In the below there is a proposition i have read in a scientific paper , i understood it but not completely just i have missed how the researcher goes here \
\begin{equation} \begin{aligned} \parallel g_{j}-K_{j} \parallel_{2,E}&=\parallel (\lambda - S)^{*}\overline{\partial}g_{j}(\lambda) \parallel+\frac{1}{\pi}\int_{E} \parallel ((\xi - S)^{*}\overline{\partial}^{2}g_{j}(\xi))(\xi - \lambda)^{-1}d_{\mu}(\xi) \parallel \\ \leq & \parallel (\lambda - S)^{*}\overline{\partial}g_{j}(\lambda)\parallel_{2,E}+4R \parallel (\lambda - S)^{*}\overline{\partial}^{2}g_{j}(\lambda) \parallel_{2,E} \end{aligned} \end{equation}
Why \begin{equation} \begin{aligned} \frac{1}{\pi}\int_{E} \parallel ((\xi - S)^{*}\overline{\partial}^{2}g_{j}(\xi))(\xi - \lambda)^{-1}d_{\mu}(\xi) \parallel \leq 4R \parallel (\lambda - S)^{*}\overline{\partial}^{2}g_{j}(\lambda) \parallel_{2,E} \end{aligned} \end{equation}
*Here is the proposition *
Let $H$ be a Hilbert space and $E$ be a bounded disk in $\mathbb{C}$, then there is a constant $C_{E}$ s.t for an
arbitrary $S \in B(H)$ and $g \in W^{2}(E,H)$, we have
$\parallel (I-P) g \parallel_{2,E} \leq C_{E}(\parallel (\lambda - S)^{*} \overline{\partial^{1}}g
\parallel_{2,E}+\parallel (\lambda - S)^{*} \overline{\partial^{2}}g \parallel_{2,E})$
where $P$ is the orthogonal projection of $L^{2}(E,H)$ onto the Bergman space $A^{2}(E,H)$.\
The proof \
Let $g_{j} \in C^{\infty}(\overline{E},H)$ be a sequence which approximates $g$ in the norm $W^{2}$. Then for
fixed $j$ we have
\begin{equation}
\begin{aligned}
\overline{\partial}(g_{j}(\lambda)-(\lambda-S)^{*}\overline{\partial}g_{j})&=\overline{\partial}g_{j}(\lambda)\{(\lambda-S)^{*}\overline{\partial}^{2}g_{j}(\lambda)+\overline{\partial}g_{j}(\lambda)\}
\\=&-(\lambda - S)^{*}\overline{\partial}^{2}g_{j}(\lambda)
\end{aligned}
\end{equation}
Now, by Cauchy-Pompeiu formula one gets,
\begin{equation}
\begin{aligned}
&g_{j}(\lambda)-(\lambda - S)^{*}\overline{\partial}g_{j}(\lambda)-\frac{1}{2 \pi i}\int_{\partial E}
(g_{j}(\xi)-(\xi - S)^{*}\overline{\partial}g_{j}(\xi))(\xi - S)^{-1} d \xi\\=& \frac{1}{\pi}\int_{E}((\xi -
S)^{*}\overline{\partial}^{2}g_{j}(\xi))(\xi - \lambda)^{-1}d_{\mu}(\xi)
\end{aligned}
\end{equation}
Let $K_{j}$ denotes the first integral in the above formula. Then $K_{j}\in A^{2}(E,H)$, therefor
\begin{equation}
\begin{aligned}
\parallel g_{j}-K_{j} \parallel_{2,E}&=\parallel (\lambda - S)^{*}\overline{\partial}g_{j}(\lambda)
\parallel+\frac{1}{\pi}\int_{E} \parallel ((\xi - S)^{*}\overline{\partial}^{2}g_{j}(\xi))(\xi -
\lambda)^{-1}d_{\mu}(\xi) \parallel \\ \leq & \parallel (\lambda -
S)^{*}\overline{\partial}g_{j}(\lambda)\parallel_{2,E}+4R \parallel (\lambda -
S)^{*}\overline{\partial}^{2}g_{j}(\lambda) \parallel_{2,E}
\end{aligned}
\end{equation}
where the second integral in Eq. was majorized as a convolution with a $L^{1}-function$ and $R$ is the
radius of $E$. So that we get the inequalities,
\begin{equation}
\begin{aligned}
\parallel g-Pg\parallel_{2,E}&\leq \parallel g- K_{j}\parallel_{2,E}\\ \leq& \parallel
g-g_{j}\parallel_{2,E}+\parallel g_{j}-K_{j}\parallel_{2,E} \\ \leq & \parallel g-g_{j}\parallel_{2,E}+\parallel
(\lambda - S)^{*}\overline{\partial}g_{j}(\lambda)\parallel_{2,E}+4R \parallel (\lambda -
S)^{*}\overline{\partial}^{2}g_{j}(\lambda) \parallel_{2,E}
\end{aligned}
\end{equation}
which prove proposition as passing to the limit.