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# 数学代写|泛函分析代写Functional Analysis代考|MATH633 First Examples

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## 数学代写|泛函分析代写FUNCTIONAL ANALYSIS代考|Finite-Dimensional Spaces

The aim of this section is to prove that every finite-dimensional normed space is a Banach space. This will be deduced as an easy consequence of the fact that every two norms on a finite-dimensional normed space are equivalent, in the sense made precise in the next definition.

Definition $1.32$ (Equivalent norms). Two norms $|\cdot|$ and $|\cdot| |$ on a vector space $X$ are equivalent if there exist constants $0<c \leqslant C<\infty$ such that for all $x \in X$ we have
$$c|x| \leqslant|x| \leqslant C|x| .$$

## 数学代写|泛函分析代写FUNCTIONAL ANALYSIS代考|Compactness

Let $X$ be a normed space. By Theorem 1.38, the collections of bounded subsets of $X$ and relatively compact subsets of $X$ coincide if and only if $X$ is finite-dimensional. Thus, in infinite-dimensional spaces, relative compactness is a stronger property than boundedness. The purpose of the present section is to record some easy but useful general results on compactness that will be frequently used. Compactness in the spaces $C(K)$ and $L^{p}(\Omega)$ will be studied in the next chapter, and compact operators, that is, operators which map bounded sets into relatively compact sets, are studied in Chapter $7 .$

By a general result in the theory of metric spaces (Theorem D.10), every relatively compact set in a normed space is totally bounded, and the converse holds in Banach spaces. This fact is used in the proof of the following necessary and sufficient condition for compactness. For sets $A$ and $B$ in a vector space $V$ we write
$$A+B:={u+v: u \in A, v \in B} .$$
Proposition 1.40. A subset $S$ of a Banach space $X$ is relatively compact if and only if for all $\varepsilon>0$ there exists a relatively compact set $K_{\varepsilon} \subseteq X$ such that $S \subseteq K_{\varepsilon}+B(0 ; \varepsilon)$.
Proof ‘If’: The existence of the sets $K_{\varepsilon}$ implies that $S$ is totally bounded and hence relatively compact, for if the balls $B\left(x_{1, \varepsilon} ; \varepsilon\right), \ldots, B\left(x_{n_{\varepsilon}, \varepsilon} ; \varepsilon\right)$ cover $K_{\varepsilon}$, then the balls $B\left(x_{1, \varepsilon} ; 2 \varepsilon\right), \ldots, B\left(x_{n_{\varepsilon}, \varepsilon} ; 2 \varepsilon\right) \operatorname{cover} S$.
‘Only if’: This is trivial (take $K_{\varepsilon}=S$ for all $\varepsilon>0$ ).
The convex hull of a subset $F$ of a vector space $V$ is the smallest convex set in $V$ containing $F$. This set is denoted by $\operatorname{co}(F)$. When $F$ is a subset of a normed space, the closure of $\operatorname{co}(F)$ is denoted by $\overline{\operatorname{co}}(F)$ and is referred to as the closed convex hull of $F$.
As a first application of Proposition $1.40$ we have the following result.

## 数学代写|泛函分析代写FUNCTIONAL ANALYSIS代考|Finite-Dimensional Spaces

$$c|x| \leqslant|x| \leqslant C|x| .$$

## 数学代写|泛函分析代写FUNCTIONAL ANALYSIS代考|Compactness

$$A+B:=u+v: u \in A, v \in B$$ $B\left(x_{1, \varepsilon} ; 2 \varepsilon\right), \ldots, B\left(x_{n_{\Omega \varepsilon} \varepsilon} ; 2 \varepsilon\right)$ cover $S$.
‘仅当’: 这是微不足道的 (平取 $K_{\varepsilon}=S$ 对所有人 $\varepsilon>0$ )。

## MATLAB代写

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