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.

Expert's answer

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

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

Since M⊂XM\subset X, we have that {an}n∈N\{a_n\}_{n\in\mathbb N} is a fundamental sequence in XX. Then there is lim⁡n→∞an=a\lim\limits_{n\to\infty}a_n=a in XX.

Since {an}n∈N⊂M\{a_n\}_{n\in\mathbb N}\subset M, we have that a∈M‾=Ma\in\overline{M}=M and so lim⁡n→∞an=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 a∈M‾a\in\overline{M}.

Then there is {an}n∈N⊂M\{a_n\}_{n\in\mathbb N}\subset M such that lim⁡n→∞an=a\lim\limits_{n\to\infty}a_n=a.

Since {an}n∈N\{a_n\}_{n\in\mathbb N} is a convergent sequence, it is a fundamental sequence, so lim⁡n→∞an=a∈M\lim\limits_{n\to\infty}a_n=a\in M.

Since we take arbitrary a∈M‾a\in\overline{M}, we have M‾⊂M\overline{M}\subset M.

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


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