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We know that any two n-dimensional normed vector spaces $X,Y$ are isomorphic, and we define the Banach-Mazur distance between $X,Y$ as $$ d(X,Y)=\inf \{ \|T\|\|T^{-1}\|:T\in GL(X,Y) \} ,$$ where $GL(X,Y)$ is the space of all linear isomorphisms. What is the motivation behind this definition? Are there other kind of metric that we can define similarly between finite dimensional normed spaces? And finally, whether there is a similar notion for infinite dimensional version? As for the context, I am reading about $\mathcal{L}_p$-spaces which utilizes this the Banach-Mazur distance.

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up vote 5 down vote accepted

For an invertible linear map $T: X \rightarrow Y$, the quantity $\|T\| \|T^{-1}\|$ is called the condition number of the operator, and measures the proportional error accrued by output in $Y$ of $T$, when an error is introduced to the argument in $X$. See here for details. The errors are measured in the respective norms of each of $X$ and $Y$, so the quantity $\|T\| \|T^{-1}\|$ measures how norm "stable" the process of passing through $X$ to $Y$ via the isomorphism $T$ is. In numerical analysis, especially, minimizing the condition number is always desired, as it means that if your computer introduces some small roundoff error or something, then this will only lead to a small error in the final result.

In a more theoretical context, by taking the infimum over all isomorphisms $T: X \rightarrow Y$, we are looking for the optimal constants used to relate the norms in each space; in other words, how close to an isometry can we get? In your situation, for example, we know that for $\mathbb{C}^n$, given the $\ell_p$ and $\ell_q$ norms and isomorphism $T$, we can find optimal estimates like $\|x\|_p \le C\|Tx\|_q$, but $C$ can never be exactly 1.

As for your last question, I don't think there's any restriction on taking either finite or infinite dimensional normed linear spaces, though I don't claim to be an expert on such matters. Of course, if the spaces are infinitely-dimensional, you have to ensure that they really are isomorphic.

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I will try to give a more geometric explanation. First note that you can always scale $T$ such that it has norm one. Hence, the Banach-Mazure distance can be rewritten as:

$$ d(X,Y)=\inf\{||T^{-1}||: T\in GL(X,Y), ||T||=1\} $$

Geometrically, $||T||=1$ means that $T(B_X)\subseteq B_Y$ and no enlargement of $T(B_X)$ will still fit inside $B_Y$ ($B_X$ and $B_Y$ represent the unit balls of the two spaces). On the other hand, we have that

$$ T^{-1}(B_Y)\subseteq||T^{-1}||B_X $$

or equivalently

$$ B_Y\subseteq||T^{-1}||T(B_X) $$


$$ T(B_X)\subseteq B_Y\subseteq||T^{-1}||T(B_X) $$

Thus, geometrically, $||T^{-1}||$ represent the smallest amount by which you must increase $T(B_X)$ such that it contains $B_Y$. The Banach-Mazur distance represents the infimum of such enlargments, taken over all linear isomorphism that send $B_X$ inside $B_Y$.

For a perhaps a better intuitive understanding, take $B_X$ to be the unit sphere. Then for any isomorphism $T$, $T(B_X)$ is going to be an ellipsoid. For the banach-Mazur distance, you are looking for the "best fit" ellipsoid. That is, you are looking for the ellipsoid that fits inside $B_Y$ (touching the boundary), such that the enlargement required for this ellipsoid to contain $B_Y$ is as small as possible.

Yes, there is a similar notion for infinite dimensional Banach spaces, with the convention that when $X$ and $Y$ are not isomorphic, $d(X,Y)=\infty$.

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