Does $\sum_{n=0}^{\infty}|f_n| r^n \leq C \|f\|_{L^{\infty}(|z| \leq r)}$ when $f(z)$ is holomorphic?

Question

Let $$f(z) = \sum_{n=0}^{\infty}f_nz^n$$ be holomorphic on $|z| < R$ and let $0 < r < R$.

I'm wondering if there exists some universal constant $C > 0$ (i.e. independent of $r,R,f$) for which $$\sum_{n=0}^{\infty} |f_n| r^n \leq C \|f\|_{L^{\infty}(|z| \leq r)}$$

I'm aware of similar, but different results

1. Cauchy's inequality : $|f_n|r^n \leq \|f\|_{L^{\infty}(|z| \leq r)}$ for all $n \geq 0$
2. Parseval's identity : $\sum_{n=0}^{\infty} |f_n|^2 r^{2n} = \frac{1}{2\pi}\int_0^{2\pi} |f(re^{i\theta})|^2 d\theta$ which implies $$\left( \sum_{n=0}^{\infty} |f_n|^2 r^{2n} \right)^{\frac{1}{2}} \leq \|f\|_{L^{\infty}(|z| \leq r)}$$
3. Hadamard Multiplication Theorem
4. Hardy's inequality

Answer 1

Such a constant $C$ does not exist. Let us assume that $C$ exits. If $f$ is a bounded holomorphic function on the unit disc, we would have, for every $r<1$, $\sum \vert f_n\vert r^n\leq C \|f\|_\infty$, so that $\sum \vert f_n\vert<+\infty$.

The set of bounded holomorphic functions on the unit disc $D$ is the Hardy space $H^\infty(D)$. The space of holomorphic functions with coefficients in $l^1$ is the Wiener space $A(D)$, which is known to be a proper subspace of $H^\infty(D)$. For instance, a Blaschke product with an inifite number oz zeroes belongs to $H^\infty(D)$ and not to $A(D)$. Good references for the Hardy space and the Wiener space are the basic books of Rudin (Real and complex analysis, Functional analysis).