Question
Suppose $T:X\rightarrow Y$ is a continuous, injective linear operator between Banach spaces. Suppose, in addition, that $T$ maps norm bounded closed sets in $X$ to closed sets in $Y$. Then the range of $T$ is closed in $Y$.
This is a problem related to one given An Invitation to Operator Theory by Abramovich and Aliprantis and I'd just like to verify my proof.
Attempt
We assume that $T$ is as above, and we shall prove that it has closed range. If $y_n = T x_n$ and $y_n\rightarrow y$, we want to show that $y=Tx$ for some $x\in X$. First, suppose $\{x_n\}_{n\geq 1}$ is unbounded. Then
$$\lim_n\, T(x_n/\|x_n\|) = \lim_n\, y_n/\|x_n\| = 0.$$
But the set $B=\{ x\in X: \| x\|=1\}$ is closed and norm-bounded, so its image under $T$ is closed. In particular, we must have $Tz=0$ for some $z\in B$. This contradicts the fact that $T$ is injective. So the sequence $\{x_n\}_{n\geq 1}$ is bounded in $X$. Since $\{x_n\}_{n\geq 1}$ is bounded, the set
$$ A=\mathrm{cl} \{ x_1, x_2, \ldots, x_n, \ldots \}$$
is closed and norm bounded. Hence $T(A)$ is closed in $Y$. In particular, $y=\lim_n\, Tx_n = Tx$ for some $x\in A \subset X$. So the range of $T$ is closed.
Thanks in advance!
Best Answer
Your proof looks good.
Here's another possible line you could follow.
Proof. For the forward direction, use the open mapping theorem. (But we don't actually need the forward direction for this problem). For the reverse direction, if $T$ is bounded below then $T^{-1}$ is bounded. So $TX = (T^{-1})^{-1} X$ is closed, being the preimage of a closed set under a continuous map.
Now for the problem: let $S$ be the unit sphere of $X$. By assumption $TS$ is closed, and by the injectivity of $T$ it does not contain 0. Hence its complement contains an open ball of some radius $c$ about 0. This means that $\|Tx\| \ge c\|x\|$ for all $x \in S$, and by linearity the same holds for all $x \in X$. So $T$ is bounded below, and by our lemma it has closed range.