Let me first fix some notations. 


If $A$ and $B$ are nonempty subsets of a Banach space $X$, we set $$d(A,B)=\inf\{\|a-b\|:a\in A,b\in B\},$$$$\widehat{d}(A,B)=\sup\{d(a,B):a\in A\}.$$ 


Let $A$ be a bounded subset of a Banach space $X$. The Hausdorff measure of non-compactness of $A$ is defined by $$\chi(A)=\inf\{\widehat{d}(A,F):F \subset X\},$$
where the infimum is taken over all finite subsets $F$ of $X$. Clearly, $A$ is relatively compact if and only if $\chi(A)=0$.


The following result is well-known (see Proposition 2.c.4 in the book of J. Lindenstrauss and L. Tzafriri):


Let $X,Y$ be infinite-dimensional Banach spaces. Assume that an operator $T: X\rightarrow Y$ is such that the restriction of $T$ to any subspace of finite co-dimension in $X$ is not an isomorphic embedding. Then, $\forall \epsilon>0$, there exists an infinite-dimensional closed subspace $M$ such that $T|\_{M}$ is compact and $\|T|\_{M}\|<\epsilon$.


I am thinking about the quantitative version of this result.


Question: Let $X,Y$ be infinite-dimensional Banach spaces and let $c>0$. Assume that $T: X\rightarrow Y$ is an operator such that $\|T\|=1$ and $\chi(TB\_{M})>c$ for every infinite-dimensional closed subspace $M$ of $X$. Is there an infinite-dimensional closed subspace $X\_{0}$ of $X$ such that $\|Tx\|\geq c\|x\|$ for all $x\in X\_{0}$?


Thank you!



A question on the quantification of compact operators
A negative answer to your question follows from a well-known result of Schlumprecht (Israel J. Math 1991): there exist arbitrarily distortable Banach spaces. Schlumprecht's result implies that for every $\lambda>1$ there exist a Banach space $W$, a Banach space $Y$ isomorphic to $W$, and an isomorphism $T:W\to Y$ such that for every infinite-dimensional subspace $Z$ of $W$ there exists $u,v\in Z$ with $||u||=||v||=1$ and $||Tu||/||Tv||\ge\lambda$. Let $\mu=\inf\_M\chi(TB\_M)$, where the infimum is over all infinite-dimensional subspaces $M$ of $W$. Let $M\_0$ be an infinite-dimensional subspace of $X$ for which $\chi(TB\_{M\_0})\le (1+\varepsilon)\mu$. Using the standard argument one can find an infinite-dimensional subspace $M\_1$ of $M\_0$ such that $||T|\_{M\_1}||\le 3(1+\varepsilon)\mu$. The statement of Schlumprecht's theorem then implies that for every infinite-dimensional subspace of $M\_1$, there is an element $w$ such that $||Tw||\le 3(1+\varepsilon)\mu/\lambda$. Picking suitable $\varepsilon$ and $\lambda$, we get the desired conclusion for $X=M\_1$ (one should normalize $T|\_X$ if you insist on $||T||=1$).


**Added on request:** Existence of $M\_1$ can be shown in the following way. One can find
a finite $((1+\varepsilon)\mu+\varepsilon)$-net $\{x\_i\}$ in
$TB\_{M\_0}$. Let $\{f\_i\}$ be norm-one functionals satisfying
$f\_i(x\_i)=||x\_i||$. Let $M\_1=\{x\in M\_0:~ \forall i~ f\_i(Tx)=0\}$.
We show that this is the desired subspace. If not there is $z\in
B\_{M\_1}$, such that $||Tz||>3(1+\varepsilon)\mu$. There is $x\_j$ such that
$||Tz-x\_j||\le(1+\varepsilon)\mu+\varepsilon$. Thus, on one hand
we have $$f\_j(x\_j)\le f\_j(Tz)+(1+\varepsilon)\mu+\varepsilon=
(1+\varepsilon)\mu+\varepsilon,$$ and on the other hand
$$f\_j(x\_j)=||x\_j||\ge||Tz||-||Tz-x\_j||>
2(1+\varepsilon)\mu-\varepsilon.$$ We get a contradiction for
sufficiently small $\varepsilon$.


