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$\}$.


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 have the following question:


Question: 
Let $T\in \mathcal{L}(X,Y)$ and $S\in \mathcal{L}(Y,Z)$. Then $$\chi(ST)\leq \chi(S)\chi(T).$$


If this question is already known, could you give a proof?


Thank you!



A question on measure of non-compactness of bounded sets
I think that the following proof should work: Let $F$ be a finite subset of $Y$ such that $T(B\_X) \subset F+(\chi (T) + \varepsilon)B\_Y $, and let $G$ be a finite subset of $Z$ such that $S(B\_Y) \subset G + ( \chi(S)+\varepsilon) B\_Z $. 
Then $ST(B\_X)\subset SF+(\chi(T)+\varepsilon)SB\_Y\subset SF+(\chi(T)+\varepsilon)G+(\chi(T)+\varepsilon) (\chi(S)+\varepsilon)B\_Z$. Observing that $SF+(\chi(T)+\varepsilon)G$ is a finite set, we get
$\chi(ST)\le (\chi(T)+\varepsilon) (\chi(S)+\varepsilon)$. Letting $\varepsilon\downarrow 0$, we get the desired inequality.


