inequality of norms 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 $\|\|_Z$?
 A: 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 $Z$.
A: 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.
