Given two Banach spaces $X$, $Y$, is always the case that a linear isometry $T: X \to Y$ exists?
I actually just want to find a map preserving the unit sphere in infinite-dimensional spaces, and an linear isometry seems to be the best option.
Bye.
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This is too general and the answer is, of course, no.
I'm assuming that you want $X, Y$ to be infinite-dimensional, although the question itself doesn't explicitly state that. But even in finite dimensions, there is at least one restriction: you need $\dim~X = \dim~Y$.
In infinite dimensions, the examples abound; take for instance $X = L^1(\mathbb{R})$ and $Y = L^2(\mathbb{R})$, with their usual norms. An isometry between $X$ and $Y$ would preserve any properties that depend only on the Banach space structure, but $X$ is not reflexive and $Y$ is. Moreover, $Y$ is Hilbertable, meaning its norm comes from an inner product, whereas in $X$ such a thing cannot be true, for otherwise $X$ would be isomorphic with its dual $X^* = L^{\infty}(\mathbb{R})$, and it's not hard to prove that this doesn't happen.
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Another way how to find out whether a norm can be derived from an inner product is checking [parallelogram law](http://en.wikipedia.org/wiki/Parallelogram_law#Normed_vector_spaces_satisfying_the_parallelogram_law), see also [this question](http://math.stackexchange.com/questions/21792/norms-induced-by-inner-products-and-the-parallelogram-law). – Martin Sleziak Nov 29 '12 at 08:35