Question #139799
Let X is Banach space and Let M be a subspace of X. Than M is itself a Banach space (using the norm from X) if and only if M is closed.
1
Expert's answer
2020-10-23T13:12:18-0400

1)Let MM be a closed subset of XX, then M=MM=\overline{M}.

Take a fundamental sequence {an}nNM\{a_n\}_{n\in\mathbb N}\subset M.

Since MXM\subset X, we have that {an}nN\{a_n\}_{n\in\mathbb N} is a fundamental sequence in XX. Then there is limnan=a\lim\limits_{n\to\infty}a_n=a in XX.

Since {an}nNM\{a_n\}_{n\in\mathbb N}\subset M, we have that aM=Ma\in\overline{M}=M and so limnan=a\lim\limits_{n\to\infty}a_n=a in MM.

By the definition of Banach space we have that MM is a Banach space.

2)Let MM be a Banach subspace of XX. Take arbitrary aMa\in\overline{M}.

Then there is {an}nNM\{a_n\}_{n\in\mathbb N}\subset M such that limnan=a\lim\limits_{n\to\infty}a_n=a.

Since {an}nN\{a_n\}_{n\in\mathbb N} is a convergent sequence, it is a fundamental sequence, so limnan=aM\lim\limits_{n\to\infty}a_n=a\in M.

Since we take arbitrary aMa\in\overline{M}, we have MM\overline{M}\subset M.

We obtain M=M\overline{M}=M, because MMM\subset\overline{M}, that is MM is closed.


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