A measure of noncompactness by a convex function 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
 A: 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.
