2
$\begingroup$

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:

  1. Is my solution correct? Are there any unauthorised assumptions in it?
  2. 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.

$\endgroup$
  • $\begingroup$ I also just realised that I've assumed $f$ is monotone increasing. But I don't think there is a loss of generality??? $\endgroup$ – Ishfaaq Feb 11 '15 at 17:36
  • $\begingroup$ $f$ is continuous at $a$ iff $-f$ is continuous at $a$, so assuming $f$ is increasing is fine. $\endgroup$ – Christopher Feb 11 '15 at 17:41
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.