Why does $X$ not Banach $\implies x\mapsto \bar x$ is not onto?

Question

We know that the map $x\mapsto \overline x $ going from $X\to X^{**}$ given by $\overline x(f)=f(x)$ is a bounded linear operator such that $||\overline x||=||x||$ where $X^{**}$ is the double dual of a NLS $X$

But it is said that in general the mapping can't be onto because $X$ may not be a Banach Space but $X^{**}$ is always Banach.

I am not getting the logic why the mapping is not onto in general.Will you kindly explain why $X$ not Banach $\implies x\mapsto \bar x$ is not onto ?

Please help me.

Answer 1

$x \mapsto \bar x$ is an isometry. If it is also surjective, then the completeness of $X^{**}$ implies the completeness of $X$, but $X$ is not complete.

Proof:

Suppose $(x_j)_{j=0}^\infty$ is Cauchy in $X$, then $(\bar x_j)_{j=0}^\infty$ is Cauchy in $X^{**}$, for which there is a limit $\bar x$. Assume that $x \mapsto \bar x$ is bijective (injectivity follows from isometry). Then $x$, the inverse image of $\bar x$, would be a limit of $(x_j)_{j=0}^\infty$, since $\|x - x_j\| = \|\overline{x - x_j}\| = \|\bar x - \bar x_j\| \to 0$, but $X$ is incomplete, so the assumption is absurd.

Answer 2

Suppose that $x \mapsto \bar{x}$ is onto. Since $X$ is not complete, there exists a cauchy sequence $\{x_n\} \subset X$ that has no limit in $X$. Since $f$ is a bounded linear operator, it is continuous, so that $f(x_n) \to y$ for some $y \in X^{**}$, as $X^{**}$ is complete. Since we assumed surjectivity, there exists some $x \in X$ such that $f(x) = y$. But by the open mapping theorem, $f^{-1}$ is continuous, which implies that $f(x_n) \to y \Rightarrow f^{-1}(f(x_n)) \to f^{-1}(y) \Rightarrow x_n \to x$. But we stated earlier that $\{x_n\}$ has no limit in $X$, a contradiction.