# Show that $f$ has infinitely many fixed points

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 $f(t)=t$.

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

## 1 Answer

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 $x=0$.

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