Can we find a banach space X on which there are two non-equivalent norms, but they induce the same topology? I am almost sure that there should be such example, otherwise it would seem to be a rather strong and surprising result. (Note that we should look for an infinite dimensional X, and the homeomorphism f, if it exists, can not be linear by the open mapping theorem.)
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1Any two infinite dimensional separable Banach spaces are topologically homeomorphic. See this. – David Mitra Mar 06 '14 at 12:50
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This is really interesting, since combined with the answers below it shows any two norms induce equivalent topologies but that only equivalent norms induce equal topologies. – Kevin Carlson Mar 06 '14 at 13:20
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In most cases, In David's comment, the two spaces are homeomorphic, but there is no linear homeomorphism. But, as the question is formulated (two toplogies on the same space) the map is the identity, and therefore linear. (Otherwise user would probably have mentioned two additive structures as well as two topologies?) – GEdgar Mar 06 '14 at 13:25
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If two different norms induce the same topology, then the identity map is a continuous linear map when $X$ is considered with one of the norms in the domain and the other one in the codomain. So it is bounded, which leads to the norms being comparable, and so equivalent as we can reverse roles.
Martin Argerami
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This also works merely for normed spaces ... we do not need to assume completeness. – GEdgar Mar 06 '14 at 13:23
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Same topology implies same neighborhoods of zero. Le be $B_1(0,r)$ a zero-centered ball in $\|\|_1$. By hypothesis is a neighborhood of zero in $\|\|_2$, so there is $B_2(0,\rho)\subset B_1(0,r)$. Repeat the reasoning exchanging the norms and we have $B_1(0,s)\subset B_2(0,\rho)\subset B_1(0,r)$. Translate the inclusions to inequalities between norms and use homogeneity.
Martín-Blas Pérez Pinilla
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