数学代写|随机过程作业代写Stochastic Processes代考|The Gelfand transform and its applications

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

数学代写|随机过程作业代写Stochastic Processes代考|Banach algebras

6.1.1 Motivating examples Let $S$ be a locally compact Hausdorff topological space, and $C_{0}(S)$ be the space of continuous functions on $S$ that vanish at infinity, equipped with the supremum norm. Until now, we treated $C_{0}(S)$ as a merely Banach space. This space, however, unlike general Banach spaces, has an additional algebraic structure: two functions on $S$ may not only be added, they may be multiplied. The product of two functions, say $x$ and $y$, is another function, a member of $C_{0}(S)$ given by
$$(x y)(p)=x(p) y(p)$$
The operation so defined is associative and enjoys the following properties that relate it to the algebraic and topological structure introduced before:
(a) $|x y| \leq|x||y|$,
(b) $(\alpha x) y=\alpha(x y)=x(\alpha y), \alpha \in \mathbb{R}$,
(c) $x\left(y_{1}+y_{2}\right)=x y_{1}+x y_{2}$,
(d) $x y=y x$.
Moreover, if $S$ is compact, then $C_{0}(S)=C(S)$ has a unit, an element $u=1_{S}$ such that $u x=x u=x$ for all $x \in C(S)$. We have $|u|=1$.

For another example, let $X$ be a Banach space and let $\mathcal{L}(\mathbb{X})$ be the space of bounded linear operators on $X$. If we define $x y$ as the composition of two linear maps $x$ and $y \in \mathbb{X}$, it will be easy to see that conditions (a) -(c) above are satisfied (use 2.3.11), and that the identity operator is a unit. Such multiplication does not, however, satisfy condition (d).
Yet another example is the space $l^{1}(\mathbb{Z})$ of absolutely summable sequences $\left(\xi_{n}\right){n \in \mathcal{Z}}$. If we define the product of two sequences $x=\left(\xi{n}\right)_{n \in \mathbb{Z}}$

and $y=\left(\eta_{n}\right){n \in \mathcal{Z}}$ as their convolution $$\left(\sum{i=-\infty}^{\infty} \xi_{n-i} \eta_{i}\right){n \in Z}$$ then conditions $(a)-(d)$ are satisfied. Also, the sequence $e{0}=\left(\delta_{n, 0}\right)_{n \geq 0}$ plays the role of the unit.

The final, very important example is the space $\mathbb{B M}(\mathbb{R})$ of (signed) Borel measures on $\mathbb{R}$ with convolution (2.14) as multiplication.

As a generalization of these examples, let us introduce the following definition.
6.1.2 Definition A Banach algebra $\mathbb{A}$ is a Banach space, equipped with an additional associative operation $\mathbb{A} \times \mathbb{A} \ni(x, y) \mapsto x y \in \mathbb{A}$, such that conditions (a) $-$ (c) above hold, and additionally
$\left(c^{\prime}\right) \quad\left(y_{1}+y_{2}\right) x=y_{1} x+y_{2} x$

数学代写|随机过程作业代写Stochastic Processes代考|The Gelfand transform

6.2.1 Multiplicative functionals As we have seen in the previous chapter, one may successfully study Banach spaces with the help of bounded linear functionals. However, to study Banach algebras, linear functionals will not suffice for they do not reflect the structure of multiplication in such algebras. We need to use linear multiplicative functionals. By definition, a linear multiplicative functional on a Banach algebra is a linear functional $F$, not necessarily bounded, such that, $F(x y)=F x F y$. In other words $F$ is a homomorphism of $\mathbb{A}$ and the algebra $\mathbb{R}$ equipped with the usual multiplication. To avoid trivial examples, we assume that $F$ is not identically equal to 0 .
6.2.2 Exercise Prove that if $F$ is a multiplicative functional on a Banach algebra with unit $u$ then $F u=1$.
6.2.3 Lemma Linear multiplicative functionals are bounded; their norms never exceed $1 .$

Proof Suppose that for some $x_{0},\left|F x_{0}\right|>\left|x_{0}\right|$. Then for $x=\frac{\operatorname{sgn} F x_{0}}{F x_{0} \mid} x_{0} \in$ A. we have that $|x|<1$ and $F x=1$. Let $y$ be defined as in 6.1.4. We have $F(x y+x-y)=0$, but on the other hand this equals $F(x y)+F x-F y=$ $(F x)(F y)+F x-F y=1$, a contradiction.
6.2.4 Remark The idea of the above proof becomes even more clear if $\mathbb{A}$ has a unit. We construct $x$ as before but then take $y$ defined in 6.1.5 to arrive at the contradiction:
$$1=F u=F y(F u-F x)=F y(1-1)=0 .$$
6.2.5 Exercise Suppose that an algebra $\mathbb{A}$ has a right or left approximate unit. Show that multiplicative functionals on $\mathbb{A}$ have norm 1 .
6.2.6 Example Let $C(S)$ be the algebra of continuous functions on a compact topological space. A linear functional on $C(S)$ is multiplicative iff there exists a point $p \in S$ such that $F x=x(p)$ for all $x \in C(S)$.

