Closed range operators Let $T$ be a linear operator between two normed spaces. I'm trying to show that an operator $T$ has closed range if and only if $\operatorname{im}(T) = (\ker{(T^*)})^{\perp}$.
Is there a way to do it without the Hahn-Banach theorem?
Thanks for any help.
 A: The first step should be to clarify what the orthogonal notation means.
Let $Y$ be normed vector space,  and let $Y^*$ denote its topological dual.
If $F$ is a subset of  $Y$, then we denote
$$
F^\perp:=\{\phi\in Y^*\;;\;\phi(y)=0\;\forall y\in F\}.
$$
This is a weak* closed subspace of $Y^*$.
Now if $G$ is a subset of $Y^*$, we denote
$$
G^0:=\{y\in Y\;;\;\phi(y)=0\;\forall \phi\in G\}.
$$
This is a normed closed subspace of $Y$. Note that it is not $G^\perp$, which is  a weak* closed subspace of the bidual $Y^{**}$.
Claim: For every subspace $F$ of $Y$, we have $(F^\perp)^0=\overline{F}$, the norm closure of $F$.
Proof: It is trivial to see that $F\subseteq (F^\perp)^0$. Since the latter is norm closed, we get $\overline{F}\subseteq (F^\perp)^0$. Now if $y$ does not belong to $\overline{F}$, Hahn-Banach gives us $\phi\in F^\perp$ such that $\phi(y)\neq 0$. So $y$ does not lie in $(F^\perp)^0$. This proves the reverse inclusion. QED.
Your question: Let $T:X\longrightarrow Y$ be a bounded linear operator between two normed vector spaces. As you observed, it is straghtforward to show that
$$
\mbox{Ker} T^*=(\mbox{Im} T)^\perp
$$
in $Y^*$. Therefore
$$
(\mbox{Ker} T^*)^0=((\mbox{Im} T)^\perp)^0=\overline{\mbox{Im} T}.
$$
Now clearly $T$ has closed range if and only if $(\mbox{Ker} T^*)^0=\mbox{Im} T$.
Note: I really can't see how we could avoid Hahn-Banach in the proof of the claim.
