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?

  • $\begingroup$ 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

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...)


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.)

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

$\newcommand{nrm}[1]{\left\lVert{#1}\right\rVert}$ 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.

  • $\begingroup$ 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

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