A question about the proof of Liouville's theorem.

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?

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

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.
• 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. – Keith Oct 2 '16 at 20:37
• @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? – quid Oct 2 '16 at 20:51
• 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. – Keith Oct 2 '16 at 20:58
• The smaller you make $r$, the worse the bound gets. – mrf Oct 2 '16 at 21:13