# Complex analysis - $\bar{z}$ cannot be uniformly approximated by polynomials in $z$ on the closed unit disc in $\mathbb{C}$.

In reference to this question, I tried to prove that $\bar{z}$ cannot be uniformly approximated by complex polynomials in $z$ on the closed unit disc $D$. I came up with a proof, but I'm not entirely sure whether it's correct.

Here's how the proof goes: Assume it can be done. Pick an $\varepsilon < 0.1$ and then you'll have a polynomial $p$ such that $|p(z) - \bar{z}| < \varepsilon$ for all $z \in D$. Multiply both sides by $z$ to get: $$|zp(z) - z\bar{z}| < |z|\varepsilon$$ Now if we just look at the points where $|z| = 1$, the inequation becomes: $$|zp(z) - 1| < \varepsilon$$

Now I'll show on some point on the unit circle $C$, $zp(z)$ has a non-positive real part, which will lead to a contradiction because then $|zp(z) - 1| > 1 > \varepsilon$. Notice that $zp(z)$ is of the form: $$zp(z) = a_nz^n + \cdots + a_1z$$ Pick $2(n!)$ points on the circle such that the angle between two adjacent points is $\frac{\pi}{n!}$. The sum of $zp(z)$ over these points will be $0$. Here's why: For each point, there exists a point $\pi$ radians away, so the $z$ term in $zp(z)$ goes to $0$. Similarly, the $z^2$, $z^3$ and so on terms also add up to $0$ hence the polynomial adds up to $0$. Which means at one of those points, the real part of $zp(z) \leq 0$. This leads to the contradiction mentioned above and completes the proof.

Is this proof correct? And if it is correct, this only works for sets containing a neighbourhood of $0$. But you'd expect a proof a global property (i.e. uniform approximation) to not depend on any local properties (i.e. a neighbourhood of $0$). It'd be nice if someone could explain how to extend this proof to any compact set in $\mathbb{C}$.

• You seem to be using $p$ to denote two different things. – Mariano Suárez-Álvarez May 22 '16 at 5:47
• FWIW, the canonical approach here is to use Cauchy's integral formula to show that a uniform limit of holomorphic functions is holomorphic, and then that $z\mapsto\bar z$ is not holomorphic. – Mariano Suárez-Álvarez May 22 '16 at 5:48
• @MarianoSuárez-Alvarez I fixed the names. – sayantankhan May 22 '16 at 5:53
• @MarianoSuárez-Alvarez I know the other proof you mentioned. I don't really know any complex analysis though, and I was told there is a proof that just requires elementary knowledge. Hence this attempt. – sayantankhan May 22 '16 at 5:55
• This does appear to be correct. Actually quite an interesting, geometric proof; I like it! – Fimpellizieri May 22 '16 at 6:04

Your proof can be translated by integrating around the unit circle (if you take $n$ big enough, you're approximating the integral). Note that $$\frac{1}{2\pi i}\displaystyle\int_{|z|=1} \bar zdz=\frac{1}{2\pi i}\int_0^{2\pi} e^{-it}ie^{it}dt =1$$ and assume that given $\varepsilon >0$ there is $p_\varepsilon(z)$ such that $|p_\varepsilon(z)-\bar z|<\varepsilon$ on $\overline B(0,1)$. Since polynomials admit primitives their integrals over closed curves are $0$, so
$$1=\left | \frac{1}{2\pi i}\int_{|z|=1} (\bar z-p(z))dz\right|\leqslant \sup_{|z|=1} |p_\varepsilon(z)-\bar z|<\varepsilon$$
by the standard estimate. This is a contradiction if $\varepsilon$ is small enough.
• +1. Matter of taste: If you leave out the final $<\varepsilon$ (and the assumption) then instead of a contradiction you get a nice lower bound of the maximal difference. – WimC May 22 '16 at 11:20