TITLE: $T$ is surjective if and only if the adjoint $T^*$ is an isomorphism (onto its image)
QUESTION [5 upvotes]: I am trying to prove the following statements: 
Let $X$ and $Y$ be normed spaces (not necessarily complete)
Let $T\in L(X,Y)$  (meaning $T:X\to Y$ is a bounded linear map). Let $T^*:Y^*\to X^*$ denote the adjoint operator. Then:

$T^*$ is sujective if and only if $T$ is an isomorphism;
$T$ is surjective if and only of $T^*$ is an isomorphism.

Here an "isomorphism" $X\to Y$ is an injective linear operator $T:X\to Y$ such that there exists $c_1,c_2>0$ with $$c_1\|x\|\leq \|Tx\|\leq c_2\|x\| \text{ for all } x\in X.$$
In particular we do not require $T$ to be surjective. 
Discussion
I have proved statement 1. and I have also proved that if $T^*$ is an isomorphism then $T$ is surjective, and that if $T$ is surjective then $T^*$ is injective. An ideal next step would be to show that $Image(T^*)$ is closed in $X^*$, at which point I could apply the open mapping theorem to conclude that $T^*$ is an isomorphism (since I already know $T^*$ is a bounded operator). However, I am struggling to show that the image of $T^*$ is closed. Any ideas would be appreciated.

REPLY [5 votes]: Let us answer the first question, hopefully the other one is similar enough: 
If $T^{\ast}$ is surjective, then it is an open map, and so $\exists r>0$ such that
$$
B(0,r) \subset T^{\ast}(B(0,1))
$$
where these denote the open balls in their respective spaces. Hence if $\varphi \in X^{\ast}, \|\varphi\| = 1, \exists \psi \in B(0,1)$ such that
$$
\frac{r}{2}\varphi = T^{\ast}(\psi)
$$
Then for any $x\in X$,
$$
\frac{r}{2}|\varphi(x)| = |\psi(Tx)| \leq \|\psi\|\|T(x)\| < \|T(x)\|
$$
Varying over all $\varphi$ as above, we see that
$$
\frac{r}{2}\|x\| \leq \|T(x)\|
$$
and so $T$ is an isomorphism.
Conversely, if $T$ is an isomorphism, then if $\varphi \in X^{\ast}$, then define $\psi : \text{Im}(T) \to \mathbb{C}$ by
$$
\psi(Tx) := \varphi(x)
$$
and this is well-defined since $T$ is injective. Furthermore
$$
|\psi(Tx)| \leq \|\varphi\|\|x\| \leq \frac{\|\varphi\|}{c_1}\|T(x)\|
$$
and so $\psi$ is bounded on $\text{Im}(T)$, and so it extends to a bounded linear functional $\psi \in Y^{\ast}$. Now clearly,
$$
T^{\ast}(\psi) = \varphi
$$
and hence $T^{\ast}$ is surjective.