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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!

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  • $\begingroup$ 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
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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.

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

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