$T$ is surjective if and only if the adjoint $T^*$ is an isomorphism (onto its image) 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.
 A: The claim that $T$ is surjective implies range of $T^*$ being closed is not true.
Take $X=c_{00}=Y$ the space of sequences with finite length. The dual space can be identified with $l^1$.
Define 
$$
Tx = (x_1, x_2/2, \dots, x_n/n,\dots).
$$
Clearly, $T:X\to Y$ is injective and surjective, however $T^{-1}$ is not bounded.
Let $g\in l^1$ be given. Then for $x\in c_{00}$,
$$
(T^*g)(x)=g(Tx) = \sum_k g_k x_k/k = \sum_k g_k x_k/k = h(x),
$$
where $T^*g=h = (g_1,g_2/2,\dots)$. Now take $e_k=(0,\dots,0,1,0,\dots)$ with the non-zero entry at position $k$. Then 
$$
\|T^*e_k\|_{l^1} = \frac1k, \ \|e_k\|_{l^1}=1,
$$
which shows that the claim is not true in the general situation.
(The example is also valid for $Y=c_0$, which is a Banach space)

If in addition $X$ and $Y$ are complete, then the claim follows with the closed range theorem.
A: 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.
A: Fix $y \in Y$. Let $z$ be a unique point in $Y$ such that $||y^*|| =  \frac{||y^* z||}{||z||}$, ie.
$$
\max_w\frac{||y^*w||}{||w||} = \frac{||y^* z||}{||z||}.
$$
(Note that we need completeness of $Y$ to be assured of the inclusion of $z$ in $Y$.)
Since $T$ is surjective, there exists a $\tilde x \in X$ such that $T \tilde x = z$. Therefore
$$
||T^* y^*|| = \max_x \frac{||y^* T x||}{||x||} \ge \frac{||y^* T \tilde x||}{||\tilde x||} = c_1\frac{||y^*T \tilde x||}{||T\tilde x||} = c_1 \frac{|| y^*z||}{||z||} = c_1 || y^*||.
$$
where we have used the inverse of the inequality $c_1 ||x|| \le ||T x||$ for general $x$. The existence of such a constant $c_1> 0$ requires the completeness of $X$.
On the other side, we have
$$
||T^* y^* || = \max_x \frac{||T^* y^* x||}{||x||} =\max_x \frac{|| y^* T(x)||}{||x||} \le || y^*|| \max_x \frac{||Tx||}{||x||} \le c_2 ||y^*||,
$$
where we have used the Cauchy-Schwarz inequality.
So $c_1 ||y^*|| \le ||T^* y^*|| \le c_2 ||y^*||$.
