Background
A function $ f: \mathbb{R}^n \rightarrow \mathbb{R} \ $ is linear if it satisfies $ (1)\;\; f(x+y) = f(x) + f(y) \ , \ and $ $ (2)\;\; f(\alpha x) = \alpha f(x) $ for all $ x,y \in \mathbb{R}^n $ and all $ \alpha \in \mathbb{R} $.
A function satisfying only (2) is not necessarily linear. For example* $ f: \mathbb{R}^2 \rightarrow \mathbb{R} \ $ defined by $ f(x) = |x| \ $ (where $ |x| \ $ is the $ L^2 $ norm) satisfies (2) but is not linear. However, a function satisfying (1) does satisfy a weaker version of (2), namely $ (2b)\;\; f(ax)=af(x) $ 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, (2b) implies (2). I want to say that continuity is a necessary condition for (1) to imply (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 (1) to imply (2) and how could I go about proving it. -What are some examples (if there are any) of a function satisfying (1) but not (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 (1') \;\; f(x+y) = f(x)+f(y) 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\rightarrow \alpha} x^q \ $ where q takes on only rational values) or by using the exponential and natual log functions, and either way proving the desired identity boils down to continuity.
I have come up with one example that satisfies (something similar to) (1) and not (2), but it doesn't quite fit the bill:
$ \ $ Define $ \phi : \mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q} \ $ defined by $ \phi(a+b\sqrt{2}) = a+b $. Then $ \phi(x+y) = \phi(x)+\phi(y) \ $ but if $ \alpha=c+d\sqrt{2} \ $ then $ \phi(\alpha(a+b\sqrt{2})) = ac+2bd + ad+bc \neq \alpha \ \phi(a+b\sqrt{2}) $. $ \ $ The problem is that even though $ \mathbb{Q}(\sqrt{2}) \ $ is a vector space over $ \mathbb{Q} $, the $ \alpha \ $ is coming from $ \mathbb{Q}(\sqrt{2}) \ $ instead of the base field $ \mathbb{Q} $.