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Let for each $n$ there is a vector space $V_n$ given which carries two Banach-space norms $\|\cdot \|_i$ ($i=1,2$) which are

1) equivalent

2) $\|x\|_1\leqslant \|x\|_2$ for each $x\in V_n$.

Consider the $\ell_\infty$-sums: $X_i=\sum_{i=1}^\infty \oplus_\infty (V_n, \|\cdot\|_i)$ ($i=1,2$).

Must the spaces $X_1$ and $X_2$ be isomorphic?

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Let each $V_n=\Bbb R$. For $x\in V_n$ let $\|x\|_1=|x|$ and $\|x\|_2=n|x|$. Then $X_1$ is just $\ell_\infty$; is $X_2$? –  Brian M. Scott Oct 7 '12 at 11:08
    
@TMK The summation index should be $n$, not $i$, no? –  fgp Oct 7 '12 at 11:41
    
Is it true that norms $\Vert\cdot\Vert_1$, $\Vert\cdot\Vert_2$ depends on $n$. Otherwise you had to state that all $V_n$ are the same –  no identity Oct 7 '12 at 14:25
    
@BrianM.Scott: the map $x\mapsto n^{-1}x$ is a linear isometry of $(\mathbb{R},\Vert \cdot \Vert_1)$ onto $(\mathbb{R},\Vert \cdot \Vert_2)$, so in the case you mention $X_2$ is also $\ell_\infty$. –  Philip Brooker Oct 12 '12 at 11:30

1 Answer 1

The answer is no in general. For a counterexample, take $V_n= \mathbb{R}^n$ with norms $\Vert \cdot \Vert_1$ and $\Vert \cdot \Vert_2$ defined by setting $\Vert (x_i)_{i=1}^n\Vert_1 = \sup_{1\leq i\leq n}\vert x_i\vert$ and $\Vert (x_i)_{i=1}^n \Vert_2 = (\sum_{i=1}^n \vert x_i\vert^2)^{1/2}$ for each $(x_i)_{i=1}^n \in \mathbb{R}^n$. Then, up to (isometric) isomorphism, $X_1 = \ell_\infty$ and $X_2=\sum_{n=1}^\infty \oplus_\infty \ell_2^n$. As $\ell_\infty$ has the Dunford-Pettis property, $X_1$ cannot have a complemented subspace isomorphic to $\ell_2$, whereas $X_2$ does have a complemented subspace isomorphic to $\ell_2$; indeed, that $\sum_{n=1}^\infty \oplus_\infty \ell_2^n$ has a complemented subspace isomorphic to $\ell_2$ is shown in Manuel Gonzalez's paper On the duality problem for weakly compact, strictly singular operators, Quaestiones Mathematicae 28(1) (2005), pp.37-38 (perhaps there are other references for this result?). In particular, $X_1$ cannot be isomorphic to $X_2$.

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