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

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.

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Yes, a measurable homomorphism between locally compact groups is automatically continuous (and a continuous homomorphism $\mathbb{R}^n \to \mathbb{R}^m$ is easily seen to be linear): see Alain Valette's answer in this MO thread for a reference for the first statement. See also: this MO thread as well as this thread here. –  t.b. Aug 24 '12 at 15:34
You are basically asking if a measurable $\mathbb{Q}$-linear function is automatically $\mathbb{R}$-linear, which intuitivelly seems obvious. I think the proof of this result in dimension 1 can easely be generalized to higher dimension. –  N. S. Aug 24 '12 at 15:38
@N.S.: Well, there are $\mathbb{Q}$-linear functions that aren't $\mathbb{R}$-linear, so my intuition is either unobvious or wrong :-) –  Xodarap Aug 24 '12 at 15:43
More generally, if $G,H$ are complete separable metric groups, then any Borel measurable homomorphism $G \to H$ must be continuous. For references, search "automatic continuity". Or contemplate the Baire category theorem. –  GEdgar Aug 24 '12 at 15:51
@Xodarap You forgot the measurability. If the function is not continuous (needed to not be $\mathbb{R}$ linear) then you can find two $\mathbb{Q}$-linearly independent elements $x_1,x_2$ so that $f(x_1)=A_1x_1$ and $f(x_2)=A_2x_2$ with $A_2 \neq A_2$. Now, write $\mathbb{R}=\mathbb{Q}x_1+\mathbb{Q}x_2+Y$, where $Y$ is a big $\mathbb{Q}$ vector space. Then for all $y \in Y$ you have $f(y+cx_1)=f(y)+cA_1x_1$ and $f(y+cx_2)=f(y)+cA_2x_2$....Such a function doesn't look measurable.... –  N. S. Aug 24 '12 at 16:02

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

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

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I'm sure you're right, but could you expand? Specifically, why does the linearity of g imply the linearity of f? –  Xodarap Aug 25 '12 at 18:53
@Xodarap I could have been more clear: the linearity of $g$ for every choice of $e$ is what implies the linearity of $f$. See the edit, and let me know if it makes sense! :) –  Sean Eberhard Aug 25 '12 at 23:05

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.

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and it is of course $\mathbb R$-linear. –  GEdgar Aug 24 '12 at 15:52
@GEdgar: Sure. The above answer is therefore not relevant, I was just taking cheap advantage of conjugation. –  André Nicolas Aug 24 '12 at 15:56