Let $f \colon \mathbb{R}^n \to \mathbb{R}^n$ be continuous with $$f(x)-f(y) = C(x,y)(x-y)$$ for all $x,y \in \mathbb{R}^n$ and a function $C = C(x,y) \colon \mathbb{R}^n \times \mathbb{R}^n \to\mathbb{R}$. Then $f(x) = ax+b$ for some $a\in \mathbb{R}$ and $ b\in \mathbb{R}^n$.
I read this statement in a proof and was wondering if this is true. Does anyone know a reference?
Starting with $y = 0$ we can see that:
$$f(x) - f(0) = C(x, 0)(x - 0) \Rightarrow f(x) = C(x, 0)(x) + f(0).$$
Let $b = f(0)$. So, we want to show that for all $x, x' \in \mathbb{R}^n$, $C(x, 0) = C(x', 0)$. Take arbitrary $x, x' \in \mathbb{R}^n$ and we know that
$$f(x) - f(x') = C(x, x')(x - x')$$
and also that
$$\begin{align*} f(x) - f(x') & = [C(x, 0)(x) + b] - [C(x', 0)(x') + b] \\ & = C(x, 0)(x) - C(x', 0)(x') \\ & = C(x, 0)(x - x') - (C(x', 0) - C(x, 0))(x'). \end{align*}$$
To get $$C(x, x')(x - x') = C(x, 0)(x - x') - (C(x', 0) - C(x, 0))(x')$$
we need $C(x', 0) = C(x, 0)$. Let $a = C(x, 0)$ and so we get $f(x) = ax + b$.