1
$\begingroup$

I am reading Stein’s complex analyze.A theorem says if {${f_n}$} is a sequence of holomorphic functions that converges uniformly to a function $f$ in every compact subset of $\Omega$,then $f$ is holomorphic in $\Omega$.

I don’t understand why compact is necessary for the theorem.

I guess it is mainly because the compactness can guarantee every functions are bounded as the situation in Morera’s theorem we will use later in the proof where a disc whose closure is in the region means the compactness.

An inappropriate analogy is the situation in real valuable functions where Weierstrass theorem says a continuous functions can be approached uniformly by polynomials in a closed interval.(However,not every continuous functions are differentiable so I mean it is not appropriate)

I don’t know whether my guess is true or not and if we remove the condition of compactness,can anyone give me a counterexample of this situation?

$\endgroup$
  • $\begingroup$ Take $\frac{1}{1-z}$ for example and the sequence of partial sums and restrict yourself to the exterior of circles closing in on $1$. ........ I think (!) $\endgroup$ – Cameron Williams Sep 6 at 4:15
  • 1
    $\begingroup$ Note that "removing the condition of compactness" in this case would be weakening the theorem, not strengthening it. Assuming uniform convergence on the whole domain, in fact, would not yield a very usable theorem in the long run. $\endgroup$ – Brian Moehring Sep 6 at 4:24
3
$\begingroup$

What we really need for this theorem is the uniform convergence of the $\{f_n\}$ to $f$ on $\Omega$, since this is what guarantees that $f$ is continuous and that $\int_Tf_n(z)dz\to\int_Tf(z)dz$. If the $\{f_n\}$ do $not$ converge uniformly, neither of those assertions (the continuity of $f$ and putting the limit inside the integral) necessarily hold anymore. So uniform convergence is the key condition, and this theorem holds whenever the $\{f_n\}$ converge uniformly on all of $\Omega$.

However, as being holomorphic is a local condition, it seems like we can get away with the $\{f_n\}$ converging to $f$ uniformly only $locally$, so that for every point in $\Omega$ there is a disc where they converge uniformly. So we can actually get the same theorem but with weaker assumptions on the $\{f_n\}$. On $\mathbb{C}$ (and any other locally compact space), local uniform continuity is equivalent to being uniformly convergent on every compact subset, which is the form the assumption takes in Stein's book.

To see that it really is a weaker assumption, take for example $f_n(z) = x^n$ on the unit disc. Then the $\{f_n\}$ converge everywhere to the zero function, but not uniformly on the disc. It does, however, converge uniformly on every compact subset of the disc.

$\endgroup$

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.