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I've been having trouble with the following problem for a few weeks now for which it seems there should be an elementary proof.

Let $U \subset \mathbb{C}^n$ be open, connected and bounded with smooth boundary. Let $A^2(U) = \mathcal{O}(U) \cap L^2(U)$, i.e., the subspace of $L^2(U)$ consisting of holomorphic functions.

Claim: $A^2(U) \cap C(\overline{U})$ is dense in $A^2(U)$

Note: the norm we want to show this subset is dense with respect to is the $L^2$ norm.

The problem comes from Jiri Lebl's text ``Tasty Bits of Several Complex Variables", exercise 5.2.6, linked here: https://www.jirka.org/scv/scv.pdf

I also managed to track down a paper by David Catlin titled ``Boundary Behavior of Holomorphic Functions on Pseudoconvex Domains" which proves a more general result to do with holomorphic Sobolev Spaces, albeit, under a greater amount of assumptions and in much less elementary manner.

I've tried a whole bunch of different ideas like patching together Taylor series from individual discs using a partition of unity, to convoluting with smooth functions and then trying to show the resulting function is holomorphic.

The Taylor series idea comes from this argument: $L^2$ convergence of Taylor series of a holomorphic function but I ran into trouble with the convergence of Taylor series.

Something that might be helpful is that the polynomials are all in $A^2(U) \cap C(\overline{U})$.

I'm also interested in the result for values other than $p=2$.

SubSpace626
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  • The theorem of Stone-Weierstrass helps you here. – Giuseppe Negro Jun 09 '19 at 08:46
  • I thought that might be the way to go and I tried various things using that theorem but I ran into some problems. You get the polynomials are dense in $C(\overline{U})$ but that doesn't tell you anything about the density in $A^2(U)$. Am I mistaken? – SubSpace626 Jun 09 '19 at 09:17
  • No you are not mistaken. What you have should be enough. Polynomials are dense in $C(\bar{U})$ with respect to the $L^\infty$ norm, which implies denseness with respect to the $L^2$ norm, because of $|f|{L^2(U)}\le C|f|{L^\infty(U)}$, nothing deep here. This should be enough to conclude, but please check carefully, I am not very focused right now. – Giuseppe Negro Jun 09 '19 at 09:38
  • After thinking about it more I don't think this is quite enough. Sure the polynomials are dense in $C(\overline{U})$ with respect to the $L^2$ norm which gives us the polynomials are dense in $A^2(U) \cap C(\overline{U})$ as that is a subset of $C(\overline{U})$ still containing the polynomials. But then we have one final step to make which is to approximate any $L^2$-integrable holomorphic function by these polynomials and I don't see how to make that connection. – SubSpace626 Jun 09 '19 at 12:37
  • You need $U$ simply connected (you can't approximate $1/z$ in $C^0_{loc}$ thus in $L^2_{loc}$). In $\Bbb{C}^n,n \ge 2$ this problem stays even with Hartogs's extension theorem. For $U$ simply connected what happens when you just take the power series at one point and approximate it termwise with an entire power series, won't it be a $L^2_{loc}$ approximation ? – reuns Jun 09 '19 at 13:25
  • But $f(z) = 1/z$ is not in $L^2(U)$ for any $U$ close enough to the origin. Outside of that though, $f(z) = 1/z$ can be approximated arbitrarily close by its Taylor series which is just polynomials. – SubSpace626 Jun 09 '19 at 13:25
  • To answer why can't you just take a power series at one point point and approximate it through that, well power series have radii of convergence so you run into issues when trying to approximate it over the whole of $U$ and not just in a polydisc. So you need some good way to piece together power series and no matter what I try there is no clean way to do that without running into convergence issues at at least some points. – SubSpace626 Jun 09 '19 at 13:33
  • The result I'm trying to prove comes straight from the textbook I've linked above and I've basically copied it word for word. See that for more details. – SubSpace626 Jun 09 '19 at 13:37
  • In retrospect, I think either 1) I was overly optimistic about how easy it is to prove or 2) it was a typo.

    Approximating via Stone-Weierstrass does not work so easily since that gets an approximation by non-holomorphic polynomials. In fact for many domains you can't do that. To use that approach you'd need to then show that projection onto holomorphic functions is still nice up to the boundary. I can't see a simple proof of that right now.

    Next version will have this changed to a ball since that's really what is required in the text (why the exercise is there to begin with).

     – Jiri Lebl Jun 28 '19 at 14:09
  • @JiriLebl How would you show it is true for a ball (if you know how)? – SubSpace626 Jun 30 '19 at 12:08
  • For the ball it is not that bad. There are actually a couple of possibilities. The easiest is to consider f(rz) for r less than but close to 1 – Jiri Lebl Jul 01 '19 at 03:11
  • @JiriLebl: Oh, so my answer was wrong in the end because polynomials need not be holomorphic? I admit that I did not consider this possibility. – Giuseppe Negro Jul 05 '19 at 14:48
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    Well, I suppose it's ok once you get polynomials being dense. But they aren't always dense. For example take n=1 and some annulus. You can't approximate the function 1/z in the annulus by holomorphic polynomials. You can approximate it by real polynomials via Weierstrass, that is polynomials in x and y if z=x+iy – Jiri Lebl Jul 15 '19 at 04:26

1 Answers1

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I am posting here because the comments are already quite cluttered.

