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Hey my fellow Banach space guys. Sorry for another elementary question, and yes I have looked for the past hour and a half to see if I can find it on Google or in my books at home.

Fix $n\in\mathbb{Z}^+$. It seems like there should exist a strictly increasing sequence of $(p_k)\in[1,2)$ such that $p_k\to 2$ and $d(\ell_{p_k}^n,\ell_2^n)\to 1$, where

$$d(X,Y)=\inf\{\|T\|\|T^{-1}\|:T\in\mathcal{L}(X,Y)\text{ is an isomorphism}\}.$$

is the usual Banach-Mazur distance between isomorphic Banach spaces $X,Y$. Probably all I need is a rough estimate of the norm of the canonical isomorphism between $\ell_p^n$ and $\ell_2^n$ and the same for its inverse. But I am unable to show this.

Supposedly there is an estimate in some 1966 paper by Gurarii/Kadets/Matsaev, but it is in Russian. Also Tomczak-Jaegermann supposedly has mentioned the estimate in some of her papers/books, but they are unavailable from my university.

Any help would be much appreciated, thanks.

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    $\begingroup$ I have trouble understanding what the problem is. The distance from $\ell_p^n$ to $\ell_2^n$ is $n^{|1/p-1/2|}$ and is given by looking at the identity operator. The upper estimate, which is what you are seeking, is very easy from Holder's inequality. The harder part is to show that the identity map gives the minimum and this is surely in Tomczak's book (but I do not have a copy here to check). $\endgroup$
    – Bill Johnson
    Commented Oct 13, 2013 at 15:39
  • $\begingroup$ Ah, okay thank you. And as for the problem, well, I'm just not quite as good at all this as some of you folks out there. I sometimes get hung up on trivial stuff like this. Not everyone has the brilliant mind of WBJ. ; ) $\endgroup$
    – Ben W
    Commented Oct 13, 2013 at 16:12
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    $\begingroup$ I am home now and checked that the result I mentioned is proved as Proposition II.E.8 in Wojtaszczyk's book "Banach spaces for analysts". For the upper estimate, he simply says that it is obvious by taking the identity operator (no hint about using Holder, I guess because what else can you do if it is obvious?). $\endgroup$
    – Bill Johnson
    Commented Oct 13, 2013 at 23:22

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