Suppose that $f$ is a continuous additive function on $\mathbb{R}$ and $f(1) = c$. Show that $f(x) = cx$.
I have an attempt at this, but the reason I am asking about it is because I did not use the hint. The hint was to first show this for rationals, and then extend it to the rest of $\mathbb{R}$. I feel I don't need this.
Let $x$ be fixed. Then, since $f$ is additive,
$f(x) = f(x – 1) + f(1) = f(x – 1) + c = f(x – 2) + f(1) + c = f(x – 2) + 2c = \dots f(x – (x – 1)) + (x-1)c$
$= f(1) + (x-1)c = c + (x-1)c = cx$
Since $x$ was fixed, I can continue the process all the way up to $x$. At the same time, $x$ was arbitrary, so this holds for all $x$.
Now clearly the same thing could be done with a rational, but I'm struggling to see why the hint was given. What am I missing here? Thanks in advance!
Best Answer
For all $x \in \mathbb{R}$ we have $f(x)=f(x+0)=f(x)+f(0)$, so $$\tag{1} f(0)=0. $$ Therefore $0=f(-x+x)=f(-x)+f(x)$, i.e.
$$\tag{2} f(-x)=-f(x) \quad \forall x \in \mathbb{R} $$
Since $f$ is additive, for all $x_1, x_2, \ldots x_n \in \mathbb{R}$ we have
$$\tag{3} f\left(\sum_{i=1}^n x_i\right)=\sum_{i=1}^n f(x_i) $$
Thanks to (2) and (3), we obtain Combining all the above, we have
$$\tag{4} f(n)=cn \quad \forall n \in \mathbb{Z}. $$
For any nonzero integer $n$, we have $$ c=f(1)=f\left(n.\frac{1}{n}\right)=\sum_{k=1}^n f\left(\frac1n\right)=nf\left(\frac1n\right), $$ i.e. $$\tag{5} f\left(\frac1n\right)=c\cdot\frac1n\quad \forall n\in \mathbb{Z}-\{0\} $$
Combining (2) - (5), we have, for all $m/n\in \mathbb{Q}$
$$ f\left(m\cdot\frac{1}{n}\right)=mf\left(\frac{1}{n}\right)=m\cdot c\cdot \frac{1}{n}, $$
i.e. $$\tag{6} f(r)=cr \quad \forall r\in \mathbb{Q}. $$
Let $x \in \mathbb{R}$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, there exists a sequence $(r_n)$ of rational numbers admitting $x$ as its limit. Because $f$ is continuous and in virtue of (6), we have
$$ f(x)=f(\lim_n r_n)=\lim_n f(r_n)=\lim_n cr_n=cx. $$
Hence $f(x)=cx$ for all $x \in \mathbb{R}$.