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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!

Nolan P
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    Your proof assumes that the last step, say the $n$th step, has $n=x-1$, i.e. $x$ is an integer – Calvin Khor Jul 11 '22 at 01:16
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    Try your argument with a specific, non-integer number, like $x = \frac{3}{2}$. Or even, something irrational like $\sqrt{2}$. How would your argument establish, in finitely many steps, that $f(x) = cx$ in these cases? – Theo Bendit Jul 11 '22 at 01:20
  • Use this post to say that continuous additive functions are linear. Then the result is a simple consequence of linearity because $f(x)=f(x\cdot 1)=xf(1)=cx$. – John Douma Jul 11 '22 at 01:37
  • " I can continue the process all the way up to x." Only if $x$ is an integer. And actually with the argument $f(x)=f(x-1) + c=f(x-2) + 2c =....... =f(x-(x-1)) + (x-1)c = f(1) + (x-1)c = cx$ it only holds if $x$ is a positive integer. – fleablood Jul 12 '22 at 17:17
  • Try it with $f(\pi)$. $\pi > 3> 1$ so we do $f(\pi) = f(\pi-1) + f(1)=f(\pi-1) + c=f(\pi-2)+f(1) +c =f(\pi-2) + 2c$ and we "do this all the way till we get to $\pi$. $\pi > 3 > \pi -1 > 2 > \pi -2 > 1 > \pi-3....$ and oops we went too far. How can you get count to $\pi$ using just $1$s? – fleablood Jul 12 '22 at 17:23

1 Answers1

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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 $x \in \mathbb{R}$.

HorizonsMaths
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