Cartesian product of Banach spaces: all norms such that the inclusion is an isometry are equivalent?

Question

Let $\mathcal{A}$ be an arbitrary (typically infinite-dimensional) Banach space with norm $\|\cdot\|_{\mathcal{A}}$ and let $\mathcal{A}^{n}$ be its Cartesian product. I came across the following statement and wonder whether it is true or not:

All norms $\|\cdot\|_{\mathcal{A}^{n}}$ on $\mathcal{A}^{n}$ such that $\|(0,...,0,\cdot,0,...,0)\|_{\mathcal{A}^{n}}=\|\cdot\|_{\mathcal{A}}$ for all $i=1,...,n$ (i.e. the inclusion is an isometry) are equivalent.

It is clear that for every $x=(x_{1},...,x_{n})\in\mathcal{A}^{n}$ it holds:

\begin{equation} \|x\|_{\mathcal{A}^{n}}\leq\sum_{i=1}^{n}\|x_{i}\|_{\mathcal{A}}=:\|x\|_{1} \end{equation}

Therefore, $\|\cdot\|_{1}$ is stronger than $\|\cdot\|_{\mathcal{A}^{n}}$.

What what about the other direction? Is it true?

Thanks for your help.

Answer 1

It is actually not true in general that such a norm $\Vert \cdot \Vert_{\mathcal{A}^n}$ must be complete, despite the fact that the contrary is presented as fact in reputable sources in the literature (see, e.g., Section B.4.11 of Albrecht Pietsch's book Operator Ideals [the version published in 1980 by North-Holland] for the case $n=2$]).

A reference for the fact that $\Vert \cdot \Vert_{\mathcal{A}^n}$ need not be complete is the paper of Eve Oja and Peeter Oja, On the completeness of Cartesian products of Banach spaces (in Russian). Acta et commentationes Universitatis Tartuensis, 661 (1984), 33−35. The English summary of the Oja-Oja paper can be read here.

I think the Oja-Oja paper is hard to come by, so when I learnt of its existence many years ago I believe I just derived my own example to satisfy myself. The Oja-Oja paper also notes that equivalence (hence also completeness) is assured when an additional condition is assumed. In particular it follows easily from the triangle inequality that if there exists a constant $c>0$ such that $$ \Vert (-x_1,\ldots,-x_{i-1},x_i,-x_{i+1},\ldots,-x_n)\Vert_{\mathcal{A}^n}\leq c \Vert (x_1,\ldots,x_{i-1},x_i,x_{i+1},\ldots,x_n)\Vert_{\mathcal{A}^n}$$ for all $x_1,\ldots, x_n\in\mathcal{A}$ then $\Vert\cdot\Vert_{\mathcal{A}^n}$ is $(1+c)$-equivalent to the norm $\Vert\cdot\Vert_1$ as defined in your question.

(If I have time I might come back to this later and give some details of how to construct a counterexample, but I recall that it's not particularly difficult).

Edit: Bill Johnson's comment below notes a nice way of obtaining a counterexample.