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I'm studying for an exam and I'm stuck on the following problem:

Does there exist a sequence of holomorphic functions $\{f_n(z)\}_{n=1}^{\infty}$ on the unit disc such that $f_n(z)\to1/z$ uniformly on $\{z\in\mathbb{C}\colon|z|=1/2\}$ as $n\to\infty$?

What tools would I use if $1/z$ were replaced with another functions, or say the $f_n$'s were replaced by polynomials instead?

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2 Answers 2

up vote 7 down vote accepted

If $f_n\to f$ uniformly on $|z|=1/2$, then $\displaystyle{\int_{|z|=1/2}f_n(z)dz\to \int_{|z|=1/2}f(z)dz}$. If each $f_n$ is holomorphic on the disk (in particular, if each is a polynomial), this implies that $\displaystyle{\int_{|z|=1/2}f(z)dz=0}$. In your example, you can compute the integral of the hypothetical limit to check whether or not this is the case.

But that method wouldn't work in case the hypothetical limit were $1/z^k$, $k>1$, for example. To adapt to that case, you could consider $z^{k-1}f_n(z)\to 1/z$.

Here is a method that would apply to a more general class of hypothetical limits. If $f_n\to f$ uniformly on $|z|=1/2$, then $(f_n)$ is uniformly Cauchy on $|z|\leq 1/2$ by the maximum modulus theorem, so there is a function $g$ such that $f_n\to g$ uniformly on the closed disk of radius $1/2$. Then $g$ must be continuous and analytic on the interior of this disk. If $f$ has constant modulus as in the examples mentioned so far, then by rescaling and applying Robert Israel's answer to your other question, we see that $g$ is a rescaled finite Blaschke product, which greatly restricts the possibilities for $f$.

In any case, $f$ must be the restriction to the boundary of a continuous function on the closed disk that is analytic on the interior, so it may be interesting to consider this class of functions. One characterization of this class is as the set of all continuous functions on the circle whose negatively indexed Fourier coefficients vanish.

Also, note that the question will always have the same answer whether polynomials or more general holomorphic functions are considered, because every holomorphic function on the unit disk is uniformly approximated by polynomials on every smaller disk (e.g, using Taylor series).

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Another approach to the problems of such type is based on the maximal modulus principle. Since $f_n\to 1/z$ on $|z|=1/2,$ then $g_n(z)=f_n (z)z\to 1$ on $|z|=1/2.$ By the MMP this implies that we have convergence on the whole set $|z|\le 1/2.$ But this is impossible, since $g_n(0)=0$ does not converge to $1$ when $n\to\infty.$

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