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# 数学代写|数学分析作业代写Mathematical Analysis代考|MATH2400 The Hahn-Banach Theorem

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## 数学代写|数学分析代写MATHEMATICAL ANALYSIS代考|The Hahn-Banach Theorem

The importance of the Hahn-Banach theorem cannot be overstated. The results following theorem 6.4.4 represent only a sample of the wide range of applications of the Hahn-Banach theorem. Unlike the three major theorems of the previous section, the Hahn-Banach theorem does not require completeness.

The Hahn-Banach theorem has many guises, and one of them is an extension theorem. The following example shows that, from the purely algebraic perspective, extending a linear functional on a subspace $M$ of a vector space $X$ is a trivial task. Compare the following example to theorem 6.4.4.

Example 1. Let $M$ be a subspace of a vector space $X$, and let $\lambda$ be a linear functional on $M$. Then $\lambda$ can be extended to a linear functional on $X$.

Let $S_1$ be a basis for $M$, and choose a subset $S_2$ of $X$ such that $S_1 \cup S_2$ is a basis for $X$. Define a function $\Lambda: S \rightarrow \mathbb{C}$ as follows:
$$\Lambda(x)= \begin{cases}\lambda(x) & \text { if } x \in S_1, \ 0 & \text { if } x \in S_2\end{cases}$$
Extend the function $\Lambda$ by linearity to a functional $\Lambda$ on $X$. The restriction of $\Lambda$ to $M$ is clearly $\lambda$

One of the corollaries of the Hahn-Banach theorem (theorem 6.4.5) is a powerful separation theorem. Earlier in the book, we saw examples of separation theorems by linear functionals, albeit in a slightly different context. See example 10 in section 4.7. The following example shows, once again, that, from the algebraic point of view, the problem of separating a subspace from a point outside it is a simple one. Compare the result below to theorem 6.4.5.

## 数学代写|数学分析代写MATHEMATICAL ANALYSIS代考|The Spectrum of an Operator

The spectrum of a square matrix $A$ is simply its set of eigenvalues, and the eigenvalues of $A$ are easy to characterize. They are exactly the complex numbers $\lambda$ for which the matrix $A-\lambda I$ is not invertible. We recall the simple fact that $A-\lambda I$ is not invertible if and only if the linear operator $T$ it generates on $\mathbb{K}^n$ is not oneto-one, and this is the case if and only if $T$ in not onto.

The definition of the spectrum of an operator $T$ on an infinite-dimensional space is exactly the same as it is for a matrix. The stark distinction here is that not every point in the spectrum of an operator on an infinite-dimensional space is an eigenvalue. This is because such an operator may be one-to-one but not onto or conversely. See example 1 . Thus the spectrum consists of two main parts: the complex numbers $\lambda$ for which $T-\lambda I$ is not one-to-one (the eigenvalues) and those for which $T-\lambda I$ is one to one but not onto. The spectrum of an operator $T$ often carries valuable information about $T$, and, in some cases, the eigenvalues of an operator and the corresponding eigenvectors completely define the operator.
Definition. A Banach algebra is a Banach space $X$ that is also an algebra with a multiplicative identity $I$ such that the norm satisfies the following additional assumptions:
(a) $|I|=1$, and
(b) $|S T| \leq|S| T |$ for all $S$ and $T$ in $X$.
We know that the set $\mathcal{L}(X)$ of bounded linear operators on a Banach space $X$ is a Banach space. In fact, $\mathcal{L}(X)$ is a Banach algebra with the composition of operators as the multiplication operation. The composition of two operators $S$ and $T$ is usually denoted by ST rather than SoT. Property (a) is obvious, and property (b) follows from the inequalities $|(S T)(x)|=|S(T(x))| \leq|S||T(x)| \leq|S \mid||T||x|$.

## 数学代写数学分析代写MATHEMATICAL ANALYSIS代考|The Hahn-Banach Theorem

Hahn-Banach 定理的重要性怎么强调都不为过。定理 $6.4 .4$ 之后的结果仅代表 Hahn-Banach 定理广泛应用的一个样本。与上 一节的三大定理不同，Hahn-Banach 定理不需要完富性。

Hahn-Banach 定理有很多形式，其中之一是扩展定理。下面的例子表明，从纯代数的角度来看，在子空间上扩展线性迄函 $M$ 向 量空间的 $X$ 是一项微不足道的任务。将以下示例与定理 6.4.4 进行比较。

$\Lambda(x)=\left{\lambda(x) \quad\right.$ if $x \in S_1, 0 \quad$ if $x \in S_2$

Hahn-Banach 定理（定理 6.4.5）的推论之一是强大的分离定理。在本书的前面部分，我们看到了线性泛函分离定理的示例， 尽管上下文略有不同。请参见第 $4.7$ 节中的示例 10。下面的例子再次表明，从代数的角度来看，将子空间与其外部的点分离是 个简单的问题。将下面的结果与定理 $6.4 .5$ 进行比较。

## 数学代写|数学分析代写MATHEMATICAL ANALYSIS代考|The Spectrum of an Operator

(a) $|I|=1$, 和
(b) $|S T| \leq|S| T \mid$ 对所有人 $S$ 和 $T$ 在 $X$.

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