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I'm trying to prove the following statement:

There doesn't exist a sequence of polynomials converging in $ \sup $ norm on $ S^1 = \{z \in \mathbb{C}: |z| = 1\}$ to the function $ z \rightarrow \overline{z} $.

I don't really know where to begin - probably I could suppose that we have such sequence $$ f_n (z) = \sum\limits_{i=0}^{N_n} a_i^{(n)}z^i $$

satisfying $$ \lim\limits_{n \rightarrow \infty} \left(\sup\limits_{|z| = 1}\left| \sum\limits_{i=0}^{N_n} a_i^{(n)}z^i - \overline{z}\right| \right) = 0$$

I can't see any obvious way to proceed from this point. I would appreciate some help (I'm rather new to complex analysis and I'd like to omit using some advanced tools)

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  • $\begingroup$ The uniform limit of holomorphic functions is holomorphic? Alternatively, there is a proof using integration theory (in reality the same proof). Does either help? $\endgroup$
    – Moya
    Mar 9, 2015 at 21:57
  • $\begingroup$ I don't yet know how to integrate complex functions - the first one might be helpful. I believe I can prove that conjugation is not differentiable, but I'm not sure if I can prove this limit theorem - the solution seems to be using some Morera's theorem, which I'm not acquainted with $\endgroup$
    – Jytug
    Mar 9, 2015 at 22:10
  • $\begingroup$ You only have the uniform limit on $S^1$, so it is hard to see how to deal with uniform limits of analytic functions. If the convergence was uniform on some open set containing $S^1$ you could note that the uniform limit is analytic but $z \mapsto \bar{z}$ is not. $\endgroup$
    – copper.hat
    Mar 9, 2015 at 22:44

1 Answer 1

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Note that for $z \in S^1$, we have $f(z)=\bar{z} = {1 \over z}$. Suppose $p_n \to f$ uniformly, then $\int_\gamma p_n dz \to \int_\gamma f dz$ where $\gamma(t) = e^{i2 \pi t}$ on $[0,1]$, but $\int_\gamma p_n dz = 0$ for all $n$, and $\int_\gamma f dz = 2 \pi i$.

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