le \|f_1-f_2\|\sum_{k=n}^{m-1}\frac{a^{k-1}}{2^{k(k-1)/2}}\le \|f_1-f_2\|\sum_{k=n}^\infty\frac{a^{k-1}}{2^{k(k-1)/2}} \end{eqnarray} From the latter inequality we obtain that $$ \lim_{m,n\to \infty}\|f_n-f_m\|=0, $$ i.e. $(f_n)$ is a Cauchy sequence, it therefore converges to some $f \in \mathcal{C}$. From the definition of $f_n$ we get for each $x \in [-a,a]$: $$ f(x)=\lim_nf_n(x)=\lim_n\Big(\frac{x}{2}f_{n-1}\Big(\frac{x}{2}\Big)+h(x)\Big)=\frac{x}{2}f\Big(\frac{x}{2}\Big)+h(x). $$

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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 $f \equiv 0$.

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