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# 数学代写|拓扑学代写TOPOLOGY代考|MATH452 THE DEFINITION AND SOME EXAMPLES

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## 数学代写|拓扑学代写TOPOLOGY代考|THE DEFINITION AND SOME EXAMPLES

We begin by restating the definition of a Banach space.
A normed linear space is a linear space $N$ in which to each vector $x$ there corresponds a real number, denoted by $|x|$ and called the norm of $x$, in such a manner that
(1) $|x| \geq 0$, and $|x|=0 \Leftrightarrow x=0$;
(2) $|x+y| \leq|x|+|y|$;
(3) $|\alpha x|=|\alpha||x|$.
The non-negative real number $|x|$ is to be thought of as the length of the vector $x$. If we regard $|x|$ as a real function defined on $N$, this function is called the norm on $N$. It is easy to verify that the normed linear space $N$ is a metric space with respect to the metric $d$ defined by $d(x, y)=|x-y|$. A Banach space is a complete normed linear space. Our main interest in this chapter is in Banach spaces, but there are several points in the body of the theory at which it is convenient to have the basic definitions and some of the simpler facts formulated in terms of normed linear spaces. For this reason, and also to emphasize the role of completeness in theorems which require this assumption, we work in the more general context whenever possible. The reader will find that the deeper theorems, in which completeness hypotheses are necessary, often make essential use of Baire’s theorem.

Several simple but important facts about a normed linear space are based on the following inequality:
$$||x|-|y|| \leq|x-y| .$$
To prove this, it suffices to prove that
$$|x|-|y| \leq|x-y| ;$$
for it follows from (2) that we also have
$$-(|x|-|y|)=|y|-|x| \leq|y-x|=|-(x-y)|=|x-y|,$$
which together with (2) yields (1). We now prove (2) by observing that $|x|=|(x-y)+y| \leq|x-y|+|y|$. The main conclusion we draw from (1) is that the norm is a continuous function:
$$x_{n} \rightarrow x \Rightarrow\left|x_{n}\right| \rightarrow|x| .$$

## 数学代写|拓扑学代写TOPOLOGY代考|CONTINUOUS LINEAR TRANSFORMATIONS

Let $N$ and $N^{\prime}$ be normed linear spaces with the same scalars, and let $T$ be a linear transformation of $N$ into $N^{\prime} .1$ When we say that $T$ is continuous, we mean that it is continuous as a mapping of the metric space $N$ into the metric space $N^{\prime}$. By Theorem $13-\mathrm{B}$, this amounts to the condition that $x_{n} \rightarrow x$ in $N \Rightarrow T\left(x_{n}\right) \rightarrow T(x)$ in $N^{\prime}$. Our main purpose in this section is to convert the requirement of continuity into several more useful equivalent forms and to show that the set of all continuous linear transformations of $N$ into $N^{\prime}$ can itself be made into a normed linear space in a natural way.

Theorem A. Let $N$ and $N^{\prime}$ be normed linear spaces and $T$ a linear transformation of $N$ into $N^{\prime}$. Then the following conditions on $T$ are all equivalent to one another:
(1) T is continuous;
(2) Tis continuous at the origin, in the sense that $x_{n} \rightarrow 0 \Rightarrow T\left(x_{n}\right) \rightarrow 0$;
${ }^{1}$ In the future, whenever we mention two normed linear spaces with a view to considering linear transformations of one into the other, we shall always assumewithout necessarily saying so explicitly-that they have the same scalars.

(3) there exists a real number $K \geq 0$ with the property that $|T(x)|$ $\leq K|x|$ for every $x \& N ;$
(4) if $S={x:|x| \leq 1}$ is the closed unit sphere in $N$, then its image $T(S)$ is a bounded set in $N^{\prime}$.

