# Bounded linear operator maps norm-bounded, closed sets to closed sets. Implies closed range?

## 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.

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Looks good to me. – J. Loreaux Aug 16 '12 at 12:21

Lemma. Suppose $X,Y$ are Banach spaces, $T : X \to Y$ is continuous and injective. Then the range of $T$ is closed if and only if there is a constant $c > 0$ such that $\|Tx\| \ge c\|x\|$ for all $x\in X$ (we say such $T$ is bounded below).
Proof. For the forward direction, use the open mapping theorem (we don't actually need it 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.