收敛性与紧集列

问题

例 4.2.6
$\mathcal{E}(\Omega)$ ：设 $\Omega$ 是 $\mathbb{R}^n$ 中的任意开集，又设 $K_m$ 是一串相对于 $\Omega$ 的紧集，适合

$K_1 \subset K_2 \subset K_3 \subset \cdots \subset K_m \subset \cdots \subset \Omega, \quad \bigcup_{m=1}^\infty K_m = \Omega,$

并令

$\varphi\|_m = \sum_{|\alpha| \leq m} \max_{x \in K_m} |\partial^\alpha \varphi(x)| \quad (m = 1, 2, \ldots)$

用 $\mathcal{E}(\Omega)$ 表示带有可数范数 ${\{\|\varphi\|_m}\}_{m=1}^\infty$ 的线性空间 $C^\infty(\Omega)$ ，则 $\mathcal{E}(\Omega)$ 是一个 $B^*$ 空间（易证）。按定义，为了 $\varphi_j \to 0 \ (\mathcal{E}(\Omega))$ ，必须

$\forall m \in \mathbb{N}, \ \exists N = N(\varepsilon, m),$ , 使得 $\max_{x \in K_m} |\partial^\alpha \varphi_j(x)| < \varepsilon \quad (\forall |\alpha| \leq m, \ \forall j > N)$ ,

或者说，在每一个相对于 $\Omega$ 的紧集 $K_m$ 上，直到 $m$ 次导数一致收敛于 $0$ 。

注
请证明： $\mathcal{E}(\Omega)$ 上的收敛性与紧集列 ${K_m}$ 的选择无关，即按不同的紧集列定义出的两组可数半范数是等价的。

思考

细腻地探讨。定义了一个子集的空间，通过序列定义的完备空间，其中每个有限范数定义在紧子集上。注意，这样的空间表现出一致收敛的属性。

界定标准

我在探讨\varphi_j \to 0在\mathcal{E}(\Omega)中的收敛性。要证明，任意\varepsilon和m，存在N使得对于所有的\alpha ≤ m和j > N，最大值\max_{x \in K_m} |\partial^\alpha \varphi_j(x)|小于\varepsilon。这提示我，在不同的紧致子集K_m的序列选择差异时，收敛的拓扑结构是一致的。

证明半范数等价性

我正在证明，来自于不同紧致子集序列的半范数是等价的。特别地，我看着任何两个递增到Ω的紧致子集序列 {K_m} 和 {K_m'}，存在函数 c_{m,m'}，使得对于所有 \varphi \in \mathcal{E}(\Omega)，存在常数 C_{m,m'} > 0 使得 |\varphi|m \leq C{m,m'} |\varphi|_{m'}'。这让我思考，如何通过这些函数来比对那些序列。

选择序列集

考虑选择紧致子集序列 {K_m} 和 {K_m'}，其中 K_m ⊂ K_{m+1}，且会合成Ω。

证明半范数族拓扑等价性

进一步细化范围

探讨範數等價性

我正在思考通过选择合适的泛函定义，如何建立範數之间的等价关系，从而定义相同的拓扑结构。

回答

To prove that the convergence in $\mathcal{E}(\Omega)$ is independent of the choice of the sequence of compact subsets $K_m$ , we need to show that any two such sequences define equivalent families of seminorms, and thus the same topology on $\mathcal{E}(\Omega)$ .

Let’s consider two sequences of compact subsets:

$\{K_m\}_{m=1}^\infty$ , with $K_1 \subset K_2 \subset \cdots \subset \Omega$ and $\bigcup_{m=1}^\infty K_m = \Omega$ .
$\{K_m'\}_{m=1}^\infty$ , with $K_1' \subset K_2' \subset \cdots \subset \Omega$ and $\bigcup_{m=1}^\infty K_m' = \Omega$ .

Define the seminorms associated with each sequence:

For $\{K_m\}$ : $\|\varphi\|_m = \sum_{|\alpha| \leq m} \max_{x \in K_m} |\partial^\alpha \varphi(x)|.$
For $\{K_m'\}$ : $\|\varphi\|_m' = \sum_{|\alpha| \leq m} \max_{x \in K_m'} |\partial^\alpha \varphi(x)|.$

Our goal is to show that these seminorms are equivalent, meaning that for each $m$ , there exist constants $C_1, C_2 > 0$ and indices $m', m''$ such that:

$\|\varphi\|_m \leq C_1 \|\varphi\|_{m'}'$ for all $\varphi \in \mathcal{E}(\Omega)$ .
$\|\varphi\|_{m'}' \leq C_2 \|\varphi\|_{m''}$ for all $\varphi \in \mathcal{E}(\Omega)$ .

Proof:

For any fixed $m$ , there exists $m'$ such that $K_m \subset K_{m'}'$ :
- Since $\bigcup_{k=1}^\infty K_k' = \Omega$ and $K_m$ is compact, it must be contained in some $K_{m'}'$ .
- Choose $m' \geq m$ so that $K_m \subset K_{m'}'$ .
Relate the seminorms using the inclusion $K_m \subset K_{m'}'$ :
- For all $\varphi \in \mathcal{E}(\Omega)$ : $\|\varphi\|_m = \sum_{|\alpha| \leq m} \max_{x \in K_m} |\partial^\alpha \varphi(x)| \leq \sum_{|\alpha| \leq m} \max_{x \in K_{m'}'} |\partial^\alpha \varphi(x)| \leq \|\varphi\|_{m'}'.$
Similarly, for any fixed $m'$ , there exists $m''$ such that $K_{m'}' \subset K_{m''}$ :
- Since $\bigcup_{k=1}^\infty K_k = \Omega$ and $K_{m'}'$ is compact, it must be contained in some $K_{m''}$ .
- Choose $m'' \geq m'$ so that $K_{m'}' \subset K_{m''}$ .
Relate the seminorms using the inclusion $K_{m'}' \subset K_{m''}$ :
- For all $\varphi \in \mathcal{E}(\Omega)$ : $\|\varphi\|_{m'}' = \sum_{|\alpha| \leq m'} \max_{x \in K_{m'}'} |\partial^\alpha \varphi(x)| \leq \sum_{|\alpha| \leq m'} \max_{x \in K_{m''}} |\partial^\alpha \varphi(x)| \leq \|\varphi\|_{m''}.$
Combining the inequalities:
- This shows that for appropriate indices $m', m''$ , the seminorms satisfy: $\|\varphi\|_m \leq \|\varphi\|_{m'}' \leq \|\varphi\|_{m''}.$
- Therefore, the seminorms $\|\cdot\|_m$ and $\|\cdot\|_m'$ are equivalent up to reindexing.

Conclusion:

Since the seminorms defined by different sequences of compact subsets are equivalent, they induce the same topology on $\mathcal{E}(\Omega)$ . Consequently, convergence in $\mathcal{E}(\Omega)$ does not depend on the specific choice of the sequence $\{K_m\}$ .

Answer:

Because any two sequences of compact subsets increasing to Ω can be used to define equivalent families of seminorms, the convergence in $\mathcal{E}(\Omega)$ is independent of the choice of compact subsets $K_m$ ; thus, the different seminorms define the same topology.

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