PRooF. (1) $\Leftrightarrow(2)$. If $T$ is continuous, then since $T(0)=0$ it is certainly continuous at the origin. On the other hand, if $T$ is continuous at the origin, then $x_{n} \rightarrow x \Leftrightarrow x_{n}-x \rightarrow 0 \Rightarrow T\left(x_{n}-x\right) \rightarrow 0 \Leftrightarrow T\left(x_{n}\right)-T(x)$ $\rightarrow 0 \Leftrightarrow T\left(x_{n}\right) \rightarrow T(x)$, so $T$ is continuous.
(2) $\Leftrightarrow$ (3). It is obvious that (3) $\Rightarrow$ (2), for if such a $K$ exists, then $x_{n} \rightarrow 0$ clearly implies that $T\left(x_{n}\right) \rightarrow 0$. To show that $(2) \Rightarrow(3)$, we assume that there is no such $K$. It follows from this that for each positive integer $n$ we can find a vector $x_{n}$ such that $\left|T\left(x_{n}\right)\right|>n\left|x_{n}\right|$, or equivalently, such that $\left|T\left(x_{n} / n\left|x_{n}\right|\right)\right|>1$. If we now put
$$y_{n}=x_{n} / n\left|x_{n}\right|,$$
then it is easy to see that $y_{n} \rightarrow 0$ but $T\left(y_{n}\right) \rightarrow 0$, so $T$ is not continuous at the origin.

## 数学代写|拓扑学代写TOPOLOGY代考|THE DEFINITION AND SOME EXAMPLES

(1) $|x| \geq 0$ ，和 $|x|=0 \Leftrightarrow x=0$;
(2) $|x+y| \leq|x|+|y|$;
(3) $|\alpha x|=|\alpha||x|$.

$$|| x|-| y|| \leq|x-y| .$$

$$|x|-|y| \leq|x-y|$$

$$-(|x|-|y|)=|y|-|x| \leq|y-x|=|-(x-y)|=|x-y|,$$

$$x_{n} \rightarrow x \Rightarrow\left|x_{n}\right| \rightarrow|x| .$$

## 数学代写|拓扑学代写TOPOLOGY代考|CONTINUOUS LINEAR TRANSFORMATIONS

$N \Rightarrow T\left(x_{n}\right) \rightarrow T(x)$ 在 $N^{\prime}$. 我们在本节中的主要目的是将连续性的要求转换为几种更有用的等价形式，并表 明所有连畦线性变换的集合 $N$ 进入 $N^{\prime}$ 本身可以以自然的方式制成规范的线性空间。

(1) $T$ 是连续的;
(2) Ti 在原点连续，即 $x_{n} \rightarrow 0 \Rightarrow T\left(x_{n}\right) \rightarrow 0$;
${ }^{1}$ 将来，每当我们提到两个范数线性空间以考虑一个到另一个的线牲变换时，我们总是会假设而不一定这么明确

(3) 存在一个实数 $K \geq 0$ 与财产 $|T(x)| \leq K|x|$ 对于每个 $x \& N$;
(4) 如果 $S=x:|x| \leq 1$ 是封闭的单位球体 $N$ ，然后它的图像 $T(S)$ 是一个有界集 $N^{\prime}$.

(2) $\Leftrightarrow(3)$ 。显然 $(3) \Rightarrow(2)$ ，因为如果这样 $K$ 存在，那 $/ x_{n} \rightarrow 0$ 明确暗示 $T\left(x_{n}\right) \rightarrow 0$. 为了表明 $(2) \Rightarrow(3)$ ， 我们假设不存在这样的 $K$. 由此得出，对于每个正整数 $n$ 我们可以从找到一个向量 $x_{n}$ 文样 $\left|T\left(x_{n}\right)\right|>n\left|x_{n}\right|$ ，或 等价地，使得 $\left|T\left(x_{n} / n\left|x_{n}\right|\right)\right|>1$. 如果我们现在把
$$y_{n}=x_{n} / n\left|x_{n}\right|,$$

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