# If f is a continuous function and $f(1)=k$, prove that $f(x)=kx$ for all $x$.

Let $f$ be a function that satisfied $f(x+ y) = f(x) + f(y).$

So the first part required me to prove that if $f$ is continuous at some point $c$, it is continuous on $\Bbb R$. I was able to do this.

The second part is as follows:

Suppose that $f$ is continuous on $\Bbb R$ and that $f(1) = k$. Prove that $f(x)=kx$ for all $x \in\Bbb R$.

I am having difficulty understand how to approach this problem. Any help whatsoever would be appreciated!

This is a standard. We prove that $f(ax)=af(x)$ firstly for $a\in\Bbb N$, secondly for integer $a$, and finally for rational $a$. Next it is enough to use continuity to prove this claim for any $a\in \Bbb R$. Now put $x=1$, that's it.
Hint: Let $n$ an integer, shows recursively that $f(n)=kn$, then show that $f({1\over n})={k\over n}$ if $n\neq 0$ by using that $f(n{1\over n})=nf({1\over n})=k$ then show that $f({p\over q})={kp\over q}, p,q$ integers, then use the fact that rational are dense in $R$ and the continuity.
• But its for x $\in$ R, not for n $\in$ N Commented Feb 26, 2017 at 19:38
• You show it for $x\in Q$, $x=lim x_n, x_n\in Q$, $f(x)=lim_nf(x_n)=lim_nkx_n=kx$. Commented Feb 26, 2017 at 19:39