Is the injectivity of the operator equivalent to the surjectivity of its adjoint Let $X$ and $Y$ be two normed linear spaces. Let $T:X \to Y^*$ be a linear operator (not necessarily continuous) and let $T^*$ be its adjoint, i.e. $T^*:Y \to X^*$ is defined by
$ \langle T^*y,x \rangle := \langle Tx,y \rangle ,$
where $\langle \cdot , \cdot \rangle$ is a proper duality pairing (i.e. between $X^*$ and $X$ on the left hand side and between $Y^*$ and $Y$ on the right hand side).
Let $R(T)$ denote the range of $T$. Is it true that:
$T$ is injective on $R(T)$ if and only if $T^*$ is surjective onto $X^*$?
 A: The full equivalence does not hold in general. Rather it holds: $T$ is injective if and only if $R(T^*)$ is weak-star dense in $X^*$.
If $T^*$ is surjective, then $T$ is injective: Suppose $Tx=0$, then
$$
0 = \langle Tx,y\rangle_{Y^*,Y} = \langle T^*y,x\rangle_{X^*,X},
$$
and since $R(T^*)=X^*$, it follows $0= \langle x^*,x\rangle_{X^*,X}$ for all $x^*\in X^*$. Hence $x=0$.
Here, it would have been sufficient to suppose that $R(T^*)$ is dense in $X^*$.
Now, let $T$ be injective, assume that $R(T^*)$ is not  weak-star dense in $X^*$. Then there is $x^*\in X^* \setminus \overline{R(T^*)}^{w^*}$, where closure is taken with respect to the weak-star topology. By Hahn-Banach separation theorem, there is $x\in X$, $t\in\mathbb R$, such that 
$$
\langle x^*,x\rangle_{X^*,X} < t \le \langle T^*y,x\rangle_{X^*,X} \quad \forall y\in Y.
$$
Since $T^*$ is linear, it follows $\langle T^*y,x\rangle_{X^*,X}=0$ for all $y\in Y$, which is equivalent to $\langle Tx,y\rangle_{Y^*,Y}=0$ for all $y$. 
Hence $Tx=0$ and by injectivity $x=0$ follows, which is a contradiction.
As a counter-example, which shows that injectivity of $T$ does not implies surjectivity of $T^*$, you can choose any compact and injective $T$, with $X,Y$ being infinite-dimensional. Then $T^*$ is compact as well, but $R(T^*)$ cannot be closed.
