$T\in\mathcal{L}(X,Y)$ maps closed bounded subsets onto closed subsets $\implies$ Range $T$ is closed. Given two normed spaces $X$ and $Y$ and let $T$ be a bounded linear operator $T:X\to Y$. Assume that $T$ maps bounded and closed subsets of $X$ onto closed subsets of $Y$. Show that the range of $T$ is closed.
Can someone give me a hint. That would be nice.
Cheers
 A: First observe that the proof given in Bounded linear operator maps norm-bounded, closed sets to closed sets. Implies closed range? does not use the fact that $X,Y$ are assumed to be complete.
It hence remains to show that we can reduce to the case of an injective operator. To this end, we "factor out the kernel":
As $T$ is bounded, the kernel $Z := \mathrm{ker}(T)$ is a closed subspace of $X$. This shows that the norm
$$
\Vert x + Z \Vert := \mathrm{d}(x, Z) = \inf \{\Vert x - z \Vert \mid z \in Z\}
$$
defines a norm on the quotient space $X/Z$ (check this, closedness is used to establish definiteness of the norm).
Let $\pi : X \to X/Z, x \mapsto x + Z$ be the canonical projection. Note that $\pi$ is bounded. It is now easy to see that the map
$$
\widetilde{T} : X/Z \to Y, x+Z \mapsto Tx
$$
is well-defined and bounded with $T = \widetilde{T} \circ \pi$. Hence, $\mathrm{range}\widetilde{T} = \mathrm{range}T$, so that it suffices to show the claim for the injective(!) map $\widetilde{T}$.
Hence, we only have to verify that $\widetilde{T}$ fulfills the requirements. To this end, let $M \subset X/Z$ be bounded and closed, e.g. $M \subset B_R (0)$ for some $R > 0$. I claim that
$$
\widetilde{T}(M) = T(\pi^{-1}(M) \cap \overline{B_R}(0)).
$$
As $\pi^{-1}(M) \cap \overline{B_R}(0)$ is closed and bounded, this suffices.
The inclusion "$\supset$" above is clear because for $x \in \pi^{-1}(M)$, we have $Tx = \widetilde{T}(\pi x) \in T(M)$. For the inclusion "$\subset$", let $x +Z \in M$ be arbitrary. Because of $M \subset B_R(0)$, the definition of the norm on $X/Z$ shows that there is some $z \in Z$ with $\Vert x - z \Vert < R$.
But then $\pi (x-z) = x+Z \in M$, i.e. $x-z \in \pi^{-1}(M) \cap \overline{B_R}(0)$ and
$$
\widetilde{T}(x+Z) = \widetilde{T}(\pi x)  = Tx = T(x-z) \in T(\pi^{-1}(M) \cap \overline{B_R}(0)).
$$
As $x+Z \in M$ was arbitrary, this completes the proof.
