I am working with an analytic function defined within the unit disk $|z| < 1$ as follows:
$$ f(z) = \sum_{n=1}^{\infty} a_{n} z^{n}, $$
where I have the condition that $\sum_{n=1}^{N} n|a_{n}| = O(\log(N))$ for the partial sums of the series' coefficients.
I am trying to establish that $f'(z)=O(\log(\frac{1}{1-r}))$ . My current approach has been to estimate $|f'(z)|$ directly via the series representation and its partial sum behavior:
$$ |f'(z)| \leq \sum_{n=1}^{\infty} n |a_{n}| r^{n-1}. $$
Applying summation by parts, I tried to decompose it further:
$$ |f'(z)| \leq (1-r) \sum_{n=1}^{\infty} S_{n} r^{n-1}, $$
where $S_{N} = \sum_{n=1}^{N} n|a_{n}|$, and I chose $M = \lfloor \frac{1}{1-r} \rfloor$ for a split in the series:
$$ (1-r)\left(\sum_{n=1}^{M} S_n r^{n-1} + \sum_{n=M+1}^{\infty} S_n r^{n-1}\right). $$
While the first part seems manageable and aligns with the O-notation related to $\log(\frac{1}{1-r})$, the challenge arises with controlling the tail, $\sum_{n=M+1}^{\infty} S_n r^{n-1}$, which I've only managed to loosely bound by $O(\frac{1}{1-r})$.
I suspect there might be a more nuanced approach or an application of a different technique to tighten this bound. Any insights on how to refine this analysis or alternative strategies to establish the bound for $f'(z)$ as $O(\log(\frac{1}{1-r}))$ would be greatly appreciated.
The trick is to take $M=\lfloor \frac{1}{1-r} \log \frac{1}{1-r}\rfloor$ or $M=[-\frac{\log \epsilon}{\epsilon}], r=1-\epsilon$ so $M\log r \approx \log \epsilon$ hence $r^M \approx \epsilon=1-r$ too when $r \to 1$ so $\epsilon$ small enough.
Then $S_n \le C\log n$ so $$(1-r)\sum_{n=M+1}^{\infty} S_n r^{n-1} \le C(1-r)\sum_{n=M+1}^{\infty} (\log n) r^{n-1}=Cr^M(1-r)\sum_{m=0}^{\infty} (\log (m+M+1)r^m$$
But $\log(m+M+1) \le \log M + \log (m+2)$ so $$Cr^M(1-r)\sum_{m=0}^{\infty} (\log (m+M+1)r^m \le C (\log M) r^M+ Cr^M(1-r)\sum_{m=0}^{\infty} (\log (m+2)r^m $$
Since $r^M \approx 1-r$ and $\log M << |\log (1-r)|$ while $\sum_{m=0}^{\infty} (\log (m+2)r^m << \sum_{m=0}^{\infty} (m+1)r^m=\frac{1}{(1-r)^2}$ we have that $$(1-r)\sum_{n=M+1}^{\infty} S_n r^{n-1}=O(1)$$ (actually with more care we can make it $o(1)$ for $r \to 1$ by majorizing $\log (m+2) <<m^{1/100}$ say)
On the other hand since $S_n \le C\log M, n \le M$ $$(1-r)\sum_{n=1}^{M} S_n r^{n-1}\le C\log M \approx C(\log \frac{1}{1-r}+\log \log \frac{1}{1-r}) \le 2C \log \frac{1}{1-r}$$ for $r$ close enough to $1$ and we are done!