If $(X,\|\cdot\|)$ is an n- dimensional normed linear space over R. Is it isomorphic to n-dimensional Euclidean space $R^n$. I know it is topologically isomorphic but what about isometry? I think if it is an inner product space then by considering some orthonormal basis we can have inner product space isomorphism but not with normed linear space structure. Am I correct?
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I am asking the converse. I know they are linearly isomorphic but do we have isometric isomorphism? – akansha Jan 22 '15 at 15:50
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1No, the spaces are not isometrically isomorphic. For instance, $({\mathbb R}^n, | \cdot|_2)$ and $({\mathbb R}^n, | \cdot|_p)$, where $1\leq p\ne 2 \leq \infty$, are not isometrically isomorphic. – Janko Bracic Jan 22 '15 at 15:51
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How to show it? – akansha Jan 22 '15 at 15:57
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1I think if we take inner product space then it is inner product space isomorphic. – akansha Jan 22 '15 at 15:57
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Try to see what is going on in the case of ${\mathbb R}^2$. – Janko Bracic Jan 22 '15 at 15:59
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Related to my previous comment: consider unit balls in $| \cdot|_p$ norms for different $p$ (the case ${\mathbb R}^2$ is enough). – Janko Bracic Jan 22 '15 at 16:01
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Exactly as you say, isometric isomorphism would preserve parallelogram law. This should help you to show that, for example, $(\mathbb R^2,|\cdot|_1)$ and $(\mathbb R^2,|\cdot|_2)$ are not isometrically isomorphic. (And, additionaly, only one of them is an inner product space.) – Martin Sleziak Jan 22 '15 at 16:30
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About finitely dimensional inner product spaces: http://math.stackexchange.com/questions/42663/why-is-it-true-that-every-finite-dimensional-inner-product-space-is-a-hilbert-sp – Martin Sleziak Jan 22 '15 at 16:31
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this paper could be interesting to you – Norbert Jan 22 '15 at 20:50