Let $E, \left \| \right \|$ be a Banach space, $\mathfrak{M}_E$ indicate a family of all nonempty bounded subset of $E$, $\mathfrak{N}_E$ the familly of all relatively compact sets, and $Ker \mu=\{X\in \mathfrak{M}_E$ such that: $\mu(X)=0\}$.

Definition: A mapping $\mu:\mathfrak{M}_E\rightarrow \mathbb R^+$ is said to be a measure of noncompactness in $E$ if it satisfies the following conditions:

1) The family $\operatorname{ker} \mu $ is nonempty and $\operatorname{ker} \mu \subset \mathfrak{N}_E$.

2) $X\subseteq Y\Rightarrow \mu (X)\leq \mu (Y)$.

3) $\mu (\bar{X})=\mu (X)$.

4) $\mu (\operatorname{Conv}(X))=\mu (X)$.

5) $\mu (\lambda X+(1-\lambda) Y)\leq \lambda \mu (X)+(1-\lambda) \mu(Y)$ for $\lambda\in [0,1]$.

6)If $\{X_n\}\subset \mathfrak{M}_E^c$, such that $X_{n+1}\subset X_n$ for $n=1,2,...$ and if $\displaystyle\lim_{n\rightarrow \infty} \mu(X_n)=0$ then $X_{\infty}=\bigcap_{n=1}^{\infty}X_n\neq \emptyset$.

Now, I have to prove this this:

Suppose $\mu_1,\mu_2$ are two measures of noncompactness in $E_1,E_2$ respectively. Moreover, assume that the function $F:\mathbb [0,\infty)^2\rightarrow \mathbb R^+$ is a convex function and $F(x_1,x_2)=0\Leftrightarrow (x_1,x_2)=0$. Then $\mu (X)=F(\mu_1(X_1),\mu_2(X_2))$ defines a measure of noncompactness in $E_1\times E_2$ where $X_i$ denote the natural projection of $X$ into $E_i$ for $i=1,2$.

It's okay with 1, 3, 4 and 6 of the definition. How can we prove 2 and 5?

EDIT: This result (without proof) comes from : Bana’s, J., Goebel, K., Measures of Noncompactness in Banach Spaces. Lecture Notes in Pure and Applied Mathematics, vol. 60, New York: Dekker 1980

  • 3
    $\begingroup$ What I mean is, how do you know it's true? It's from homework, or an article, or …? If an article, which one? $\endgroup$
    – LSpice
    Apr 24, 2018 at 15:45
  • $\begingroup$ What are $\mathfrak{M}_E$ and $\mathfrak{N}_E$, and what is $\ker \mu$? $\endgroup$ Apr 24, 2018 at 17:08

1 Answer 1


Property 2 and 5 hold, provided $F:\mathbb{R}_+^2\to\mathbb{R}_+$ is increasing in each variable (which I think the authors implicitly assumed). Indeed:

(2) $X\subset Y$ implies $X_1\subset Y_1$ and $X_2\subset Y_2$, thus $\mu_1(X_1)\le \mu_1(Y_1)$ and $\mu_2(X_2)\le \mu_2(Y_2)$ and finally $F(\mu_1(X_1),\mu_2(X_2))\le F(\mu_1(Y_1),\mu_2(Y_2))$.

Moreover, by the monotonicity and by the convexity of $F$:

(5) $\mu(\lambda X+(1-\lambda)Y)=F\big(\mu_1(\lambda X_1+(1-\lambda)Y_1),\mu_2(\lambda X_2+(1-\lambda)Y_2)\big)$ $\le F\big(\lambda\mu_1( X_1)+(1-\lambda)\mu_1(Y_1),\lambda\mu_2( X_2)+(1-\lambda)\mu_2(Y_2)\big)$ $\le \lambda F\big(\mu_1( X_1),\mu_2( X_2)\big)+(1-\lambda)F\big(\mu_1(Y_1),\mu_2(Y_2)\big)=\lambda\mu(X)+(1-\lambda)\mu(Y).$

About the necessity of this condition. Note that a convex positive function on $\mathbb{R}_+^2$, vanishing exactly at the origin need not to be increasing in each variable: e.g. $F(x,y)=(x-y)^2+ y^2$ has $F(0,1)=2>F(1,1)=1$. Also note that the definition implies that the range of a measure of non-compactness on a Banach space of infinite dimension is a set of positive real numbers that accumulates at $0$. Therefore, apart the trivial case of $E_1$ and $E_2$ both finite dimensional, the above monotonicity condition on $F$ is necessary in order that $F(\mu_1,\mu_2)$ satisfy 2.

Details. (In case it wasn't clear enough) Suppose you have a pair $\mu_1$ and $\mu_2$ of measures of noncompactness on infinite dimensional Banach spaces resp. $E_1$ and $E_2$, with unit balls $B_1$ and $B_2$, and suppose your 2-variables convex $F:\mathbb{R}_+^2\to\mathbb{R}_+$, with $F^{-1}(0)=(0,0)$, produces via your construction a measure of non-compactness $\mu$ at least for this pair $\mu_1,\mu_2$. Then by 2, $$\mathbb{R}_+^2\ni (x,y)\mapsto \mu(xB_1\times yB_2):=F(\mu_1(xB_1),\mu_2(yB_2))$$ is increasing separately wrto $x$ and wrto $y$. But both
$\mathbb{R}_+ \ni t\mapsto \mu_i(tB_i),$ for $i=1$ and $i=2$, are increasing (maybe with jumps), not identically zero, and $o(1)$ for $t\to0$. This implies that $F(x,y)$ itself must be increasing separately in both variables everywhere in the rectangle $[0,\sup \mu_1]\times[0,\sup\mu_2]$, being convex.

  • $\begingroup$ Thanks for your feedback @Pietro Majer, I agree with you in all what you have written, PROVIDED THAT $F$ is partially increasing which is not supposed by the author in the paper. I do not know if we can use the convexity to deduce a particular type of increase ! $\endgroup$
    – Motaka
    Apr 27, 2018 at 9:58
  • $\begingroup$ "I do not know if we can use the convexity to deduce a particular type of increase": what do you mean? Il a convex function $F$ on $\mathbb{R}^2_+$, for just one pair of measures of non-compactness $\mu_1$ and $\mu_2$ on infinite dimensional spaces, gives rise to a $F( \mu_1,\mu_2)$ satisfying (2), then $F$ must be increasing in both variables. $\endgroup$ Apr 27, 2018 at 11:22
  • $\begingroup$ I mean that, for example IF $F$ was like that: $F:\mathbb R^+\rightarrow \mathbb R^+$ with the conditions above , we will have the result ($F$ will be increasing here), but in $ \mathbb R^2_+$ I can't see why it 's okay. . If I get your idea: "we have to suppose that $F$ is be increasing in $ \mathbb R^2_+$ to have the result, otherwise we can't" $\endgroup$
    – Motaka
    Apr 27, 2018 at 14:27
  • 1
    $\begingroup$ I'm saying, $F$ must be increasing in both variables, otherwise it is false that it produces a measure of non compactness. Added details. $\endgroup$ Apr 27, 2018 at 21:11
  • $\begingroup$ I get it! Thank you very much Sir Even if it seems very strange because this definition and this theorem was from a classical paper of Jonas, which is used by others.. $\endgroup$
    – Motaka
    Apr 28, 2018 at 10:43

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