Let $X$ and $Y$ be two Banach spaces with norms $\|\|_X$ and $\|\|_Y$ respectively. If $Z=X\times Y$ is also a Banach space with norm $\|\|_Z$ then what is the relation between $\|\|_X,\:\|\|_Y$ and $\|\|_Z$?
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3$\begingroup$ The question is ill-posed. Just putting a norm on the Cartesian product somehow is a random thing to do. See en.wikipedia.org/wiki/… for the usual direct sum. $\endgroup$– Charles MatthewsCommented Mar 24, 2011 at 9:41
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3$\begingroup$ "The" relation - why do you think there is a unique one? This question in its current form is just casting a net behind a trawler... $\endgroup$– Yemon ChoiCommented Mar 24, 2011 at 9:48
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2 Answers
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Maybe I'll risk an answer. It is perhaps natural to expect that the projection maps $$ P_X: Z\rightarrow X; \qquad P_Y:Z\rightarrow Y $$ be bounded. If so, then $Y \cong \ker P_X$ is closed in $Z$, and $X\cong\ker P_Y$ is closed in $Z$. By the Open Mapping Theorem, $Z/Y \cong X$ and $Z/X\cong Y$, which means that the norm on $Z$ is equivalent to the norm $\|(x,y)\|_1 = \|x\|+\|y\|$. Conversely, of course, if $Z$ has a norm equivalent to $\|\cdot\|_1$, then the projections will be bounded.
So: the natural projections are bounded if and only if you have a "nice" norm on $Z$.
answered Mar 24, 2011 at 12:27
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$\begingroup$ It suffices to make the following assumption: If $x_n\to x$ in X, $y_n\to y$ in Y and $(x_n,y_n)\to z$ in Z, then z=(x,y). If we assume this, then Z becomes a Banach space under the norm $\|(x,y)\|=\|(x,y)\|_Z+\|x\|_X+\|y\|_Y$. Since this norm is stronger than either $\|.\|_Z$ or $\|.\|_1$, it must be equivalent to both by the open mapping theorem. $\endgroup$ Commented Mar 24, 2011 at 14:04
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$\begingroup$ I think, by the Closed Graph Theorem, your condition on sequences is equivalent to $P_X, P_Y$ being bounded. $\endgroup$ Commented Mar 24, 2011 at 16:30
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OK, here is an example of what one has to expect: the two Banach spaces of convergent sequences $c$ and of square summable sequences $\ell^2$ are Banach spaces ($\ell^2$ is even Hilbert. As vector spaces they have the same dimension with a vector spaces basis (Hamel basis) of the same cardinality as $\mathbb{R}$ (maybe we need some set-theoretic stuff here as a continuum hypothesis, who knows...?) Thus there is a linear isomorphism $c \longrightarrow \ell^2$ which allows to transport the Banach norm of $c$ to $\ell^2$. This way, $\ell^2$ becomes a Banach space in two ways. One is Hilbert the other one is not even reflexive. So the topologies can not be comparable at all.
Note that this is not in contradiction to the open mapping theorem.
For your question, this means that it may happen that the three norms have no relation whatsoever.
answered Mar 24, 2011 at 12:23
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$\begingroup$ Stefan, it is provable in ZFC that every separable infinite dimensional Banach space has Hamel dimension the continuum. $\endgroup$ Commented Mar 24, 2011 at 13:35
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$\begingroup$ @Bill: thanks, that was the missing link. At some point I heared something like that, but I couldn't remember where. $\endgroup$ Commented Mar 24, 2011 at 13:44