数学代写|随机过程作业代写Stochastic Processes代考|Examples of Gelfand transform

6.3.1 Probability generating function The space $l^{1}=l^{1}\left(\mathbb{N}{0}\right)$ of absolutely summable sequences $x=\left(\xi{n}\right)_{n \geq 0}$ is a Banach algebra with con-

volution $(6.5)$ as multiplication. We already know (see $5.2 .3$ or $5.2 .16$ ) that a linear functional on $l^{1}$ has to be of the form
$$F x=\sum_{n=0}^{\infty} \xi_{n} \alpha_{n}$$
where $\alpha_{i}=F e_{i}=F\left(\delta_{i, n}\right){n \geq 0}$, and $\sup {n \geq 0}\left|\alpha_{n}\right|<\infty$. To determine the form of a multiplicative functional, observe first that $e_{2}=e_{1} * e_{1}$ and, more generally, $e_{n}=e_{1}^{n *}, n \geq 0$ where the last symbol denotes the $n$th convolution power of $e_{1}$. Also, $F e_{0}=1$. Hence $\alpha_{n}=\alpha^{n}, n \geq 0$, where $\alpha=\alpha_{1}$. Therefore, if $F$ is multiplicative, then there exists a real $\alpha$ such that $F x=\sum_{n=0}^{\infty} \alpha^{n} \xi_{n}$. This $\alpha$ must belong to the interval $[-1,1]$ for otherwise the sequence $\alpha_{n}$ would be unbounded. Conversely, any functional of this form is multiplicative.

This proves that a multiplicative functional on $l^{1}$ may be identified with a number $\alpha \in[-1,1]$. Moreover, one may check that the weak* topology restricted to the set of multiplicative functionals is just the usual topology in the interval $[-1,1]$. In other words, the image via the Gelfand transform of the vector $x \in l^{1}$ is a function $\hat{x}$ on $[-1,1]$ given by $\hat{x}(\alpha)=\sum_{i=0}^{\infty} \alpha^{i} \xi_{i}$. However, in this case, the space of multiplicative functionals is “too rich”; the information provided by $\alpha \in[-1,0)$ is superfluous. Therefore, in probability theory it is customary to restrict the domain of $\hat{x}$ to the interval $[0,1]$.

We note that the interval $[-1,1]$ is compact, and that this is related to the fact that $l^{1}$ has a unit. It is also worth noting that the image via Gelfand transform of a sequence in $l^{1}$ is not just “any” continuous function on $[-1,1]$; this function must be expandable into a series $\hat{x}(\alpha)=$ $\sum_{n=0}^{\infty} \xi_{n} \alpha^{n}$ and in particular be infinitely differentiable and analytic. This illustrates the fact that the image of a Banach algebra via the Gelfand transform is usually a subset of the algebra $C_{0}(\mathcal{M})$.

数学代写|随机过程作业代写Stochastic Processes代考|Banach algebras

6.1.1 激励例子让小号是局部紧致 Hausdorff 拓扑空间，并且C0(小号)是连续函数的空间小号消失在无穷远，配备了至上规范。直到现在，我们对待C0(小号)作为一个单纯的 Banach 空间。然而，这个空间与一般 Banach 空间不同，它有一个额外的代数结构：两个函数小号不仅可以相加，还可以相乘。两个函数的乘积，比如说X和是, 是另一个函数，是C0(小号)由
(X是)(p)=X(p)是(p)

（a）|X是|≤|X||是|,
(b)(一种X)是=一种(X是)=X(一种是),一种∈R,
(c)X(是1+是2)=X是1+X是2,
(d)X是=是X.

6.1.2 定义 A Banach 代数一种是一个 Banach 空间，配备了一个附加的关联操作一种×一种∋(X,是)↦X是∈一种, 使得条件 (a)−(c) 上述保留，另外
(C′)(是1+是2)X=是1X+是2X

数学代写|随机过程作业代写Stochastic Processes代考|The Gelfand transform

6.2.1 乘性泛函 正如我们在前一章中所看到的，借助有界线性泛函可以成功地研究巴拿赫空间。然而，要研究 Banach 代数，线性泛函是不够的，因为它们不能反映此类代数中的乘法结构。我们需要使用线性乘法泛函。根据定义，Banach 代数上的线性乘法泛函是线性泛函F，不一定有界，这样，F(X是)=FXF是. 换句话说F是一个同态一种和代数R配备了通常的乘法。为了避免琐碎的例子，我们假设F不完全等于 0 。
6.2.2 练习证明如果F是带单位的 Banach 代数上的乘法泛函在然后F在=1.
6.2.3 引理 线性乘法泛函是有界的；他们的规范从不超过1.

6.2.4 备注 上述证明的思想变得更加清晰，如果一种有一个单位。我们构建X像以前一样，然后采取是在 6.1.5 中定义以得出矛盾：
1=F在=F是(F在−FX)=F是(1−1)=0.
6.2.5 练习假设一个代数一种有一个右或左的近似单位。证明乘法泛函一种有范数 1 。
6.2.6 示例让C(小号)是紧拓扑空间上的连续函数的代数。线性泛函C(小号)当存在一个点时是乘法的p∈小号这样FX=X(p)对全部X∈C(小号).

数学代写|随机过程作业代写Stochastic Processes代考|Examples of Gelfand transform

6.3.1 概率生成函数空间l1=l1(ñ0)绝对可和序列X=(Xn)n≥0是具有 con- 的 Banach 代数

FX=∑n=0∞Xn一种n

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