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Let $X,Y$ be two isomorphic Banach spaces, and

$$ d(X,Y):=\inf\{\|T\|\|T^{-1}\|: T \textrm{ is an is an isomorphism from $X$ to $Y$} \} $$

the Banach Mazur distance between $X$ and $Y$. Assuming that $\|T\|=1$, it is easy (see this post) that this distance is the infimum of the numbers $r\geq 1$ such that

$$ T(U_{X}) \subset U_{Y} \subset r T(U_{X}), $$

where $U_{X}$ and $U_{Y}$ are, respectively, the closed unit balls of $X$ and $Y$.

I have seen here, that the Banach Mazur distance can be interpreted as the infimum of the numbers $r\geq 1$ such that

$$ U_{Y} \subset T(U_{X}) \subset r U_{Y}, $$

Why are true the above inclusions? Maybe it's obvious, but I can't prove it formally.

Thanks!

user123043
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1 Answers1

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$U_Y \subseteq T(U_X)$ says for every $y \in Y$ with $\|y\| \le 1$, there is $x \in X$ with $y = Tx$ and $\|x\| \le 1$. Since $x = T^{-1} y$, that is equivalent to $\|T^{-1}\|\le 1$.

$T(U_X) \subseteq r U_Y$ says $\|Tx \| \le r$ for every $x \in X$ with $\|x\| \le 1$, i.e. $\|T\| \le r$.

Thus if there exists isomorphism $T$ with $U_Y \subseteq T(U_X) \subseteq r U_Y$, this isomorphism has $\|T\| \|T^{-1}\| \le r$, and the Banach-Mazur distance $\le r$.

Conversely, if the Banach-Mazur distance is $d$, for any $\epsilon > 0$ there is an isomorphism $T$ with $\|T\| \|T^{-1}\| \le d+\epsilon$. We may assume wlog that $\|T^{-1} \| = 1$, and we have $U_Y \subseteq T(U_X) \subset (d+\epsilon) U_Y$. So the infimum of the numbers $r$ as described is at most $d$.

Robert Israel
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