Let $\mathcal{P}$ be the space of all polynomials endowed with the norm $$\|p\|:=\sup_{|z|\leq1}|p(z)|=\sup_{|z|=1}|p(z)|.$$ Let $S$ be the forward shift operator on $\ell^{2}$, that is, $$S(x_0,x_1,\ldots):=(0,x_0,x_1,\ldots).$$ For $p\in\mathcal{P}$, let $p'\in\mathcal{P}$ denote the (formal) derivative.
Is the operator $$\mathcal{P}\ni p\mapsto p'(S)\in B(\ell^{2})$$ bounded?
If this is true, then the proof must rely on the structure of $S$. Namely, it is in general not true that $$\mathcal{P}\ni p\mapsto p'(A)\in B(H)$$ is bounded for arbitrary Hilbert spaces $H$ and operators $A\in B(H)$. For example, if $H=\mathbb{C}$ and $A=1$. Indeed, if $p_{n}=z^{n}$, then $|p_{n}'(1)|=n$ and $\|p_{n}\|=1$ for all $n\in\mathbb{N}$.
Assume the spectral radius $r(A)$ of $A$ is greater or equal $1.$ Then the mapping $p\to p'(A)$ is unbounded. Indeed for $p(z)=z^n$ we have $$\|p'(A)\|=n\,\|A^{n-1}\|\ge n\,r(A^{n-1})=n\,r(A)^{n-1}\ge n$$ In particular the mapping is unbounded for the shift operator, whose spectral radius is equal $1.$
There are operators for which the mapping is bounded. Let $A$ be a normal operator so that $r(A)<1.$ Then $p'(A)$ is a normal operator. Thus $$\|p'(A)\|=\max_{z\in \sigma(A)}|p'(z)|\le \max_{|z|\le r(A)}|p'(z)|$$ By the Cauchy integral formula we have $$p'(z)={1\over 2\pi i}\int\limits_{|w|=1}{p(w)\over (w-z)^2}\,dw, \quad |z|\le r(A)$$ Hence $$|p'(z)|\le (1-r(A))^{-2}\max_{|w|\le 1}|p(w)|$$ Thus we finally obtain $$\|p'(A)\|\le (1-r(A))^{-2}\max_{|w|\le 1}|p(w)|$$