# 数学代写|泛函分析作业代写Functional Analysis代考|The Continuous Functional Calculus for Normal Operators

## 数学代写|泛函分析作业代写Functional Analysis代考|The Continuous Functional Calculus for Normal Operators

Every polynomial in the real variables $x$ and $y$ can be written as a polynomial in the variables $z$ and $\bar{z}$ by substituting $z=x+i y, \bar{z}=x-i y$. For example, $x^2+y^2=z \bar{z}$. For polynomials $p(z, \bar{z})=\sum_{i, j=0}^k c_{i j} z^i \bar{z}^j$ and normal operators $T \in \mathscr{L}(H)$, we define
$$p\left(T, T^{\star}\right):=\sum_{i, j=0}^k c_{i j} T^i \bar{T}^{* j} .$$
The crucial result that enables us to extend the continuous functional calculus to normal operators is the following spectral mapping theorem.
Proposition 8.21. If $T \in \mathscr{L}(H)$ is normal and $p$ is a polynomial in $z$ and $\bar{z}$, then
$$\sigma\left(p\left(T, T^{\star}\right)\right)={p(\lambda, \bar{\lambda}): \lambda \in \sigma(T)}$$
Proof By Proposition 8.14, every $\lambda \in \sigma(T)$ is an approximate eigenvalue of $T$, that is, there exists a sequence $\left(x_n\right){n \geqslant 1}$ of norm one vectors such that $\lim {n \rightarrow \infty} T x_n-\lambda x_n=0$. Then $\lim {n \rightarrow \infty} T^{\star} x_n-\bar{\lambda} x_n=0$ by Proposition 8.13. This implies $\lim {n \rightarrow \infty} p\left(T, T^{\star}\right) x_n-$ $p(\lambda, \bar{\lambda}) x_n=0$, so $p(\lambda, \bar{\lambda})$ is an approximate eigenvalue for $p\left(T, T^{\star}\right)$. In particular, $p(\lambda, \bar{\lambda}) \in \sigma\left(p\left(T, T^{\star}\right)\right)$. This proves the inclusion ‘ $\supseteq$ ‘.

For the inclusion ‘ $\subseteq$ ‘, fix an arbitrary $\mu \in \sigma\left(p\left(T, T^{\star}\right)\right)$. We wish to prove the existence of a $\lambda \in \sigma(T)$ such that $p(\lambda, \bar{\lambda})=\mu$.

Step 1 – Fixing $\varepsilon>0$, we claim that there is a nonzero closed subspace $Y$ of $H$, invariant under both $T$ and $T^{\star}$, such that
$$\left|\left.\left(p\left(T, T^{\star}\right)-\mu I\right)\right|_Y\right|<\varepsilon$$
To prove the claim, let $S:=p\left(T, T^{\star}\right)-\mu I$. This operator is normal and we have $0 \in$ $\sigma(S)$. Let $R:=S^{\star} S$. Arguing as above, we find that $0 \in \sigma(R)$. Consider the continuous function $f:[0, \infty) \rightarrow[0,1]$ given by
$$f(t):= \begin{cases}1, & 0 \leqslant t \leqslant \varepsilon / 2 \ 2(1-t / \varepsilon), & \varepsilon / 2 \leqslant t \leqslant \varepsilon \ 0, & t \geqslant \varepsilon\end{cases}$$
and let $f(R)$ be the selfadjoint operator obtained from the continuous functional calculus for selfadjoint operators (Theorem 8.20 ).

## 数学代写|泛函分析作业代写Functional Analysis代考|Applications of the continuous functional calculus

We now turn to some applications of the continuous functional calculus.
Proposition 8.27 (Square roots). If $T \in \mathscr{L}(H)$ is positive, there exists a unique positive operator $S \in \mathscr{L}(H)$ such that $S^2=T$.
Henceforth, this operator $S$ will be denoted by $T^{1 / 2}$.
Proof Since $T$ is positive, $T$ is selfadjoint and its spectrum is contained in $[0, \infty)$ by Theorem 8.11. Hence, $f(t)=\sqrt{t}$ is a well defined continuous function on $\sigma(T)$. The operator $S:=f(T)$ is positive by Theorem 8.20 , and it satisfies $S^2=T$ by the properties of the continuous functional calculus. It remains to prove uniqueness. Suppose $\widetilde{S}$ is another positive operator with the property that $\widetilde{S}^2=T$. With $f(t)=\sqrt{t}, g(t)=t^2$, and $h(t)=t$ we have, by the properties of the continuous functional calculus and Theorem 8.25
$$f\left(\tilde{S}^2\right)=f(g(\widetilde{S}))=(f \circ g)(\widetilde{S})=h(\widetilde{S})=\widetilde{S}$$
It follows that $S=f\left(S^2\right)=f(T)=f\left(\widetilde{S}^2\right)=\widetilde{S}$. This completes the uniqueness proof.
Definition 8.28 (Modulus of an operator). The modulus of an operator $T \in \mathscr{L}(H)$ is the positive operator $|T|:=\left(T^{\star} T\right)^{1 / 2}$.

