泛函分析证明(1)

证明0:

令 $C = \{\sum_{j = 1}^{n} \lambda_{j}\ x_{j}| x_{1}, x_{2}, ..., x_{n} \in K, \lambda_{1}, \lambda_{2}, ..., \lambda_{n} \in F, n \in N \}$

令 $lin(K) = \cap A_{i}, K \subset A_{i}$,其中 $A_{i}$ 是线性子空间

易知 $K \subset C$ 且 C 是线性空间,故 $C \in \{A_{i}\}$

所以 $lin(K) = C = \{\sum_{j = 1}^{n} \lambda_{j} x_{j}| x_{1}, x_{2}, ..., x_{n} \in K, \lambda_{1}, \lambda_{2}, ..., \lambda_{n} \in F, n \in N \}$

证明1:

先证明 $S \subset C = \{\sum_{j = 1}^{n} \lambda_{j} x_{j}| x_{1}, x_{2}, ..., x_{n} \in K, \sum_{j = 1}^{n} \lambda_{j} = 1, n \in N \}$ 中的 C 是凸集

任取 $c_{1}, c_{2} \in C, c_{i} = \sum_{j = 1}^{k} \lambda_{ij} x_{j}, \sum_{j = 1}^{k} \lambda_{ij} = 1, i = 1, 2$,则有

$\lambda c_{1} + (1 - \lambda) c_{2} = \lambda \sum \lambda_{1j} x_{j} + (1 - \lambda) \sum \lambda_{2j} x_{j}$

$= \sum (\lambda \lambda_{1j} + (1 - \lambda) \lambda_{2j}) x_{j}$,由

$\sum (\lambda \lambda_{1j} + (1 - \lambda) \lambda_{2j}) = \lambda \sum \lambda_{1j} + \sum \lambda_{2j} - \lambda \sum \lambda_{2j}$

$= \lambda + 1 - \lambda = 1$

所以 C 是凸集,再由证明0的思路可证

$co(S) = C = \{\sum_{j = 1}^{n} \lambda_{j} x_{j}| x_{1}, x_{2}, ..., x_{n} \in S, \sum_{j = 1}^{n} \lambda_{j} = 1, n \in N \}$


证明2:

假设结论不成立,即 $x \notin C, x \in X\C$

由于 $C$ 是闭集,所以 $X\C$ 是开集,即 $\exists \epsilon > 0$,满足 $B(x, \epsilon) \subset X\C$

但是 $\forall \epsilon > 0, \exists k > N$,使得 $\left \| x_{N} - x \right \| < \epsilon$,即 $x_{k} \in B(x, \epsilon)$,而 $x_{k} \notin X\C$

故矛盾,所以 $x \in C$

证明3:

对于柯西序列 $(x_{n})_{n = 1}^{\infty}$ 上的收敛于 $x$ 的子序列 $(x_{nk})_{k = 1}^{\infty}$,有

  • $\left \| x_{n} - x_{nk} \right \| < \frac{\epsilon}{2}, n > N_{1}$

  • $\left \| x_{nk} - x \right \| < \frac{\epsilon}{2}, nk > N_{2} > N_{1}$

故取 $n > N_{2}$ 时,有

$\left \| x_{n} - x \right \| \leq \left \| x_{n} - x_{nk} \right \| + \left \| x_{nk} - x \right \| = \epsilon$

所以 $\underset{n \to \infty}{lim} x_{n} = x$,结论成立


证明5:

若有 $A \subset \underset{x \in F_{\epsilon}}{\cup} B(x, \epsilon), F_{\epsilon} = \{x_{i}, i = 0, 1, 2, ..., n\}$,对于任意 ${x}' \in B(x_{i}, \epsilon)$,有

$\left \| x_{0} - {x}' \right \| \leq \left \| x_{0} - x_{i} \right \| + \left \| x_{i} - {x}' \right \| = \left \| x_{0} - x_{i} \right \| + \epsilon$

取 $r = max(\left \| x_{0} - x_{i} \right \|)$

易知 ${x}' \in B[x_{0}, r]$,故结论成立。

要证明反之不成立,则只需考虑无穷维空间的单位基底作为子集,易知该集合有界,但取 $\epsilon < \frac{\sqrt{2}}{2}$ 时,$F_{\epsilon}$ 中元素为无穷个,则知反之不成立

xiehao

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