这是一份ucl伦敦大学学院 math0048作业代写的成功案
According to $5.6 .9$, for $|\lambda|>|T|$ there is an operator $(1-T / \lambda)^{-1}$ presenting the sum of the Neumann series; i.e.,
$$
R(T, \lambda)=\frac{1}{\lambda}\left(1-{ }^{T} / \lambda\right)^{-1}=\frac{1}{\lambda} \sum_{k=0}^{\infty} \frac{T^{k}}{\lambda^{k}}
$$
It is clear that
$$
|R(T, \lambda)| \leq \frac{1}{|\lambda|} \cdot \frac{1}{1-|T | /| \lambda \mid} . \triangleright
$$
MATH0048 COURSE NOTES ：
This entails sesquilinearity:
$$
\begin{aligned}
(y, x)=&(y, x){\mathbb{R}}-i(i y, x){\mathbb{R}}=(x, y){\mathbb{R}}-i(x, i y){\mathbb{R}} \
&=(x, y){\mathbb{R}}+i(i x, y){\mathbb{R}}=(x, y)^{*}
\end{aligned}
$$
since
$(x, i y){\mathbf{R}}=1 / 4\left(|x+i y|^{2}-|x-i y|^{2}\right)$ ${ }^{1}{ }^{1} / 4\left(|i||y-i x|^{2}-|-i||i x+y|^{2}\right)=-(i x, y){\mathbb{R}}$.