M. Schechter introduced an operational quantity characterizing strictly singular operators as follows:

For an operator $T:X\rightarrow Y$, we set $$\tau(T)=\sup_{M}\inf_{x\in S_{M}}\|Tx\|,$$ where $M$ represent an infinite-dimensional closed subspace of $X$.

If $A$ and $B$ are two nonempty subsets of a Banach space $X$, we set $$d(A,B)=\inf\{\|a-b\|:a\in A,b\in B\},$$$$\widehat{d}(A,B)=\sup\{d(a,B):a\in A\}.$$ Thus, $d(A,B)$ is the ordinary distance between $A$ and $B$, and $\widehat{d}(A,B)$ is the non-symmetrized Hausdorff distance from $A$ to $B$.

Let $A$ be a bounded subset of a Banach space $X$. The Hausdorff measure of non-compactness of $A$ is defined by $\chi(A)=\inf\{\widehat{d}(A,F):F\subset X$ finite subset $\}$.

Then $\chi(A)=0$ if and only if $A$ is relatively norm compact. For an operator $T: X\rightarrow Y$, $\chi(T)$ will denote $\chi(TB_{X})$.

I prove the following result:

Theorem. Let $T:X\rightarrow Y$ be an operator. Then $$\tau(T)\leq 2\chi(T).$$

I have two questions about this theorem.

Question 1. Is this theorem new? I am not sure that this theorem has already appeared somewhere.

Question 2. Is the constant 2 in the theorem optimal?

Thank you!


It seems to me that one can show that $\tau(T)\leq \chi(T)$ as follows. The case $\tau(T)=0$ is clear. Let $\tau(T)>0$, $\varepsilon\in(0,\tau(T))$, and let $M\subset X$ be an infinite dimensional subspace for which $\inf_{x\in S_{M}}\|Tx\|\ge \tau(T)-\varepsilon$. Then $T|M$ is an isomorphism and $T(B_M)$ contains a $(\tau(T)-\varepsilon)$-multiple of a ball of an infinite-dimensional subspace of $Y$. By the lemma which goes back to M. Krein, M. Krasnoselskii, and D. Milman (see Lemma 2.c.8 in Lindenstrauss-Tzafriri, volume I) we get that $\chi(TB_{X})\ge \tau(T)-\varepsilon$. Letting $\varepsilon\downarrow 0$, we get the desired conclusion.


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