Background
A function $ f: \mathbb{R}^n \to \mathbb{R} \ $ is linear if it satisfies
$$ f(x+y) = f(x) + f(y) \ \text {, and} \tag 1 \label 1 $$
$$ f(\alpha x) = \alpha f(x) \tag 2 \label 2 $$
for all $ x,y \in \mathbb{R}^n $ and all $ \alpha \in \mathbb{R} $.
A function satisfying only \eqref{2} is not necessarily linear. For example* $ f: \mathbb{R}^2 \to \mathbb{R} \ $ defined by $ f(x) = |x| \ $ (where $ |x| \ $ is the $ L^2 $ norm) satisfies \eqref{2} but is not linear. However, a function satisfying \eqref{1} does satisfy a weaker version of \eqref{2}, namely
$$ f(ax)=af(x) \tag {2b} \label {2b} $$
for all $ a \in \mathbb{Q} $.
*Edit: As pointed out in the comments this example doesn't quite work since $|ax|=|a||x|$.
When $ f $ is continuous it's relatively straight-forward to show that under the extra hypothesis that $ f $ is continuous, \eqref{2b} implies \eqref{2}. I want to say that continuity is a necessary condition for \eqref{1} to imply \eqref{2}, or at least (worst) there is some extra hypothesis required (possibly weaker than continuity), but I'm not sure how to show it.
My question is therefore two-fold:
- Is continuity a necessary condition for \eqref{1} to imply \eqref{2} and how could I go about proving it.
- What are some examples (if there are any) of a function satisfying \eqref{1} but not \eqref{2}?
This can be stated in a slightly more general context as follows:
Suppose $ V\ $ is a vector space over $ \mathbb{R}\ $ and $ f: V \rightarrow \mathbb{R}\ $ satisfies
$$ f(x+y) = f(x)+f(y) \tag {1'} $$
for all $ x,y \in V $.
Under what conditions is $ f\ $ a vector space homomorphism?
The reason I believe continuity is necessary is because of the similarity to the fact that $ x^{\alpha} x^{\beta} = x^{\alpha + \beta} $ for all $ \alpha,\beta \in \mathbb{R} $. Irrational powers can be defined either via continuity (i.e. if $ \alpha \ $ is irrational, then $ x^{\alpha}:= \lim_{q\to \alpha} x^q \ $ where q takes on only rational values) or by using the exponential and natural logarithm functions, and either way proving the desired identity boils down to continuity.
I have come up with one example that satisfies (something similar to) \eqref{1} and not \eqref{2}, but it doesn't quite fit the bill:
$ \ $ Define $ \phi : \mathbb{Q}\left(\sqrt{2}\right) \to \mathbb{Q} \ $ defined by $ \phi\left(a+b\sqrt{2}\right) = a+b $. Then $ \phi(x+y) = \phi(x)+\phi(y) \ $ but if $ \alpha=c+d\sqrt{2} \ $ then $ \phi\Big(\alpha\left(a+b\sqrt{2}\right)\Big) = ac+2bd + ad+bc \neq \alpha \ \phi\left(a+b\sqrt{2}\right) $.
$ \ $ The problem is that even though $ \mathbb{Q}\left(\sqrt{2}\right) \ $ is a vector space over $ \mathbb{Q} $, the $ \alpha \ $ is coming from $ \mathbb{Q}\left(\sqrt{2}\right) \ $ instead of the base field $ \mathbb{Q} $.
Best Answer
It is not true that $|ax|=a|x|$; the correct identity is $|ax|=|a||x|$.
Whether or not adding the hypothesis of continuity is necessary for additive functions to be linear depends on the axiom of choice. Using a Hamel basis $B$ for $\mathbb{R}^n$ over $\mathbb{Q}$ together with one of its countable subsets $A=\{x_1,x_2,\ldots\}$, you can construct a discontinuous $\mathbb{Q}$ linear map from $\mathbb{R}^n$ to $\mathbb{R}$ by taking the unique $\mathbb{Q}$ linear extension of the function $f:B\to\mathbb{R}$ such that $f(x_k)=k|x_k|$ and $f(B\setminus A)=\{0\}$. Since $\mathbb{R}$ linear maps between finite dimensional real vector spaces are continuous, such a map cannot be linear. However, it is consistent with ZF that all additive functions from $\mathbb{R}^n$ to $\mathbb{R}$ are continuous (I am however not knowledgeable in the set theoretic background needed to show this).