Operator $T^*$ surjective iff $T$ topologically injective Let $T : E \to F$ operator between Banach spaces, and $T^* : F^* \to E^*$ - adjoint operator. I want to proof next proposition: $T$ is topologically injective (or equivalently: injective with closed image) iff $T^*$ surjective.
It is simple to proof that if $T$ topologically injective than $T^*$ surjective. In fact, if $T$ is topologically injective then $T$ is some embedding of space $E$ in space $F$. So if $f \in F^*$ than $T^* f$ it is just restriction of functional $f$ on space $T(E)$. So we need to proof that every functional $f$ on $T(E)$ can be continued to functional $\hat{f}$ on $F$. But it is just Han Banach theorem.
It is simple to proof that if $T^*$ surjective then $T$ injective. Suppose $T$ is not injective. We can take some two distinct points $x,y \in E$ such that $T(x)=T(y)$, and we can build functional $f \in E^*$ such that $f(x) \neq f(y)$ (it is again Han Banach theorem). So, for any functional $g \in F^*$ we have $T^* g \neq f$ because right part separate points $x,y$ but left part not.
But I cant proof that if $T^*$ surjective then $T(E)$ - closed. How I can do it?
 A: Consider $X := \overline{T[E]}$, which is a closed subspace of $F$, and the operator $S\in L(E,X)$, given by $Sx = Tx$ for $x \in E$. We now have for $f^* \in F^*$ and $x \in E$ that 
$$ (T^*f^*)x = f^*(Tx) = f^*|_X(Sx) = S^*(f^*|_X)x $$
that is $S^*[X^*]\supseteq T^*[F^*] = E^*$, $S^*$ is onto. On the other side, as $S$ has dense image, $S^*$ is one-to-one: If $x_1^*, x_2^* \in X^*$ with $S^*x_1^* = S^*x_2^*$, we have that $x_1^*|_{S[E]} = x_2^*|_{S[E]}$, which implies $x_1^* = x_2^*$. So $S^*$ is a continuous bijection between the Banach spaces $X^*$ and $E^*$, hence an isomorphism by the open mapping theorem. That is, there is an $c > 0$ such that 
$$ \def\norm#1{\left\|#1\right\|} c\norm{x^*} \le \norm{S^*x^*}, \qquad x^* \in X^*. $$
We will now show, that $S$ is onto: It suffices to prove that $T[U_E]$ ($U_E$ denotes the open unit ball) contains $c U_X$, we will first show that $c U_X \subseteq \overline{T[U_E]}$. Let $x \in cU_X$, that is $\norm x < c$. If $x \not\in \overline{T[U_E]}$, then there is $x^* \in X^*$ such that 
$$ \alpha := \sup_{y \in \overline{T[U_E]}} \Re x^*(y) < \Re x^*(x) $$
We have $\alpha \ge 0$ and may assume that $\alpha > 0$ (if $\alpha = 0$, replace $\alpha$ by $\frac 12 \Re x^*(x)$. Replacing $x^*$ with $\alpha^{-1}x^*$, gives that wlog we may assume $\alpha = 1$. As $\overline{T[U_E]}$ is circular, we conclude (multiplying with the right $\lambda \in S^1$) that 
$$ \sup_{y \in T[U_E]} |x^*(y)| \le 1 < |x^*(x)| $$
That is,
\begin{align*} \norm{S^*x^*} &= \sup_{e \in U_E} |x^*(Te)|\\
   &= \sup_{y \in T[U_E]} |x^*(y)|\\
   &\le 1 \\
   & < |x^*(x)|\\
   &\le \norm{x^*}\norm x\\
   &< c\norm{x^*}
\end{align*}
contradiction. So $x \in \overline{T[U_E]}$. 
To show that even $cU_X \subseteq T[U_E]$ we will use the completeness of $E$, so let $x_0 \in cU_X$ and $\epsilon > 0$ with $\epsilon < c$. By the above, there is $e_0 \in U_E$ with $$ \norm{x_0 - Te_0} < \epsilon $$
or $$ \frac c{\epsilon} \norm{x_0 - Te_0} < c $$
hence we may choose $e_1 \in U_E$ with 
$$ \norm{\frac c{\epsilon} x_0 - T\left(\frac c{\epsilon} e_0 + e_1\right)} < \epsilon \iff \norm{cx_0 - T(ce_0 + \epsilon e_1)} < \epsilon^2 $$
Inductively we find $e_i \in U_E$ with 
$$ \norm{x_0 - T\left(\sum_{i=0}^n \frac{\epsilon^i}{c^i} e_i\right)} < \frac{\epsilon^n}{c^n} $$
As $E$ is complete, and $\sum_{i=0}^\infty \frac{\epsilon^i}{c^i} < \infty$, $e := \sum_{i=0}^\infty \frac{\epsilon^i}{c^i}e_i$ exists, and by the above $Se =Te = x_0$. As $e \in U_E$, we are done. 
