Consider the following question:

Let $X$ be a normed space and $T: X\longrightarrow X''$ is the canonical mapping. Prove the range of T, denoted $\mathscr{R}(T)$, is closed iff $X$ is complete.

Where I want to stick with the "$\implies$" part of the of the proof. Here I'm concerned with—as the title implies—whether I can get a away with the following proof.


Let $x \in X$ be arbitrary, then $Tx \in \mathscr{R}(T)$. Since $\mathscr{R}(T)$ is closed, $Tx$ must be one of its limit points. Since $Tx$ is a limit point of $\mathscr{R}(T)$ then there is a sequence $(x_n'')$ in $\mathscr{R}(T)$ such that $x_n'' \longrightarrow Tx$. Since each element $x_i''$ of $(x_n'')$ is in $\mathscr{R}(T)$, it can be written $x_i'' = Tx_i$. In other words, $Tx_n \longrightarrow Tx$ in $\mathscr{R}(T)$. Formally, for any given $\epsilon > 0$ there exists an integer $N(\epsilon)$ such that for any $n > N$ we have

$$\|Tx_n - Tx\| < \epsilon$$

and since canonical mappings are norm preserving (and linear) we have

$$\|x_n - x\| = \|T(x_n - x)\| < \epsilon$$

So $x_n \longrightarrow x$ in $X$.

Since $x'' \in \mathscr{R}(T)$ was arbitrary, every point in $X$ has a convergent sequence in $X$. Convergent sequences in metric spaces are Cauchy, and therefore $X$ is complete.


Here I didn't directly prove that a Cauchy sequence in $X$ implies that it is converges in $X$. But I think I proved that every vector in $X$ has a convergent sequence to it, which I believe is equivalent. Is this true and valid?

  • $\begingroup$ No, every point in $\mathbb{Q}$ has a sequence converging to it, but $\mathbb{Q}$ is not complete. $\endgroup$ – Floris Claassens Apr 3 at 16:04
  • $\begingroup$ In general, I agree. But what about in the application above? $\endgroup$ – Zduff Apr 3 at 16:18
  • $\begingroup$ I can't come up with a counter-example of the top of my head. But I would still say no. I would suggest just proving this using Cauchy sequences. $\endgroup$ – Floris Claassens Apr 3 at 16:26

Your proof is wrong, and it misses the point. You start with $x''=Tx$, and then choose a sequence $\{x''_n\}$ with $x''\to Tx$. What prevents you from taking $x_n''=x''$ for all $n$? Then you are proving nothing.

Besides the above, the argument is unnecessarily complicated, if you already know that $T$ is an isometry.

You want to show that $X$ is complete, so you should start with a Cauchy sequence $\{x_n\}\subset X$, not with a point in the range of $T$. As $$ \|Tx_n-Tx_m\|=\|x_n-x_m\|$$ for all $n,m$, the sequence $\{Tx_n\}$ is Cauchy, so there exists $x$ with $Tx=\lim Tx_n$. Then $$ \|x-x_n\|=\|Tx-Tx_n\|\to0. $$ And that's it. And, by the way, this proves that the range of any isometry with complete domain is complete, this has nothing to do with double duals or anything.

  • $\begingroup$ Yeah I see your point now. But previously I started the way you suggest, but I'm hung on some assumption in your proof. You say $(Tx_n)$ is Cauchy , so there exists a $Tx$ that it converges to. Is this because the range of $T$ is closed? Because I'm under the impression that this is only true if the range of $T$ is finite dimensional. $\endgroup$ – Zduff Apr 3 at 17:35
  • 1
    $\begingroup$ Yes, it's because the range of $T$ is closed, and it that has nothing to do with dimension: it's topology of metric spaces. $\endgroup$ – Martin Argerami Apr 3 at 18:09
  • $\begingroup$ So every closed metric space is complete? $\endgroup$ – Zduff Apr 3 at 18:36
  • 1
    $\begingroup$ Not in abstract. But here your metric space is a subset of a complete space (the double dual) so closed is the same as complete. $\endgroup$ – Martin Argerami Apr 3 at 18:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.