On linear, continuous, injective map $T:V \to W$ with closed/ dense image Let $V,W$ be normed linear spaces and $T:V \to W$ be a linear continuous, injective map .
Then is $T^{**}: V^{**} \to W^{**}$ is injective ? If this is not true in general, what if we also assume $Im T$ is closed/ dense in $W$ ? 
My try: $g\in \ker T^{**} \implies  g\circ T^*=T^{**}(g)=0 \implies g(T^*(f))=0, \forall f \in W^*$. 
So $Im (T^*) \subseteq \ker g$. 
Now $\{0\}=\ker T^{**} $ if and only if $g\circ T^*=0,\forall g\in \ker T^{**}$ if and only if $\ker g=V^*, \forall g \in \ker T^{**}$. 
Now if $Im(T^*)$ is dense in $V^*$, then $Im (T^*) \subseteq \ker g$ and $\ker g$ is closed would readily imply $\ker g=V^*$, but unfortunately we don't know that $Im (T^*)$ is dense in $V^*$. 
I don't know how to proceed further. 
 A: I am almost sure that this is not true in general without assuming that $T$ has closed image. I cannot think of a counter example now, but I am almost sure.
However, if one assumes that $T$ has closed image then the result follows; we prove the following two claims:

*

*If $S:X\to Y$ is a bounded operator that is injective and has closed range, then the adjoint operator $S^*:Y^*\to X^*$ is surjective.

*If $S: X\to Y$ is a bounded operator that is surjective, then the adjoint operator $S^*:Y^*\to X^*$ is injective.

Proof of the claims:
For (1), since $S$ is injective and has closed range, we conclude (from the open mapping theorem) that $S$ is bounded below, so there exists a constant $\delta>0$ so that $\delta\|x\|\leq\|Sx\|$ for all $x\in X$. Let $\varphi\in X^*$ be a bounded linear functional on $X$. We define a functional $\psi_o:\text{Im}(S)\to\mathbb{C}$ as $\psi_o(Sx)=\varphi(x)$ for all $x\in X$. Note that this is well-defined, since $S$ is injective. Moreover, this is bounded, since $S$ is bounded below: we have that
$$|\psi_o(Sx)|=|\varphi(x)|\leq\|\varphi\|\cdot\|x\|\leq\|\varphi\|\cdot\delta^{-1}\cdot\|Sx\|$$
for all $x\in X$. By the Hahn-Banach theorem we may extend $\psi_o$ to a bounded functional (of the same norm) $\psi\in Y^*$. Now we have that $S^*(\psi)=\psi\circ S=\psi_o\circ S=\varphi$, so $S^*$ is indeed surjective.
For (2), Let $\varphi\in Y^*$ be a functional so that $S^*(\varphi)=0$, i.e. $\varphi\circ S=0$. Since $S$ is surjective, we have that $\varphi(z)=0$ for all $z\in Y$, so $\varphi=0$.
Combining these two: we have $T:X\to Y$ an injective operator with closed range. By (1) the adjoint operator $T^*:Y^*\to X^*$ is surjective. Applying (2) to $T^*:Y^*\to X^*$ we conclude that $T^{**}:X^{**}\to Y^{**}$ is injective. I hope this answers your question.
