# Supremum of Banach Spaces

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.

• 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 6 '12 at 15:47
• @JacobSchlather. There is nothing pointwise here – Norbert Dec 6 '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 6 '12 at 15:51
• 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 6 '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 6 '12 at 16:41

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.
• 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 6 '12 at 17:21
• Do you mean $(X_\infty,\lVert\cdot\rVert_\infty$ at the first line (hence at the third)? – Davide Giraudo Dec 6 '12 at 17:40
• Okay. I guess now my question is why $(X_\infty,||x||_i)$ is a Banach space. – JSchlather Dec 6 '12 at 18:12