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证明$f\mapsto f'$是连续线性泛函

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hbghlyj Posted at 2022-9-19 23:16:20 |Read mode
Last edited by hbghlyj at 2022-9-20 13:20:00$C[a,b]$ 表示(具有 sup 度量的) $[a,b]$ 上的连续实值函数的空间,
$C^1[a,b]$ 表示(具有 $f\mapsto \|f\| + \|f'\|$ 度量的) $[a,b]$ 上的连续可导实值函数的空间,
证明 $f(x) \mapsto f'(x)$ 是 $C[a,b]\to C^1[a,b]$ 的连续函数.

来自原回答的最后一句话
If you make the $\mathcal{C}^1$ norm $f\mapsto \|f\| + \|f'\|$, then differentiation is continuous in that norm.

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 Author| hbghlyj Posted at 2022-9-19 23:17:49
建议修改帖子标题80个字符的限制

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一个汉字占3个字符  Posted at 2022-9-19 23:20
“字符”指ascii字符  Posted at 2022-9-19 23:22

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Czhang271828 Posted at 2022-9-20 15:32:31
题目有误, 应该是证明
\[
\begin{align*}
\mathcal L:C^1[a,b]&\to C[a,b],\\
f&\mapsto f',
\end{align*}
\]
是连续线性泛函 (functional).

这是显然的. 记 $\|\cdot\|_0$, $\|\cdot\|_{1}$ 为 $C[a,b]$ 与 $C^1[a,b]$ 间的范数, 则
\[
\sup_{f\in C[a,b]}\dfrac{\|\mathcal L(f)\|_0}{\|f\|_1}=\sup_{f\in C[a,b]}\dfrac{\|f'\|}{\|f\|+\|f'\|}\leq 1.
\]

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函数是 C1 到 C0 吧, 题目调整一下@hbghlyj  Posted at 2022-9-22 19:30

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2025-4-21 21:58 GMT+8

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