Show that $f(x)=\alpha x$ for some $\alpha \in R$

Question

Suppose $f(x+y)=f(x)+f(y)$ for some $f: R \rightarrow R$ such that $f$ is continuous at $0$.Show that $f(x)=\alpha x$ for some $\alpha \in R$. Hint: show that $f(nx) = nf(x)$, then show $f$ is continuous on $R$. Then show that $f(x)/x = f(1)$ for all rational $x$.

How can I show $f(nx)=nf(x)$?? I am stuck with the first hint, so I can't show any attempt.

Thank you in advance!

Answer 1

By mathematical induction.

(1) It is obviously true for $n=1$.

(2) If $f(kx)=kf(x)$, then $$f((k+1)x)=f(kx+x)=f(kx)+f(x)=kf(x)+f(x)=(k+1)f(x)$$

Answer 2

Notice if we let $y=x$, then

$$ f(2x) = 2 f(x) $$

Now, let $x=2x'$, then

$$ f(4x') = 2 f(2x') = 4 f(x') $$

Continuing in this fashion youll get $f(nx) = nf(x) $

Answer 3

This is what do you want, $$f(m)=f(1+1+...+1)=f(1)+f(1)+...+f(1)=mf(1),m\in\mathbb{z^+}.$$ Deduce, $$f(1)=f(\frac{n}{n})=f(\frac{1}{n})+f(\frac{1}{n})+...+f(\frac{1}{n})=nf(\frac{1}{n}),n\in\mathbb{z^+}.$$ Hence, $$f(\frac{m}{n})=\frac{m}{n}f(1)$$ Hence, use $\mathbb{Q}$ dense in $\mathbb R$ and $f$ is continuous. we get the desired result.