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Suppose $u$ is a holomorphic function in the unit disk and $\lvert u(z)\rvert\leq 1$. It follows from Cauchy's inequality that $$ \left\lvert u(z) - \sum_0^{2N-1}\frac{u^{(k)}(0)z^k}{k!}\right\rvert \leq (2N+1) $$

The author claims that from the maximal principle we obtain $$ \left\lvert u(z) - \sum_0^{2N-1}\frac{u^{(k)}(0)z^k}{k!}\right\rvert \leq (2N+1)\lvert z\rvert^{2N} $$ How?

I don't see how it follows. If anyone can show me how to obtain this from the maximum modulus principle I would appreciate it.

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I am not sure how the first inequality follow's from Cauchy's inequality (nor what that is specifically referring to), but we know that since $u$ is holomorphic in the unit disk, that it has a power series at zero. Now, when you subtract away the first $2N$ terms of the power series you get a function $\tilde u(z) = z^{2N}h(x)$, where $h(x)$ is holomorphic in the unit disk. Since $|z|^{2N} = 1$ on the boundary of the disk, we see that $|h(z)| = |\tilde u(z)|\leq 2N + 1$ on the boundary... and hence in the entirety of the disk by the maximum principle.

However, this is actually a slightly flawed proof, since we do not know that we can apply the maximum principle to an open set where the function does not extend continuously to the boundary. So, what we do is apply the maximum principle to the disk of radius $1-\epsilon$, $D_{1-\epsilon}(0)$. We then get that since $|z|^{2N} = (1-\epsilon)^{2N}$ of the boundary of this disk that $|h(z)| \leq (2N + 1)(1-\epsilon)^{-2N}$ in the interior of $D_{1-\epsilon}(0)$. However, if we pick a particular point $z_0$ in the unit disk, it will be in this smaller disk for all $\epsilon > 0$ small enough and so if we take $\epsilon \to 0$ in this inequality, we obtain the desired inequality.

The reason that we were able to apply the maximum principle to this inequality even though $\tilde u$ might not extend continuously to the boundary is that $|z|^{2N}$ did extend continuously to the boundary (and beyond).

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  • $\begingroup$ How can you guarantee that $h(z)$ is holomorphic? Take for instance the holomorphic constant function $u(z) = 1$, then if $\tilde{u}(z) = z^{2N}h(z)$ the function $h$ must have a pole at the origin $\endgroup$ – Quoka Feb 25 '18 at 18:42
  • $\begingroup$ $h$ is holomorphic, which can been seen if you write the Taylor series of $u$ at the origin (which converges to $u$ in the whole of the interior of the disk, since $u$ is holomorphic there): If you write $u(z) = \sum_{n = 0}^\infty a_nz^n$, then $\tilde{u}(z) = \sum_{n=2N}^\infty a_nz^n$ because you subtracted off the first $2N-1$ terms. We then see that we can factor out $z^{2N}$ to get $\tilde{u}(z) = \sum_{n = 2N}^\infty a_nz^n = z^{2N}\sum_{n = 0}^\infty a_{n+2N}z^n$ and $h(z) = \sum_{n = 0}^\infty a_{n+2N}z^n$ is holomorphic can converges where the Taylor series of $u$ converges. $\endgroup$ – user357980 Feb 26 '18 at 0:18
  • $\begingroup$ Ah! Thanks you! That's what I was missing :-) $\endgroup$ – Quoka Feb 26 '18 at 0:42

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