Is there any entire function, such that the above statement holds true? The idea seems to be that I want an entire function that can approximate the conjugate that is not analytic in the unit circle, but I have no idea on how to begin...
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No. You can get a contradiction from using Parseval on $\int_0^{2\pi}|e^{-it}-f(e^{it})|^2,dt$. – David C. Ullrich Sep 26 '20 at 13:56
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$|z|=1$ implies $\bar z =1/z$ so the inequality translates to $|1/z-f(z)|<1, |1-zf(z)|<|z|=1$, but now $1-zf(z)$ is analytic in the unit disc so the inequality must hold there by maximum modulus, while it plainly doesn't hold for $z=0$.
Contradiction so no such $f$ (even analytic on the open unit disc and continuos on the boundary only say, not to speak of entire)
Conrad
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Compare the magnitude of
$\int_{\text{CCW unit circle}}(\overline {z}-f(z)) dz$
with the length of the path which is $2\pi$.
Clearly if $f$ is entire, then it contributes zero to the contour integral around the closed curve. And along the contour, we have $\overline {z} =1/z$. So the above integral reduces to
$\int_{\text{CCW unit circle}}(\overline {z}-f(z)) dz=\int_{\text{CCW unit circle}}(1/z) dz=i(2\pi)$
The magnitude of the integral matches the length of the curve exactly and therefore by the triangle inequality, the maximum absolute value of $\overline {z}-f(z)$ has to be at least $1$.
If we allow a pole inside the unit circle, then $f(z)$ makes a nonzero contribution to the contour integral and we can get around the above contradiction. For instance, $f(z)=1/z$ gives $|\overline {z}-f(z)|=0$ everywhere on the unit circle.
Oscar Lanzi
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