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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!

  • 2
    Looks good to me.2012-08-16

2 Answers 2

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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.

  • 0
    @Greywhite: The open mapping theorem only applies to maps between Banach spaces. So in order to apply it, you would need for ran$T$to be a Banach space, which only happens if it is a closed subspace of Y. Hence your argument is circular. (Indeed, your argument would appear to prove that every injective continuous linear map between Banach spaces has closed range, which is certainly not true.)2018-12-09
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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 $\lim_{k\to\infty}\frac{y_{n_{k}}}{\|x_{n_{k}}\|}=0$.