Continuity proof using $f(x+y) = f(x) + f(y)$ ($3$ parts, last part!)

Question

Suppose that $f : \mathbb{R} \rightarrow \mathbb{R}$ satisfies $f(x+y) = f(x) + f(y)$ for each $x, y \in \mathbb{R}$.

$1)$ I've proven that $f(qx)=qf(x)$ for all $x \in \mathbb{R}$ and $n \in \mathbb{Q}$

$2)$ I've proven that $f$ is continuous at $0$ if and only if $f$ is continuous on $\mathbb{R}$

$3)$ Question: We want to prove that if $f$ is continuous at $0$, then there is an $m \in \mathbb{R}$ such that $f(x)=mx$ for all $x \in \mathbb{R}$

I was able to show that if $x \in \mathbb{Q}$, then $m=f(1)$ using what I've proven earlier: $f(x)=f(1\cdot x)=xf(1)$

But $x \in \mathbb{R}$ so this proof is incomplete. My intuition says my $m$ is correct, but I haven't used continuity yet; played around with the fact that continuous at $0$ implies continuity over all real numbers (using what I proved in $2$). I feel as though the proof should go through that road and not my first line of reasoning. I'm unsure how to bridge this gap.

Answer 1

Hint: Consider the function $g(x)=f(x)-mx$, which is continuous by hypothesis. We know that $g|_{\mathbb{Q}}=0$ and that $\mathbb{Q}$ is dense in $\mathbb{R}$. What can we say about $g$ then?

If $g(x)\neq0$ for some $x\in\mathbb{R}$, by continuity there would exist $\delta>0$ such that $g(y)\neq0$ for $y\in(x-\delta,x+\delta)$. But $(x-\delta,x+\delta)$ contains (infinitely many) rational points (and at that points $g$ should be zero).

Answer 2

Let $(r_n)$ be a sequence of rationals.

You proved that

$$(\forall n\in\Bbb N) \;\; f(r_n)=r_nf(1)$$

Now take a real $x$.

we can find $(r_n)$ such that

$$\lim_{n\to+\infty}r_n=x$$

thus

$$f(x)=f(\lim_{n\to+\infty}r_n)=$$

by continuity,

$$\lim_{n\to+\infty}f(r_n)=\lim_{n\to+\infty}r_nf(1)=xf(1)$$