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This may be a silly question but every two norms on $\mathbb{R}^n$ are equivalent and $\Vert\cdot\Vert_2$ comes from the usual dot product so $(\mathbb{R}^n,\Vert\cdot\Vert_2)$ is (at least) an inner product space (pre-Hilbert).

Why can't we say that $(\mathbb{R}^n,\Vert\cdot\Vert_{\infty})$ is also pre-Hilbert? I know the uniform norm doesn't come from an inner product but two spaces are (topologically) the same if their norms are equivalent and up until now (at least in my course) it is only the topology that gives the space its "uniqueness".

The fact that a norm comes from an inner product doesn't change the topology nor the algebraic (vector space) structure. I get the product may be useful but what unique, defining, core property ... does it have??

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    $\begingroup$ Because the $\infty$ norm is not induced by an inner product. $\endgroup$ Commented Jan 4, 2019 at 20:26
  • $\begingroup$ @ncmathsadist the question is why would $(\mathbb{R}^n,\Vert\cdot\Vert_2)$ and $(\mathbb{R}^n,\Vert\cdot\Vert_{\infty})$ be any different under any classification if they are essentially the same. What does the inner product have to make the difference (apart from being an inner product) $\endgroup$ Commented Jan 4, 2019 at 20:30
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    $\begingroup$ They are topologically equivalent but, for instance, they are not isometric as Banach spaces. $\endgroup$ Commented Jan 4, 2019 at 20:35

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The question is how do you interpret $\mathbb R^n$. Indeed,

  • if you consider it a real vector space, then it carries a natural inner product.
  • if you consider it a normed space equipped with the Euclidean distance, then yes, it carries an inner product that can be recovered from the norm
  • if you consider it a normed space with some fixed norm, then no. For example, note that the norm coming from an inner product is strictly convex, so it rules out the max norm. For different reasons, the $\ell_p$-norms for $p\neq 2$ are counterexamples too.
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