Cauchy's functional equation for $\mathbb R^n$

Question

Suppose $f(x+y)=f(x)+f(y)$. If $f:\mathbb R\to \mathbb R$ and is measurable, then $f(x)=cx$. This is referred to as Cauchy's functional equation.

Suppose $f:\mathbb R^n\to \mathbb R^n$ instead. Does it still hold that $f$ is linear?

Wikipedia says that Hilbert's fifth problem is a generalization of this functional equation, but I can't parse that page well enough to understand how it relates.

Answer 1

I think we can deduce the $n$-dimensional result from the $1$-dimensional result. We know $$f(x+y)=f(x) + f(y)$$ for all $x,y\in\mathbf{R}^n$ and we want to prove that in fact $$f(\lambda x + \mu y) = \lambda f(x) + \mu f(y)$$ for all $x,y\in\mathbf{R}^n$ and $\lambda,\mu\in\mathbf{R}$. It clearly suffices to prove that $$f(\lambda x) = \lambda f(x)$$ for all $x\in\mathbf{R}^n$ and $\lambda\in\mathbf{R}$. But, for fixed $x,e\in\mathbf{R}^n$, $g:\lambda\mapsto \langle f(\lambda x), e\rangle$ is a measurable function $\mathbf{R}\to\mathbf{R}$ satisfying $g(\lambda + \mu) = g(\lambda) + g(\mu)$, from which it follows by the $1$-dimensional result that $g(\lambda) = \lambda g(1)$. In other words, for all $x,e\in\mathbf{R}^n$ we have $\langle f(\lambda x),e\rangle = \langle \lambda f(x),e\rangle$. Since $e$ is arbitrary, this implies $f(\lambda x) = \lambda f(x)$ for all $x\in\mathbf{R}^n$ and $\lambda\in\mathbf{R}$, so

$$f(\lambda x + \mu y) = f(\lambda x) + f(\mu y) = \lambda f(x) + \mu f(y).$$

Answer 2

This is not an answer, but we don't quite have linearity. For example, $f(z)=az+b\bar{z}$ is a continuous solution of the Cauchy functional equation of $\mathbb{C}$. (And all continuous solutions are of this form.) In this case, differentiability implies linearity.