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It is well known how to define standard product topology on a product space $\prod_{i \in I} X_i$.

Assume now that $(X,\lVert \, \cdot \, \rVert_{X})$ and $(Y,\lVert \, \cdot \, \rVert_{Y})$ are normed spaces and that the space $X \times Y$ is also equipped with a norm $\lVert \, \cdot \, \rVert_{X \times Y}$.

Is it true that all norms on $X \times Y$ are equivalent?

It is quite easy to prove this if $\lVert \, \cdot \, \rVert_{X \times Y}$ is one of the p-norms, i.e. $\lVert (x,y) \rVert_p = (\lVert x \rVert_X^p + \lVert y \rVert_Y^p)^{1/p}$. All such norms are equivalent. We only need to know that all norms on a finite dimensional space are equivalent (in this case we use it for $\mathbb{R}^2$).

How it is general case?

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If $X$ or $Y$ is infinite dimensional, then $X\times Y$ will be, and there will be inequivalent norms. You probably want to add some condition such as that $(x,y)\mapsto x$ and $(x,y)\mapsto y$ are continuous. –  Jonas Meyer Mar 29 '11 at 17:54
    
Thanks. I thought so but I was confused after reading part of a book. –  xen Mar 29 '11 at 18:02
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No. $X\times Y$ could pretty much be any vector space. There are vector spaces with different norms such that the corresponding normed spaces are not isomorphic. For example, the Banach spaces $c_0$ and $\ell^1$ are isomorphic as vector spaces but not as normed spaces. –  Jonas Meyer Mar 29 '11 at 18:18
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@xenom: I think I saw what you're talking about, on page 49. Carothers says "it doesn't matter what norm we take", but then only really mentions the $p$ norms as being equivalent on finite products. I don't know exactly what was intended. –  Jonas Meyer Mar 29 '11 at 18:48
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@xenom: That is correct; the statement is false if taken literally, because you could do silly things like induce new norms via vector space isomorphisms with normed spaces that are not isomorphic as normed spaces. Carothers doesn't make explicit which general properties should be satisfied, but it looks like only the $p$ norms on the product are intended, $1\leq p \leq \infty$, at least for that section, in which case the norms are far from arbitrary. –  Jonas Meyer Mar 29 '11 at 19:04
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up vote 4 down vote accepted

It is not true in general. When the vector space if finitely dimensional all norms are always equivalent, but when the space has infinite dimension (for example the space of continuous functions on the reals) then norms don't have to be equivalent.

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Now I see. As Jonas Meyer said in infinite dimensional space there will always be inequivalent norms. –  xen Mar 29 '11 at 18:04
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