这是一份GLA格拉斯哥大学MATHS5074_1作业代写的成功案例
If $f:[a, b] \rightarrow \mathbb{R}$ is continuous, then the function $(f \circ \phi) \phi^{\prime}:[\alpha, \beta] \rightarrow \mathbb{R}$ is integrable and
$$
\int_{\phi(\alpha)}^{\phi(\beta)} f(x) d x=\int_{\alpha}^{\beta} f(\phi(t)) \phi^{\prime}(t) d t
$$
If $f:[a, b] \rightarrow \mathbb{R}$ is integrable and $\phi^{\prime}(t) \neq 0$ for every $t \in(\alpha, \beta)$, then the function $(f \circ \phi)\left|\phi^{\prime}\right|:[\alpha, \beta] \rightarrow \mathbb{R}$ is integrable and
$$
\int^{b} f(x) d x=\int^{\beta} f(\phi(t))\left|\phi^{\prime}(t)\right| d t
$$
MATHS5074_1/MATHS4108_1 COURSE NOTES ：
Now take $\lambda \in \operatorname{Sp}(T)$. Put
$$
\lambda \neq z \Rightarrow g(z):=\frac{f(\lambda)-f(z)}{\lambda-z} ; \quad g(\lambda):=f^{\prime}(\lambda) .
$$
Clearly, $g$ is a holomorphic function (the singularity is “removed”). obtain
$$
g(T)(\lambda-T)=(\lambda-T) g(T)=f(\lambda)-f(T) .
$$
Consequently, if $f(\lambda) \in \operatorname{res}(f(T))$ then the operator $R(f(T), f(\lambda)) g(T)$ is inverse to $\lambda-T$. In other words, $\lambda \in \operatorname{res}(T)$, which is a contradiction. Thus,
$$
f(\lambda) \in \mathbb{C} \backslash \operatorname{res}(f(T))=\operatorname{Sp}(f(T))
$$
i.e., $f(\operatorname{Sp}(T)) \subset \operatorname{Sp}(f(T)) . \triangleright$