When is the image of a linear operator closed? Let $X$, $Y$ be Banach spaces. Let $T \colon X \to Y$ be a bounded linear operator.
Under what circumstances is the image of $T$ closed in $Y$ (except finite-dimensional image).
In particular, I wonder under which assumptions $T \colon X \to T(X)$ is a bounded linear bijection between Banach spaces, so it is at least an isomorphism onto its image by bounded inverse theorem.
 A: An answer to your last question is that a bounded linear map $T$ between Banach spaces is injective with closed range if and only if it is bounded below, meaning that there is a constant $c>0$ such that for all $x$ in the domain, $\|Tx\|\geq c\|x\|$.  You can read more about this in Chapter 2 of An invitation to operator theory by Abramovich and Aliprantis.
A: Thrm 1: Suppose $X$ is a  Banach space, $Y$ is a normed vector space, and $T:X\to Y$ is a bounded linear operator.
Then the range of $T$ is closed in $Y$ if $T$ is open.
Proof:  Suppose $\mathrm{ran}(T)$ is not closed in $Y$. Let $\delta>0$ be given. The goal is to show that there exists $x\in X$ such that $\|T(x)\|/\|x\|<\delta$. Since $\delta$ is arbitrary this will demonstrate that $T$ is not open.
Since $\mathrm{ran}(T)$ is not closed there is a sequence $\{y_n\}$ in $\mathrm{ran}(T)$ and point $y\in Y\setminus\mathrm{ran}(T)$ such that $y_n\to y$. This means that there are corresponding $x_n\in X$ such that $y_n=T(x_n)$. Since $T$ is continuous it cannot be that $\{x_n\}$ is a convergent sequence or $y$ would be in the range of $T$.
Since $\{x_n\}$ does not converge it is not Cauchy. So there exists an $\epsilon>0$ such that $\forall N\in\mathbb{N} \ \ \exists n,m \ge N $ s.t. $\|x_n-x_m\|>\epsilon$. On the other hand, since $y_n\to y$, there is an $M\in\mathbb{N}$ such that $\forall k\ge M \ \ \ \|T(x_k)-y\|<\delta\frac{\epsilon}{2}$. By choosing $N=M$, there exist $n,m\ge N$ such that $\|x_n-x_m\|>\epsilon$, $\ \|T(x_n)-y\|<\delta\frac{\epsilon}{2}$, and $\|T(x_m)-y\|<\delta\frac{\epsilon}{2}$. By the Triangle Inequality, $\|T(x_n)-T(x_m)\|=\|T(x_n-x_m)\|<\delta \ \epsilon$.
Let $x=(x_n-x_m) \in X$. Then 
\begin{align*}
      \frac{\|T(x)\|}{\|x\|} &< \frac{\delta \ \epsilon}{\epsilon} \\
                         &= \delta.
\end{align*} 
Thrm 2: Suppose $X$ and $Y$ are both Banach spaces, and $T:X\to Y$ is a bounded linear operator. Then $\mathrm{ran}(T)$ is closed in $Y$ if and only if $T$ is open.
Proof: If $\mathrm{ran}(T)$ is closed then $Y$ is a surjective map onto this subspace so by the Open Mapping Theorem $T$ is open.
The other direction is just the first Theorem.
EDIT:  The so called "Thrm2" is false, as per the counterexample provided in comments. The "proof" makes the false assumption that $Y=T(X)$.
