Two Banach spaces, if and only if criterion for range of closed unbounded operator to be closed? Let $E$ and $F$ be two Banach spaces. Let $A: D(A) \subset E \to F$ be a closed unbounded operator. How do I see that $R(A)$ is closed if and only if there exists a constant $C$ such that$$\text{dist}(u, N(A)) \le C\|Au\| \text{ for all }u \in D(A)?$$Here, $D$ denotes domain, $R$ denotes range, $N$ denotes kernel.
Idea. We probably want to consider the operator $T: E_0 \to F$, where $E_0 = D(A)$ with the graph norm and $T = A$ in some regard? But I am not quite sure on what to do next.
 A: The first part of the answer was wrong, so I have removed it. I have written a new version and put it at the end. Also the second part (which is now the first part) has been rewritten.
Suppose $R$ is closed (with this $R$ becomes a Banach space).
In general $D$ with the graph norm $||u||_G := ||u|| + ||A(u)||$ is a Banach space. This is so because it is a closed subset of $E \oplus F$ since $A$ is a closed operator.
$A$ is clearly bounded on this space. Taking the quotient space $D/N$ the map $\tilde A([u]):=A(u)$ is well defined, bounded, injective and has the same range as $A$. As such $\tilde A: D/N \to R$ is a bijective bounded map between two Banach spaces. The inverse $\tilde A^{-1} : R \to D/N$ is then also a bounded linear operator.
The norm on $D/N$ is given by: $||[u]||_{D/N}=\text{dist}(u,N)_G=\text{dist}(u,N)+||A(u)||$.
Finally:
$$\text{dist}(u,N)=||[u]||_{D/N}-||A(u)||=||\tilde A^{-1}(\tilde A([u]))||_{D/N}-||A(u)||≤(||\tilde A^{-1}||-1)\ ||A(u)||$$
The other direction works as follows:
Suppose $\text{dist}(u,N)≤C\ ||A(u)||$ $\forall u \in D$.
As seen before $D/N$ is a Banach space if $D$ is given graph norm. On this space we can consider the bijective bounded function $\tilde A: D/N \to R$.
The inverse of $\tilde A$ is also bounded since:
$$\frac{||\tilde A^{-1}(\tilde A([u]))||_{D/N}}{||\tilde A([u])||}=\frac{\text{dist}(u,N)+||A(u)||}{||A(u)||}≤C+1$$
Now let $A(u_n)$ be Cauchy. Then $||[u_n]-[u_m]||_{D/N}≤||\tilde A ^{-1}||\cdot ||A(u_n)-A(u_m)]]$ and $[u_n]$ is Cauchy. But since $D/N$ is a Banach space the limit $[u]$ exists and $||A(u_n)-A(u)||≤||\tilde A||\cdot ||[u_n]-[u]||_{D/N}$, so as a result $A(u_n)$ converges to $A(u)$.
