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) $X\subseteq Y\Rightarrow \mu (X)\leq \mu (Y)$.
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> 
> 3) $\mu (\bar{X})=\mu (X)$.
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> 
> 4) $\mu (\operatorname{Conv}(X))=\mu (X)$.
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> 5) $\mu (\lambda X+(1-\lambda) Y)\leq \lambda \mu (X)+(1-\lambda) \mu(Y)$ for $\lambda\in [0,1]$.
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> 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 $i=1,2$.




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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



A measure of noncompactness by a convex function
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.


