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Let $X$ be a linear space with a family of complete norms $(\| \circ \|_I)_{I \in \mathcal{I}}$ on $X$, i.e. for every $I \in \mathcal{I}$ the tuple $(X,\|\circ\|_I)$ is a Banach space. Now define $$\| x \|_\infty := \sup_{I \in \mathcal{I}} \| x \|_I \in [0,\infty]$$ for every $x \in X$ and $X_\infty = \{ x \in X ~:~ \|x\|_\infty < \infty \}$. It is easy to see that $(X_\infty,\|\circ\|_\infty)$ is a normed space. In particular, $X_\infty$ is a linear space.

Is $(X_\infty,\|\circ\|_\infty)$ a Banach space, i.e. is it complete?

My guess would be negative, but I can't come up with an (counter-)example.

What I have tried so far: It is easy to see that if $X$ is finite-dimensional and $\mathcal{I}$ is finite, then $(X_\infty,\|\circ\|_\infty)$ is again complete. Therefore I asumme that either I have to find many "incompatible" norms on a finite-dimensional space, or only "a few" norms on a "complicated" space. I would appreciate hints for either direction.

x_Y_z
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  • The natural thing to try would be to take a cauchy sequence $x_n$ in $X$. Then take a pointwise limit in each projection, then show that the sequence converges in norm to the pointwise limit. Have you tried to do this? – JSchlather Dec 06 '12 at 15:47
  • @JacobSchlather. There is nothing pointwise here – Norbert Dec 06 '12 at 15:49
  • @Norbert, I was being sloppy, by pointwise I meant pointwise in the sense of the product. Rather what I meant was a component wise limit. If $x_n$ is a Cauchy sequence in $X$ then for each $I \in \mathcal I$ we have a Cauchy sequence $(x_n)_I$ in $X_I$ and since each $X_I$ is a Banach space, we can look at its limit $x_I$. Then define $x$ this way. – JSchlather Dec 06 '12 at 15:51
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    Obvious observation. To construct a counterexample it is necessary to find at least two complete norms on $X$ which give different limits for the same sequence ${x_n}$. Wlog we can assume that for all $i\neq j$ operator $\mathrm{id}:(X,\Vert\cdot\Vert_i)\to(X,\Vert\cdot\Vert_j)$ is unbounded. – Norbert Dec 06 '12 at 15:54
  • Maybe we should first try some examples on which we have many different norms. This already seems difficult for me. – Hui Yu Dec 06 '12 at 16:41
  • Also as Norbert has pointed out, this forces the identity mapping to be unbounded from between two norms, which also seems strange to me. – Hui Yu Dec 06 '12 at 16:43
  • Another point is that if your banach space has banach algebra structure and an involution, then the identity is a -(algebraic)-homomorphism, so it automatically becomes continuous. So to find such an example we either forget all -algebras (which covers almost everything I know), or we construct some norms on them that make them Banach but the multiplication fails to be continuous. – Hui Yu Dec 06 '12 at 16:49

1 Answers1

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The answer is no.

If $I$ is finite, then $X_\infty=X$ as linear spaces, so I assume $I$ is finite. Using open mapping theorem one can show that if $(X_\infty,\Vert\cdot\Vert_\infty)$ is Banach space then all the spaces $(X_\infty,\Vert\cdot\Vert_i)$ are isomorphic. Indeed, for all $i\in I$ the map $$ \mathrm{Id}:(X_\infty,\Vert\cdot\Vert_\infty)\to(X_\infty,\Vert\cdot\Vert_i):x\mapsto x $$ is bounded and bijective, hence isomorphism. Since for all $i\in I$ we have $(X_\infty,\Vert\cdot\Vert_\infty)\cong(X_\infty,\Vert\cdot\Vert_i)$, the spaces $(X_\infty,\Vert\cdot\Vert_i)$ for $i\in I$ are pairwise isomorphic.

Now to construct a counterexample it is sufficient to introduce finitely many non-isomorphic Banach space structures on the same linear space $X$. I'll construct two structures.

Let $(X,\Vert\cdot\Vert)$ be a separable Banach space non-isomorphic to $\ell_2$. For example take $X=(\ell_p,\Vert\cdot\Vert_p)$ with $p\neq 2$. Now consider norm $\Vert\cdot\Vert'$ given in this answer which turns $X$ into a Hilbert space. By construction $(X,\Vert\cdot\Vert)$ and $(X,\Vert\cdot\Vert')$ are not isomorphic. So norms $\Vert\cdot\Vert$ and $\Vert\cdot\Vert'$ gives the desired counterexample.

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Norbert
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  • Could you expand a little bit on how the open mapping theorem gives that result? It seems to me that if I take $X_1=\mathbb R$ and $ X_2=\mathbb R^2$ then $X_\infty \cong \mathbb R^3$ is a banach space but $X_1$ is not isomorphic to $X_2$. – JSchlather Dec 06 '12 at 17:21
  • Could you detail the part about open mapping theorem? – Davide Giraudo Dec 06 '12 at 17:22
  • @JacobSchlather As far as I understand the question linear space is alwas the same. The only thing is changing is the norm. – Norbert Dec 06 '12 at 17:31
  • Do you mean $(X_\infty,\lVert\cdot\rVert_\infty$ at the first line (hence at the third)? – Davide Giraudo Dec 06 '12 at 17:40
  • @DavideGiraudo Now I see mistake in my proof. because $X_\infty\neq X$ in general so $\mathrm{Id}$ is not well defined. – Norbert Dec 06 '12 at 17:47
  • But in your example, $X_{\infty} = \ell_p$ so it seems it goes through. In fact, you show that if $X$ is a vector space and $||;||_i$ for $i = 1, 2$ are two complete incomparable norms, then the norm $\max_i ||;||_i$ can't be complete. – levap Dec 06 '12 at 17:48
  • @levap Yes you are right one can salvage this proof by minor modifications. – Norbert Dec 06 '12 at 17:49
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    Okay. I guess now my question is why $(X_\infty,||x||_i)$ is a Banach space. – JSchlather Dec 06 '12 at 18:12
  • @JacobSchlather, what is wrong with my proof? – Norbert Dec 06 '12 at 19:07
  • In order to apply the Open mapping theorem, you need that both spaces are Banach Spaces. So you need to deduce that $(X_\infty, ||\cdot ||\infty)$ is a banach space implies that $(X\infty, ||\cdot ||i)$ is also a banach space. Otherwise you could have the issue that the map is not bounded from below. I think you're still right, this just doesn't quite get you there yet. It also seems to me that there could be elements $x \in X$ such $x \notin X\infty$ in which case your map is not surjective. – JSchlather Dec 06 '12 at 19:18
  • I assume that $(X_\infty,\Vert\cdot\Vert_\infty)$ is Banach and deduce necessary conditions. Then I show that this conditions are not always satisfied, hence $(X_\infty,\Vert\cdot\Vert_\infty)$ is not Banach in general. Also in my counterexample $I$ is finite, hence $X_\infty=X$. – Norbert Dec 06 '12 at 20:03
  • Thanks a lot! I think the idea of constructing a new Hilbert space out of a separable Banach space is pretty neat. Do you think we can find more elementary examples if we maybe allow $\mathcal{I}$ too be countable or even bigger? – x_Y_z Dec 07 '12 at 11:56
  • @x_Y_z I think (this is just my opinion) that we can avoid usage of axiom of choice in this construction. Axiom was used when we introduces Hilbert norm on $X$. – Norbert Dec 07 '12 at 18:02