Corollary 8.29. If $T \in \mathscr{L}(H)$ is normal operator, then $|T|=f(T)$, where $f(z):=|z|$.
Proof Let $g(z):=\bar{z} z$. Then $f^2=g$ and therefore $|T|^2=T^{\star} T=g(T)=f^2(T)=$ $(f(T))^2$ by the multiplicativity of the continuous functional calculus. Since by Corollary 8.24 the operator $f(T)$ is positive, the result follows by taking square roots.

We continue with a polar decomposition result. In view of future applications we phrase it for bounded operators $T \in \mathscr{L}(H, K)$, where $H$ and $K$ are Hilbert spaces. The modulus of such an operator is the positive operator $|T|:=\left(T^{\star} T\right)^{1 / 2}$ on $H$. An operator $U \in \mathscr{L}(H, K)$ will be called unitary if it is isometric and surjective. A partial isometry is a bounded operator $V \in \mathscr{L}(H, K)$ for which there exists orthogonal direct sum decomposition $H=H_0 \oplus H_0^{\perp}$ such that $V$ is isometric from $H_0$ into $K$ and zero on $H_0^{\perp}$.

# 泛函分析代考

## 数学代写|泛函分析作业代写Functional Analysis代考|The Continuous Functional Calculus for Normal Operators

$z=x+i y, \bar{z}=x-i y$. 例如， $x^2+y^2=z \bar{z}$. 对于多项式 $p(z, \bar{z})=\sum_{i, j=0}^k c_{i j} z^i \bar{z}^j$ 和普通运营商 $T \in \mathscr{L}(H)$ ，我们定义
$$p\left(T, T^{\star}\right):=\sum_{i, j=0}^k c_{i j} T^i \bar{T}^{* j} .$$

$$\sigma\left(p\left(T, T^{\star}\right)\right)=p(\lambda, \bar{\lambda}): \lambda \in \sigma(T)$$

$$\left|\left(p\left(T, T^{\star}\right)-\mu I\right)\right|_Y \mid<\varepsilon$$ 对于包含 ‘ $\subseteq^{\prime}$ ，修复一个任意的 $\mu \in \sigma\left(p\left(T, T^{\star}\right)\right)$. 我们想证明存在一个 $\lambda \in \sigma(T)$ 这样 $p(\lambda, \bar{\lambda})=\mu$. 第 1 步一修复 $\varepsilon>0$ ，我们声称存在一个非零闭子空间 $Y$ 的 $H$, 在两者下不变 $T$ 和 $T^{\star}$ ，这 样
$$\left|\left(p\left(T, T^{\star}\right)-\mu I\right)\right|_Y \mid<\varepsilon$$

$$f(t):={1, \quad 0 \leqslant t \leqslant \varepsilon / 22(1-t / \varepsilon), \quad \varepsilon / 2 \leqslant t \leqslant \varepsilon 0, \quad t \geqslant \varepsilon$$

## 数学代写|泛函分析作业代写Functional Analysis代考|Applications of the continuous functional calculus

$f(t)=\sqrt{t}$ 是定义明确的连续函数 $\sigma(T)$. 运营商 $S:=f(T)$ 由定理 8.20 为正，并且满足 $S^2=T$ 根据连续泛函的性质。有待证明唯一性。认为 $\widetilde{S}$ 是另一个具有以下属性的正算 子 $\widetilde{S}^2=T$. 和 $f(t)=\sqrt{t}, g(t)=t^2$ ， 和 $h(t)=t$ 根据连续泛函的性质和定理 8.25， 我们有
$$f\left(\tilde{S}^2\right)=f(g(\widetilde{S}))=(f \circ g)(\widetilde{S})=h(\widetilde{S})=\widetilde{S}$$

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