Let $\phi$ be a continuous homomorphism from from the group $(\mathbb{R}^2,+)$ to the group $(\mathbb{R}, +)$. The, how could one prove that $\phi$ is also a linear transformation?
Given that $\phi$ is a homomorphism, we know that $\phi(x+y) = \phi(x) + \phi(y)$, so, one of the criteria for being a linear transformation is met.
What's left to prove is that $\phi(ax) =a\phi(x)$ (for $a\in \mathbb{R}$). Obviously $a\phi$ is also continuous for all $a\in \mathbb{R}$ and it's also obvious that any linear transformation from $\mathbb{R}^2$ to $\mathbb{R}$ is a continuous homomorphism. So, how could I prove that last step? Using the fact that $\lim_{x\to x_0} \phi(x) = \phi(x_0)$? (I've tried that and couldn't get anywhere.)
Hint: Let $(e_1,e_2)$ be a basis of $\mathbb{R}^2$, show that:
Let $n,m, p,q$ integers $f(ne_1+me_2)=nf(e_1)+mf(e_2)$
$f({1\over q}e_1)={1\over q}f(e_1)$,
$f({p\over q}e_1+{m\over n}e_2)={p\over q}f(e_1)+{m\over n}f(e_2)$ use continuity to show that $f(xe_1+ye_2)=xf(e_1)+yf(e_2)$.