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Let $D$ denote the closed complex unit disk and let $f:D \rightarrow \mathbb C$ be any continuous function. Suppose we wish to approximate $f$ by a holomorphic function $g$ in the uniform metric: we seek to minimize $\sup\{z\in D : |g(z) - f(z)|\}$, subject to the condition that $g$ be holomorphic on the interior of the disk.

The uniform limit of holomorphic functions is holomorphic, so this quantity cannot be zero if $f$ is not holomorphic. Is there any general result about how small it can be for some class of choices of $f$?

In particular, how closely can we approximate the conjugate function $z \mapsto \bar z$ by a holomorphic function?

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up vote 8 down vote accepted

I don't know the answer in general, but if $g$ is holomorphic on $D$ and $f(z)=\overline z$, then $\sup\{|f(z)-g(z)|:z\in D\}\geq 1$, and this lower bound is achieved when $g=0$. To see this, note that $\int\limits_{|z|=1}g(z)dz = 0$, and $\int\limits_{|z|=1}f(z)dz=2\pi i$. So $$2\pi =\left|\int_{|z|=1}(f(z)-g(z))dz\right|\leq \int_{|z|=1}|f(z)-g(z)||dz|\leq 2\pi \,\sup\{|f(z)-g(z)|:|z|=1\}.$$

This gives $\sup\{|f(z)-g(z)|:z\in D\}\geq\sup\{|f(z)-g(z)|:|z|=1\}\geq 1$ as claimed.


Similar lower bounds can be given in some cases using the nonzero Fourier coefficients of $f$ with negative index, but they will not typically be sharp (and such Fourier coefficients may not exist; see the added comments below). If $\int\limits_{|z|=1}z^kf(z)dz=a\neq 0$ for some $k\geq 0$ and if $h$ is holomorphic, then $$|a|=\left|\int_{|z|=1}(z^kf(z)-h(z))dz\right|\leq \int_{|z|=1}|z^kf(z)-h(z)||dz|\leq 2\pi \,\sup\{|z^kf(z)-h(z)|:|z|=1\}.$$ So if we take $h(z)=z^k g(z)$ for some holomorphic function $g$, this implies that $\sup\{|f(z)-g(z)|:|z|=1\}\geq\frac{|a|}{2\pi}$.


Added: I removed an incorrect remark at the end of my answer that was made in part because I was confusing approximation on the disk with approximation on the circle. Note that there exist nonanalytic $f$ such that $\int_{|z|=1}z^kf(z)dz= 0$ for all $k$; e.g., you could have nonzero continuous (even real-analytic) $f$ such that $f|_{\partial D}\equiv 0$. The example $f(z)=1-|z|^2$ is one that is even constant on all circles centered at $0$. In general you could get similar lower bounds by finding a closed curve in $D$ on which the integral of $f$ is nonzero, but I don't know how useful that is in practice.

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