The following is from Bartle's Elements of Real Analysis.
A function on $\Bbb R$ is additive if $f(x + y) = f(x) + f(y)$ for
each $x, y \in \Bbb R$. Prove $(a)$ An additive function that is
continuous at $x = 0$ is continuous everywhere and $(b)$ A monotone
additive function is continuous on $\Bbb R$.
I proved $(a)$. This is a sketch of what I came up with for $(b)$.
$$ f(0) = f( 0 + 0 ) = f(0) + f(0) \implies f(0) = 0 $$
$$ 0 = f(0) = f(x – x) = f(x) + f(-x) \implies \forall x \in \Bbb R \; f(-x) = -f(x) $$
For $n \in \Bbb N$ we have $f(nx) = f(x) + … + f(x) = nf(x)$ and $n f(\frac{x}{n}) = f( \frac{nx}{n}) = f(x) $ which implies that $ f(\frac{x}{n}) = \frac 1 n f(x) $. We also have that $a \gt 0 \implies f(a) \ge 0$.
So from these properties we can conclude that $f(q) = q \cdot f(1)$ for any rational number $q$.
Now suppose $(x_n)$ is any sequence that tends to $0$. Then we can find a rational sequence $(q_n)$ – by looking in $(x_n – \frac 1 n, x_n + \frac 1 n)$ – which tends to $0$ such that, $$ 0 \le |x_n| \le |q_n| $$
Whence by the monotony of the function , $$ \forall n \in \Bbb N \;\;\;\; 0 \le f(|x_n|) \le f(|q_n|) = |q_n| \cdot f(1) \implies f(|x_n|) \to 0$$
Now we can show by considering cases that $|f(x_n)| = f(|x_n|)$ and hence we have that $f(x_n) \to 0$. So $f$ is continuous at $0$ and hence from part $(a)$ we can conclude that it is continuous on $\Bbb R$.
I have a couple of questions:
- Is my solution correct? Are there any unauthorised assumptions in it?
- There has to be a prettier way of solving this. Can someone please give me a hint?
Sorry for the long prose. Thanks for taking the time to read and for responding.
Best Answer
Your proof seems fine. For a nicer way:
Assume wlog that $f$ is increasing.
Suppose $f$ is not continuous at $0$. Then there exists $\epsilon > 0$ such that for all $\delta > 0$ there is $0 < x < \delta$ with $f(x) > \epsilon$. (This uses $f(-x) = -f(x)$ to force $0 < x < \delta$). But this implies $f(y) > \epsilon$ for all $y > x$, and so $f(y) > \epsilon$ for all $y > 0$.
But $f(y) = f(y/2 + y/2) = f(y/2) + f(y/2) > \epsilon + \epsilon = 2\epsilon$. So $f(y) > 2\epsilon$ for all $y > 0$. Repeating this, we get $f(y) > 2^n \epsilon$ for all $y > 0 $ and $n \in \mathbb{N}$. But this is clearly impossible.
So $f$ is continuous at $0$, and so by part (a) $f$ is continuous everywhere.