Prove that $c_0$ is not reflexive

Question

I read a simple proof of showing $c_0$ is not reflexive. It says that $c_0$ is separable and $l^\infty$ is not separable so the canonical map from $c_0$ to $c_0^{**}=l^\infty$ is not an isomorphism. And the proof ends here. I would like to ask how does it imply the map is not surjective. Could the map be not injective?

Answer 1

The canonical map from $X$ to $X^{**}$ is isometric. If it is surjective the unit ball $B_X$ in $X$ is mapped isometrically onto the unit ball $B_{X^{**}} $ of $X^{**}.$ The unit ball of $c_0$ does not admit extreme points unlike the unit ball in $\ell^\infty=c_0^{**}.$ Hence the canonical map cannot be surjective.

Remark The argument based on separability gives a stronger conclusion: $c_0$ and $\ell^\infty$ are not isomorphic. In particular the canonical map (isometry) is not surjective.

Answer 2

If $X$ and $Y$ are Banach spaces and $T : X \to Y$ is continuous, there exists a constant $M$ satisfying $\|Tx\|_Y \le M \|x\|_X$ for all $x \in X$. Thus if $T$ is surjective and $D \subset X$ is dense, then $TD \subset Y$ is also dense. In particular, if $X$ is separable so would be $Y$.

Answer 3

Here is another short proof: notice that the canonical map $\iota:c_0\to \ell^\infty$ (i.e. the inclusion map) preserves pointiwse multiplication of sequences. The constant sequence $(1,1,\dots)\in\ell^\infty$ is an identity element for pointwise multiplication in $\ell^\infty$. If the canonical inclusion was surjective, then $c_0$ would admit an identity element for pointwise multiplication, which is not the case (easy exercise).