4
$\begingroup$

Let $\mathbb{X}$ be a normed space that is complete and $\mathbb{Y}$ be another normed space which is not complete. Then can a bounded linear map $A:\mathbb{X} \to \mathbb{Y}$ be bijective or not?

$\endgroup$
  • $\begingroup$ What did you try? $\endgroup$ – sinbadh Feb 4 '16 at 0:09
  • $\begingroup$ What do you mean? $\endgroup$ – Bruno Sterner Feb 4 '16 at 0:12
  • $\begingroup$ Do you have some conjecture, counterexample? something? $\endgroup$ – sinbadh Feb 4 '16 at 0:13
  • $\begingroup$ I'm inheritantly trying to show that for a normed space $\mathbb{X}$ a map from it's dual space (a complete space) to the space of continuously differentiable functions on $[0,1]$ (not a complete space) cannot be bijective $\endgroup$ – Bruno Sterner Feb 4 '16 at 0:16
  • $\begingroup$ But i'm just curious if it's true generally $\endgroup$ – Bruno Sterner Feb 4 '16 at 0:17
5
$\begingroup$

Edit: In fact there's a very simple theorem here that gives the whole truth: Given a bounded linear bijection $T:X\to Y$, where $X$ is complete, $Y$ is complete if and only if $T^{-1}$ is bounded. (If $Y$ is complete the open mapping theorem shows that $T^{-1}$ is bounded. On the other hand if $T^{-1}$ is bounded it's trivial to show that $Y$ is complete: A Cauchy sequence in $Y$ comes from a Cauchy sequence in $X$, which converges...)


Original:

Yes, it's possible. This surprises me; I thought the answer was no. The reason I thought the answer was no was something like this:

Let's agree that an isomorphism in the present context is a bounded linear bijection whose inverse is also bounded. Now (i) a bounded linear bijection between Banach spaces must be an isomorphism, (ii) if $X$ and $Y$ are isomorphic normed spaces and $X$ is complete then $Y$ is complete. Of course I never thought that was actually a proof here; all it proves is that $Y$ is complete if $Y$ is complete. But those facts in my head made me think the answer was no.

Anyway, here's an example. Let $X=\ell^2$, the usual space of square-summable sequences. Define $T:X\to X$ by $$Tx=(x_1,x_2/2,x_3/3,\dots).$$

Then $T$ is certainly bounded and injective. Now let $Y=T(X)$, and give $Y$ the norm it inherits from $\ell^2$. Regard $T$ as a map from $X$ to $Y$. It's still bounded and injective, and now it's surjective.

So $T:X\to Y$ is a bounded linear bijection. And $Y$ is not complete. (Proof: If $Y$ were complete then the open mapping theorem would show that $T^{-1}:Y\to X$ was bounded, but $T^{-1}$ is certainly not bounded.)

$\endgroup$
  • $\begingroup$ This is nice, thanks for sharing. $\endgroup$ – Bruno Sterner Feb 4 '16 at 0:49
  • 1
    $\begingroup$ No, thanks for asking! I did think the answer was no - learned something today. $\endgroup$ – David C. Ullrich Feb 4 '16 at 0:51
1
$\begingroup$

$\newcommand{nrm}[1]{\left\lVert{#1}\right\rVert}$ Long story short: yes, and it happens quite often.

For instance, let $I=(0,1)$. Consider the map \begin{align}\psi:W^{1,p}(I)&\hookrightarrow L^p(I)\\u&\mapsto u\end{align}

Since $\nrm{u}_{W^{1,p}}=\nrm{u'}_p+\nrm{u}_p$, it holds $\nrm\psi\le1$.

But $\psi\left(W^{1,p}(I)\right)=F$ contains $C^\infty_c(I)$, therefore it is dense in $L^p(I)$. Obviously, $F\ne L^p(I)$ because all functions in $F$ have finite $\operatorname{supess}$.

Hence $(F,\nrm{\bullet}_p)$ is not Banach and $\psi:W^{1,p}(I)\to F$ is continuous and bijective.

$\endgroup$
  • $\begingroup$ I believe that one. Really the same as my example - if you have any bounded linear injective $T:X\to Z$ which is not bounded below you get an example by setting $Y=T(X)$. $\endgroup$ – David C. Ullrich Feb 4 '16 at 0:46
  • $\begingroup$ Yes, the trick is finding a proper dense subspace which is Banach w.r.t. a different norm, and such that the inclusion is continuous. another way would be with the inclusions $L^p(0,1)\subsetneq L^q(0,1)$ for $q<p$. I was finishing writing this when you posted yours, and I deemed it worth mentioning because the maps involved are very well-known. $\endgroup$ – user228113 Feb 4 '16 at 0:54
  • $\begingroup$ Certainly worth mentioning - happens all the time that one post appears while another is being typed.... $\endgroup$ – David C. Ullrich Feb 4 '16 at 1:03
  • $\begingroup$ Nice example.The larger norm is complete.Since it is larger, its topology is stronger so the identity embedding to the weaker-normed space is bounded (and continuous) and the weaker norm is incomplete, as it sits inside a Banach space as a dense proper subset. $\endgroup$ – DanielWainfleet Feb 4 '16 at 10:04

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.