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!

  • $\begingroup$ It is proved in Corollary 3 of "Measures of non-compactness of operators on Banach lattices" (Troitsky, 2002) that this is true when $X=Y=Z$. I am not sure about the general case. $\endgroup$ – Ben W Nov 2 '16 at 17:45
  • $\begingroup$ I guess "Semigroups of Operators and Measures of Noncompactness" (Lebow/Schechter, 1971) mentions that the the answer is yes in (3.5) of that paper. They do not give a proof, and their reference is in Russian. But presumably the proof is routine. $\endgroup$ – Ben W Nov 2 '16 at 18:15
  • $\begingroup$ Thanks, Ben. "Semigroups of Operators and Measures of Noncompactness" (Lebow/Schechter, 1971) mentioned the question. But they do not give a proof. Since their reference is in Russian, I can not download the paper. I want to know the proof. $\endgroup$ – Dongyang Chen Nov 3 '16 at 0:30

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.

  • $\begingroup$ Thanks, August. The proof is standard, but I can not realize it. $\endgroup$ – Dongyang Chen Nov 3 '16 at 13:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.