If a real additive function f is monotonic, then it is linear.

I need to show that the monotonic function f that satisfies caushy additives functional equation is linear


Here is the outline of a proof:

Using the fact that $f$ is additive, show that

  • $f(0)=0$
  • $f(x)=x\cdot f(1)$ for $x=0,1,2,\ldots$
  • $f(x)=x\cdot f(1)$ for $x\in \mathbb Z$
  • $f(x)=x\cdot f(1)$ for $x\in \mathbb Q$

Then use the fact that $f$ is monotonic to show that $\lim_{x\to a}f(x)=f(a)$ for all real $a$.

Then show that $f(x)=x\cdot f(1)$ for $x\in \mathbb R$.

That shows that $f$ is linear.

There are variations on this proof, of course. Let us know if you need more details.

  • 1
    $\begingroup$ The proof can be simplified further noting that, as it is a group homomorphism, it is enough to show continuity at $0$. $\endgroup$ – Bernard Feb 28 '15 at 14:41
  • $\begingroup$ @Bernard: True, but the OP did not mention groups or homomorphisms. I tried to keep my answer at the level of the question. $\endgroup$ – Rory Daulton Feb 28 '15 at 14:42
  • $\begingroup$ @Rory Daulton thanks for the outlines $\endgroup$ – M'smary Feb 28 '15 at 15:07

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