WARNING: This is not really an answer, because the holomorphic polynomials are not always dense in $C(\overline U)$; see Jiri Lebl's comment.

If you can prove that the set of all polynomials $P$ is dense in $C(\overline U)$ with respect to the $L^2(U)$ norm, as you claim in the comments, then you are done. Use sequences. You need to show that, for each $f\in A^2(U)$, there exists a sequence $p_n\in P$ such that $$\tag{1}\lVert p_n-f\rVert_2\to 0.$$ Now, since $f\in A^2\subset L^2$, there exists a sequence $g_n\in C(\overline U)$ such that $$\tag{2} \lVert g_n-f\rVert_2\to 0.$$ For each $n$, you can construct an approximation $p_{n, k}\in P$ to $g_n$; $$\tag{3} \lim_{k\to\infty} \lVert p_{n, k}-g_n\rVert_2=0.$$ Ok, now let $p_n:=p_{n,n}$.

Giuseppe Negro
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  • I may be wrong but doesn't this only show the polynomials are dense in $A^2(U) \cap C(\overline{U})$, not in $A^2(U)$ as required? – SubSpace626 Jun 09 '19 at 22:31
  • If we take $Y = L^2(U)$ and $X = A^2(U)$ then our argument doesn't work because we don't have that the polynomials form a dense subset of $L^2(U)$. All we know is that they are dense in $C(\overline{U})$ with respect to the $L^2(U)$ norm. – SubSpace626 Jun 10 '19 at 06:20
  • I have updated my comment, forget about $X$s and $Y$s, it should be clearer now. – Giuseppe Negro Jun 10 '19 at 07:09
  • Ah yes, all makes sense now. I have one final query though. Is the reason we can choose $g_n$ as above because of the fact that $L^2$ can be defined in terms of the completion of $C^{\infty}_c(\overline{U})$? – SubSpace626 Jun 10 '19 at 07:50
  • To say that "$L^2$ is the completion of $C^\infty_c(\overline U)$" is exactly the same thing as "$C^\infty_c(\overline U)$ is dense in $L^2$". – Giuseppe Negro Jun 10 '19 at 08:18
  • Yep exactly what I thought. Thanks for all help. I really do appreciate it! – SubSpace626 Jun 10 '19 at 08:21
  • Pleasure. On this site, if you find an answer to be useful, you can upvote it. Also, when you feel that your question has been solved, you should accept the best answer. – Giuseppe Negro Jun 10 '19 at 08:23
  • After some further thought, I don't quite understand why we can pick $(g_n)$ as stated. $L^2(U)$ is the completion of $C_c^{\infty}(U)$ (not $\overline{U}$) in the $L^2$ norm. But the set $C_c^{\infty}(\overline{U}) \subset C_c^{\infty}(U)$ so we don't immediately get the density as I first thought. Any reason as to why this is possible? – SubSpace626 Jun 13 '19 at 12:14
  • It's the same, L2 does not see the boundary since it is a measure zero set. Thus, L2(U) is the same as L2(over line U). – Giuseppe Negro Jun 13 '19 at 12:16
  • So that is what I thought at first but https://math.stackexchange.com/questions/613408/when-does-the-boundary-have-measure-zero?noredirect=1&lq=1 seems to suggest otherwise. Is the smooth boundary condition imposed on U coming in to play here? – SubSpace626 Jun 13 '19 at 12:20
  • Of course.This is because the boundary is smooth. A basic fact is that a codimension 1 manifold of $\mathbb R^n$ has measure zero. – Giuseppe Negro Jun 13 '19 at 12:33
  • I'm not so sure that $p_n$ actually necessarily converges to $g_n$. Say $g_n$ is a non-zero polynomial. Take the sequence $(p_{n,k}) = (0,0,\ldots,0, 0, g_n, g_n , \ldots )$ where the first $g_n$ occurs in the $k+1$th spot in the sequence. Then $p_{n,n} = (0,0,0,\ldots)$ which does not converge to $g_n$ but each $p_{n,k} \to g_n$ as $k \to \infty$. – SubSpace626 Jun 14 '19 at 09:47
  • I am sure it works. These are standard density arguments. Maybe there is a mistake in my answer, but it MUST work in the end. If a set (here $P$) is dense in a big space (here $C(\overline U)$), and it is also contained in a subspace of such big space (here $A^2$), then the set is dense in the subspace also. It is obvious if you think in terms of approximating sequences. – Giuseppe Negro Jun 14 '19 at 10:27