2
$\begingroup$

I have a question about the proof of Liouville's theorem as given on Wikipedia.

If I assume that $f(z)$ is entire and bounded only on $B_r(0)$ for some $r$ > 0, then using the same idea of the proof above, why can't I conclude that $f$ is constant on the whole domain?

$\endgroup$
  • $\begingroup$ can you write the usual proof of Liouville's theorem ? $\endgroup$ – reuns Oct 2 '16 at 20:48
  • $\begingroup$ what do you mean the usual proof? $\endgroup$ – Keith Oct 2 '16 at 20:48
  • $\begingroup$ the proof that a bounded entire function is constant.. $\endgroup$ – reuns Oct 2 '16 at 20:49
  • $\begingroup$ I do not really know what is the usual proof, can you just spot the problem of this proof?@user1952009 $\endgroup$ – Keith Oct 2 '16 at 21:59
2
$\begingroup$

If you only assume that $f$ is bounded on some specific disc, you can't let $r \to \infty$ to get $a_k = 0$ (with notation as in the linked proof).

Note that if $f$ is entire, then $f$ is bounded on every disc since it is continuous. The bound on $f$ depends on the radius of the disc though in general.

$\endgroup$
  • $\begingroup$ Suppose that I have $R>0$ then $f = \sum_0 ^\infty z^n \int _{B_R(0)}\frac{f(w)}{w} dw$, and $ \int _{B_r(0)}\frac{f(w)}{w} dw$, am I right? If this is right(which I do think is right), then I can sill use the same techniques as before. $\endgroup$ – Keith Oct 2 '16 at 20:37
  • $\begingroup$ @Keith as written that's not right at all. Likely you forgot a power of $n$. But the problem persists. For some $R>0$ you get a bound for the $n$ coefficient of $M/R^{n}$ and that's it. How would you continue? $\endgroup$ – quid Oct 2 '16 at 20:51
  • $\begingroup$ @Keith as I said, write the usual proof of Liouville's theorem, and it will become obvious when it applies and when it doesn't $\endgroup$ – reuns Oct 2 '16 at 20:54
  • $\begingroup$ Sorry I do forget (n+1). I would not work on $R>0$. My thought is to use that $\int_{B_R(0)} \frac{f(w)}{w^{n+1}}dw = \int _{B_r(0)} \frac{f(w)}{w^{n+1}}dw$, then I can make $r$ arbitrarily small. $\endgroup$ – Keith Oct 2 '16 at 20:58
  • $\begingroup$ The smaller you make $r$, the worse the bound gets. $\endgroup$ – mrf Oct 2 '16 at 21:13

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.