2
$\begingroup$

Let $f:ℂ→ℂ$ be an entire function. Assume that the equation $f(s)=a$ has infinitely many real solutions for all reals $a$. Show that $f$ has infinitely many fixed points, i.e., there exists infinitely many reals $t$ such that $f(t)=t$.

$\endgroup$
  • 2
    $\begingroup$ $f(z) = i+\sin z$, and $a = i$. There must be some further conditions to make it true. $\endgroup$ – Daniel Fischer May 23 '14 at 14:29
  • 2
    $\begingroup$ I mean $1 + \sin z$ also satisfies that it has infinitely many zeros, but definitely has finitely many fixed points (at least real ones). $\endgroup$ – Patrick Da Silva May 23 '14 at 14:31
  • $\begingroup$ @DanielFischer: The condition is: the equation $f(s)=a$ has infinitely many real solutions for all reals $a$. $\endgroup$ – DER May 23 '14 at 14:31
  • $\begingroup$ @DER : Where did you get this problem? $\endgroup$ – Patrick Da Silva May 23 '14 at 14:31
  • 2
    $\begingroup$ $f(x) = x\sin(x)/2$ doesn't have many real fixpoints. $\endgroup$ – mercio May 23 '14 at 14:43
2
$\begingroup$

It is still false after your update. Take $$f(z) = \frac12 z \sin z.$$ It is clear that $f(x) = a$ has infinitely many real solutions for every real $a$, but the only real solution to $f(x) = x$ is $x=0$.

$\endgroup$
  • $\begingroup$ I'm curious : why did you put that $\frac 12$? It didn't help me to see it "clearly" $\endgroup$ – Patrick Da Silva May 23 '14 at 14:46
  • $\begingroup$ @ mrf: I have edited the question. $\endgroup$ – DER May 23 '14 at 14:46
  • 1
    $\begingroup$ @PatrickDaSilva Without the $\frac12$, there are lots of fixpoints. $\endgroup$ – mrf May 23 '14 at 14:47
  • 1
    $\begingroup$ @mrf : Right, didn't think about that. haha! I only thought about the computations in the "proof". $\endgroup$ – Patrick Da Silva May 23 '14 at 15:11

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.