Are polynomials dense is L2 of the unit disk?

Question

Let $D$ be the unit disk in the complex plane, and let $X$ be the subset of $L^2(D)$ consisting of polynomials in the complex variable $z=x+iy$ with complex coefficients. My question is, is $X$ dense in $L^2(D)$?

If not, does anyone know of a function in $L^2(D)$ which cannot be written as an $L^2$ limit of polynomials?

@CountIblis So then do polynomials in $z$ not separate points? Becausethis answer seems to be saying polynomials are not dense in the continuous functions for the sup norm. — Keshav Srinivasan, Apr 05 '19 at 02:30

score 5 · Accepted Answer · answered Apr 05 '19 at 03:47

5

No. For instance, $g(z)=\overline{z}$ is orthogonal to every polynomial (proof sketch: if $f(z)=z^n$, then $\langle f,g\rangle=\int_D f\overline{g}=\int_D z^{n+1}=0$ by either direct computation in polar coordinates or using the symmetries of $z^{n+1}$). It follows that $g$ is not in the closure of the polynomials.

answered Apr 05 '19 at 03:47

Eric Wofsey

330,363

Are all the L2 limits of polynomials holomorphic? – Keshav Srinivasan Apr 05 '19 at 04:14
I believe that's true but don't recall a proof off the top of my head. – Eric Wofsey Apr 05 '19 at 04:19
I posted a follow-up question on that: https://math.stackexchange.com/q/3175501/71829 – Keshav Srinivasan Apr 05 '19 at 04:32

score 0 · Answer 2 · answered Apr 05 '19 at 02:17

0

Yes. $X$ is dense in $L^2$.

Certainly $X$ is dense in the set of all the continuous functions, $C(D)$. $C(D)$ is also dense in $L^2$, so $X$ is dense in $L^2$.

answered Apr 05 '19 at 02:17

Ma Joad

7,420

1

Why is $X$ dense in $C(D)$? This answer seems to be sa^ying the polynomials are not dense in $C(D)$ with respect to the sup norm. Why would the $L^2$ case be any different? – Keshav Srinivasan Apr 05 '19 at 02:24

Are polynomials dense is L2 of the unit disk?

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