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$\set{\sum_k a_k e_k:\abs{a_k}≤c_k}$ 在 $H$ 中紧致

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hbghlyj Posted at 2024-2-11 03:43:25 |Read mode
at.yorku.ca/b/ask-an-analyst/2020/5255.htm#5255(October 23, 2018)
设 $\{e_k\}_{k\ }$是 Hilbert 空间 $H$ 中的正交规范基。如果 $(c_k)_k$ 是一正实数列,使得 $\sum_k c_k{}^2<\infty$,那么集合 $A=\{ \sum_k a_k e_k: |a_k| \leq c_k \}$ 在 $H$ 中是紧致的。

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 Author| hbghlyj Posted at 2024-2-11 04:08:22
The set $A$ can be described in a more convenient way as\[A=\left \{\sum_{k=1}^\infty x_kc_ke_k\,:\,|x_k|\le 1\right \}\]
設\[A\ni u_n:=\sum_{k=1}^\infty x_k^{(n)}c_ke_k\]需要找出一個$u_n$的收敛子列:

$\abs{x_k^{(n)}}\le1$有界$\xRightarrow{\text{Bolzano-Weierstrass}} x_k^{(1)},x_k^{(2)},\dots$ 存在收敛子列
$\xRightarrow{\text{diagonalization}}$存在子列$u_{n_1},u_{n_2}\dots$ 使 $x^{(n_1)}_k,x^{(n_2)}_k,\dots$對每個$k$均收敛(到$x_k$)
設 $u:=\sum_{k=1}^\infty x_kc_ke_k$,就能證明 $u_{n_j}\to u$(見math.stackexchange.com/a/4707695/

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2025-4-20 22:25 GMT+8

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