Additivity & one-point Continuity of $f \in \mathbb{R}^\mathbb{R}$ imply there is $\alpha \in \mathbb{R}$ s.t. $f(\alpha x) = \alpha x$

Question

I am looking for hints/help concerning a proposition I found self-studying Carothers:

Suppose that $f : \mathbb{R} \to \mathbb{R}$ satisfies $f(x+y) = f(x)+f(y)$ for every $x, y \in \mathbb{R}$. If $f$ is continuous at some point $x_0 \in \mathbb{R}$, prove that there is some constant $\alpha \in \mathbb{R}$ such that $f(\alpha x) = \alpha x$ for all $x \in \mathbb{R}$.

My initial ideas to prove the result move along the following lines.

1. Notice by additivity that $f(0) = 0$.
2. Take the point $x_0 \in \mathbb{R}$ where $f$ is continuous by assumption.
3. Obtain that if $|x_0| > \delta_\varepsilon$, then $|f(x_0)|<\varepsilon$
4. Take a $\varepsilon >0$ large enough to get a $\delta_{\varepsilon}>0$ such that $|x_0|<\delta_{\varepsilon}$
5. etc...

The final idea is to construct $\alpha$ as the result of $\tan \theta$, where $\theta$ is the angle between the $x$-axis, and the line that passes through $x_0$ from the origin.

Does this all make sense?

Thank you for your time.

Answer 1

Hint: $f(x)=f(x+x_0-x_1)-f(x_0-x_1)$. Use that to show that $f(x)$ is continuous at any $x_1$ if it is continuous at $x_0$.

Next, let $\alpha = f(1)$ and show that $f(x)=f(1)x$ first for $x$ an integer, then $x$ a rational, and then, by continuity, all the reals.

Answer 2

Hint: $f(1)=f(r\frac{1}{r})=rf(\frac{1}{r})$ where $r\in\mathbb{Z}$ thus $f(\frac{1}{r})=\frac{1}{r}f(1)$ use this and show that $f(x)=f(x.1)=xf(1)$ for rati0nal number then use density of ratinal number in $\mathbb{R}$ and continuty of $f$