Answer to Question #139799 in Functional Analysis for anjali

Question

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.

Answer 1

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

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

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

Since $\{a_n\}_{n\in\mathbb N}\subset M$, we have that $a\in\overline{M}=M$ and so $\lim\limits_{n\to\infty}a_n=a$ in $M$.

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

2)Let $M$ be a Banach subspace of $X$. Take arbitrary $a\in\overline{M}$.

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

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

Since we take arbitrary $a\in\overline{M}$, we have $\overline{M}\subset M$.

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