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.


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!

  • 2
    $\begingroup$ Looks good to me. $\endgroup$ – J. Loreaux Aug 16 '12 at 12:21

Your proof looks good.

Here's another possible line you could follow.

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. (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.

  • $\begingroup$ Thanks Nate. I actually wanted to prove a certain injective operator was bounded below -- hence my interest in the closedness of its range. The lemma I posted was particularly straightforward to prove for my operator, so I just wanted to check my tracks! $\endgroup$ – Tom Aug 17 '12 at 7:54
  • $\begingroup$ Hi Nate. For the forward direction in your Lemma, how do you show that if the range of T is closed, then T is bounded below "without" using the open mapping theorem? $\endgroup$ – mononono Mar 2 '15 at 23:24
  • $\begingroup$ @Dejon: The forward direction essentially is the open mapping theorem; you can't avoid it unless you re-prove it. But what I meant was that we don't need the forward direction of my lemma to prove the statement in the question, only the reverse direction. $\endgroup$ – Nate Eldredge Mar 3 '15 at 1:48
  • $\begingroup$ @NateEldredge : I think there is some confusion for the reverse direction ; as $T$ is injective , $T^{-1}$ is defined ( i.e. its domain ) on range of $T$ i.e. $T(X)$ ; and then $T(X)=(T^{-1})^{-1}(X) $ would obviously be closed in $T(X)$ itself ... still I can prove that if $T$ is bounded from below and $X$ is Banach , then $T(X)$ is closed in $Y$ ( though the proof is considerably different than working with just inverse images ) $\endgroup$ – user228169 Feb 27 '16 at 14:17
  • $\begingroup$ Since T is continuous and if we view it as an operator from X to ranT it is surjective, then by the open mapping theorem it is open. And as a bijection it is thus closed. So it is naturally bounded below thus ranT is closed. Is my statement right or which step goes wrong? Thanks! $\endgroup$ – Greywhite Dec 9 '18 at 8:34

You are assuming that $(x_{n})$ unbounded implies $\left(\frac{1}{\|x_{n}\|}\right)$ goes to $0$, so that you can then conclude $\lim_{n\to\infty}\frac{y_{n}}{\|x_{n}\|}=0$. This is not necessarily true.

For example, the sequence $(x_{n})$ in $\mathbb{R}$ given by $x_{n}=\begin{cases}n & \mbox{if }n\mbox{ is even}\\1 & \mbox{if }n\mbox{ is odd}\end{cases}$ is unbounded. However, $\frac{1}{\|x_{n}\|}=\begin{cases}\frac{1}{n} & \mbox{if }n\mbox{ is even}\\1 & \mbox{if }n\mbox{ is odd}\end{cases}$ does not converge.

Fortunately, $\left(\frac{1}{\|x_{n}\|}\right)$ does contain a subsequence converging to $0$. So you can get around the problem with a minor change to your proof:

Assume $(x_{n})$ is not bounded. Then there is a subsequence $(x_{n_{k}})$ of $(x_{n})$ such that $x_{n_{k}}\not=0$ for all $k$ and $\lim_{k\to\infty}\|x_{n_{k}}\|=\infty$. Then $\lim_{k\to\infty}\frac{y_{n_{k}}}{\|x_{n_{k}}\|}=0$.


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