Question

A normed space which is isometric to a Banach space is itself a banach space

Answer 1

Problem #6345 A normed space which is isometric to a Banach space is itself a Banach space. Prove.

Solution Let $(X, \| \cdot \|_X)$ be a Banach space, $(Y, \| \cdot \|_Y)$ is a normed linear space, which is isometric to $X$, that is exists $T \colon X \to Y$, which is bijective, linear and "preserves" the distance that is $\| Tx \|_Y = \| x \|_X$. We are to prove that every fundamental sequence $\{y_n\}_n \subset Y$ converges in $Y.T$ is bijective, hence there exists the sequence $\{x_n\}_n \subset X$, such that $y_n = Tx_n$. From isometric we obtain $\| y_n - y_m \|_Y = \| Tx_n - Tx_m \|_Y = \| x_n - x_m \|_X$, thus $\{x_n\}_n$ is fundamental in $X$, and due to $X$ is Banach space, it converges in $X$. Denote by $x = \lim x_n$, what is equivalent to $\| x - x_n \|_X \to 0$, $n \to \infty$. Denote by $y = Tx$, then $\| y - y_n \|_Y = \| Tx - Tx_n \|_Y = \| x - x_n \|_X \to 0$, $n \to \infty$. Hence $\{y_n\}_n$ converges to $y$. To sum it over, we proved that every fundamental sequence in $Y$ converges to some element in $Y$. We are done.

Answer