5
$\begingroup$

Let $f:\mathbb{C}\to \mathbb{C}$ be a complex function such that $$|f(x-y)|=|f(x)-f(y)|,\qquad x,y\in\mathbb{C}.$$ Is it true that $$f(x+y)=f(x)+f(y),\qquad x,y\in\mathbb{C}?$$

The answer is affirmative when $f:\mathbb{R}\to\mathbb{R}$ is a real function, and the proof is not difficult. The above generalization seems harder: I inferred it from a conjecture that I saw in a paper.

Improve this question
$\endgroup$
9
  • 2
    $\begingroup$ What about $f(x)=x+a$? In any case your question is not appropriate for this site, please use MathStackExchange. $\endgroup$
    – abx
    Commented Oct 13, 2018 at 12:09
  • 1
    $\begingroup$ @abx,if $f(x)=x+a$ not such condition,and this problem I think it's a research question. It's not simple. $\endgroup$
    – math110
    Commented Oct 13, 2018 at 12:32
  • 7
    $\begingroup$ @CarloBeenakker: Actually, there are many (non-continuous) functions besides $f(x)=cx$ that satisfy the second equation (hence also the first equation). $\endgroup$
    – GH from MO
    Commented Oct 13, 2018 at 12:52
  • 1
    $\begingroup$ I voted to close, but after the edit it seems like a reasonable question (with the $\|$ corrected to $|$). $\endgroup$
    – Nik Weaver
    Commented Oct 13, 2018 at 13:34
  • 3
    $\begingroup$ I edited the question greatly (for clarity and language), and I voted to reopen it. At the same time, I ask the OP to make a bigger effort (e.g. next time) to formulate his/her question nicely. In particular, the post refers to some paper, hence a reference would be necessary. This site is for professionals, so participants should act like professionals. $\endgroup$
    – GH from MO
    Commented Oct 14, 2018 at 13:36

1 Answer 1

Reset to default
13
+25
$\begingroup$

counterexample $$f(x)=1-e^{\Re(x) i}$$

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .