WLOG, let $$f(z)=\sum_{n=0}^\infty a_nz^n$$ be a holomorphic function on the unit disc with radius of convergence $1$ around $z=0$. My question is about a sense in which we can say that the rate of convergence of the taylor expansion of $f$ around $0$ must be worse everywhere on the boundary of the disc of convergence than in the interior.
If we define $$E_n(z) = \left|f(z) – \sum_{k=0}^n a_k z^k\right|$$ to be the error resulting from the $n$-term truncation of the Taylor series. Then more specifically, the question can be stated as follows:
Given some $z_1$ in the interior of the unit disc, and some $z_2$ on the boundary, and perhaps some conditions on $f$, do we know that for all sufficiently large $n$, $E_n(z_1) \leq E_n(z_2)$?
I believe that this holds automatically if the Taylor series of $f$ diverges at $z_2$, but either way the interesting case seems to me to be the one where the Taylor series converges at $z_2$, so that can be assumed if you like.
By Cauchy-Hadamard, we can know that for all $z$ in the interior of the unit disc (with $ 0<|z|=r < 1$), we have that $$\limsup_{n \to \infty} |a_nz^n|^{1/n} \leq r < 1 $$ and that we can not obviously say that about the boundary. But it seems a lot less trivial to show lower bounds on the error.
Best Answer
Edited from previous answer.
Counterexample: Define
$$f(z) = \sum_{k=0}^{\infty}\frac{z^{2k}-z^{2k+1}}{(k+1)^2}.$$
The radius of convergence of this power series is $1.$ Let $z_1=-1/2, z_2=1.$ Then $E_n(-1/2)>0$ for all $n$ and $E_n(1)=0$ for all odd $n.$
If we don't insist on a finite radius of convergence, we can use my previous answer:
$$f(z) = \sum_{k=0}^{\infty}\frac{z^{2k}-z^{2k+1}}{k!}.$$
This $f$ is entire. Again let $z_1=-1/2, z_2=1$ to see $E_n(-1/2)>0$ for all $n$ and $E_n(1)=0$ for all odd $n.$