On sort-of-linear functions: does $f(x+y) = f(x) + f(y)$ imply $f(\alpha x) = \alpha f(x)$? 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} $.
 A: 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).  
A: Yes, continuity is necessary.  In one variable the counterexamples are known as pathological solutions to Cauchy's functional equation.  They require the axiom of choice to construct: given the axiom of choice, we can pick a basis for $\mathbb{R}$ as a vector space over $\mathbb{Q}$, and any function $f : \mathbb{R} \to \mathbb{R}$ which is only required to be $\mathbb{Q}$-linear can be specified by arbitrarily specifying its behavior on this basis.
Pathological solutions are very strange (for example their graph is dense in the plane) and can be ruled out by almost any "niceness" hypothesis, as described in the wiki article.
