# Sequence of analytic functions and the limit function has same number of zeros

Let $\{ f_n \}$ be a sequence of analytic functions on the closed ball $\overline{B}(0,R)$ that converges uniformly on this close ball to an analytic function $f.$ Assume that $f$ has no zeros on $|z|=R.$ Prove that for $n$ large $f_n$ has the same number of zeros in $B(0,R)$ as $f.$ My approach: Intutively this is clear. For a proof I tried, from the definition, for any $\epsilon >0,$ there exists $N > 0$ such that $|f_n(z)-f(z)| < \epsilon$ for $n > N$ and for all $z \in B(0,R).$ Since elements in the sequence and the limit function are analytic they both have Taylor expansions about $0,$ as well as two expressions can be written using the Argument principle. But I wasn't successful in showing the desired result. Any help is appreciated.

• You are assuming each $f_n$ is holomorphic on some $B(0,R'),$ where $R'>R,$ and $f_n \to f$ uniformly on $\overline {B(0,R)}?$ – zhw. Aug 25 '16 at 20:05
• with the argument principle, you can show that the position of the zeros converge. – reuns Aug 26 '16 at 0:10
• I might well be wrong. Just mentioned my approach. Thanks. – user358174 Aug 26 '16 at 0:11
• you need to bound $\frac{f'(z)}{f(z)} - \frac{f_n'(z)}{f_n(z)}$ on $|z-a| = r$ where $a$ is a zero. since $f(z) \ne 0$, $|f(z)|$ has a minimum on $|z-a| = r$, and since $f_n(z) \to f(z)$ uniformly, the same is true for $|f_n(z)|$. And also (by the Cauchy integral formula, say) $f_n'(z)$ converges uniformly to $f'(z)$. – reuns Aug 26 '16 at 0:15
• @user1952009 thank you. I considered that difference being the integrands appeared in the integral of the Argumen't principle, but I didn't think of a bound. From the uniform convergence it does follow that $f'_n(z) \rightarrow f'(z)$ uniformly as well. So if the bounds for this is arbitrarily small (in terms of $\epsilon$) we can conclude that both have the same number of zeros. Isn't that what you're suggesting ? Thank you for your time. – user358174 Aug 26 '16 at 0:34

You may use Rouché's theorem: If $|g(z)-f(z)|<|f(z)|$ on the boundary of a compact set then $f$ and $g$ has the same number of zeros inside.
Since $f$ is continuous and has no zeros on the (compact) boundary $|z|=R$ there is $r>0$ so that $|f(z)|\geq r>0$ on the boundary. Now pick $N$ so that for $n\geq N$ we have $|f_n(z)-f(z)|\leq r/2< r \leq |f(z)|$ on the boundary and apply Rouché.
• In any case you need to consider $n$ large enough so that $|f(z)-f_n(z)|<|f(z)|$ or else $f_n(z)$ may have a zero on the boundary which would bring havoc to any proof. So probably better using Rouché directly. – H. H. Rugh Aug 25 '16 at 19:58