Let $\alpha$ to be the [Kuratowski measure of non-compactness](https://en.wikipedia.org/wiki/Measure_of_non-compactness), in a Banach space $E$.


It's very easy to prove that $\alpha (D\_1\times D\_2)\leq \alpha (D\_1)+\alpha (D\_2)$, where $D\_1$ and $D\_2$ are bounded subsets in $E$.


Let $A:E\times E\rightarrow E$ be a mapping. We define $A^0(x,y)=x$ and $A^n(x,y)=A\Big(A^{n-1}(x,y),A^{n-1}(y,x)\Big)$ for all $x,y\in E$, $n\in \mathbb{N}^\*$.



> 
> $A$ is said to satisfy the **condensing condition** if for any $D\_1, D\_2\subset E $ if either $\alpha (D\_1)$ or $\alpha (D\_2)$ is greater than $0$, then we have $$\alpha \Big(A(D\_1,D\_2)\Big)<\max\{\alpha (D\_1),\alpha (D\_2)\}$$
> 
> 
> 


*It is very important for me to construct an operator $T:E\times E\rightarrow E$, which is not a condensing operator, but there is an $n\in \mathbb N^\*$ such that $T^n$ is condensing.* I'd be very thankful if someone can give very easy and simple examples of such an operator.



A non-condensing operator with a power condensing
**Example 1.** The following kind of nilpotent construction satisfies the properties require in the question:


Let $F$ be an infinite dimensional Banach space and let $E = F \times F$ (say, with the maximum norm). Let $I\_F: F \to F$ denote the identity operator and define $T: E \to E$ as the operator matrix
$$
\begin{pmatrix}
 0 & I\_F \\
 0 & 0
\end{pmatrix}.
$$
Note that $T^2 = 0$.


Finally, we define $A: E \times E \to E$ by the formula $A(x,y) = T(x+y)$ for all $x,y \in E$. Then:


* $A^2(x,y) = A(T(x+y), T(x+y)) = 2T^2(x+y) = 0$ for all $x,y \in E$. In particular, $A^2$ is condensing.
* On the other hand, $A$ is not condensing: Let $B \subseteq F$ denote the closed unit ball in $F$ and let $D := \{0\} \times B \subseteq E$. Then $D$ is not compact, so $\alpha(D) > 0$. However,
$$
 A(D,D) = T(D) + T(D) = B \times \{0\} + B \times \{0\} = 2(B \times \{0\}),
$$
so $\alpha(A(D,D)) = 2\alpha(B \times \{0\}) = 2\alpha(\{0\} \times B) = 2\alpha(D) \ge \alpha(D)$.


*Remark.* The equality $\alpha(B \times \{0\}) = \alpha(\{0\} \times B)$ used above follows from the fact that $E \ni (f,g) \mapsto (g,f) \in E$ is an isometry.


**Example 2.** We can simply adjust the nilpotent construction from Example 1 to obtain a non-nilpotent example:


Let $E$ and $A$ be as in Example 1, let $G$ be another Banach space and let $K: G \to G$ be any compact operator.


We define $\tilde E = E \times G$ (with the maximum norm) and $\tilde A: \tilde E \times \tilde E \to \tilde E$ as $\tilde A((x,g), \, (y,h)) = (A(x,y),\, K(g+h))$ for $x,y \in E$ and $g,h \in G$. Then:


* $\tilde A$ is not condensing. Indeed, for the set $D \subseteq E$ from Example 1 we have $\tilde A(D \times \{0\}, D \times \{0\}) = A(D,D) \times \{0\}$, so
$$
\alpha(\tilde A(D \times \{0\}, D \times \{0\})) = \alpha(A(D,D)) = 2\alpha(D) = 2\alpha(D \times \{0\}).
$$
* $\tilde A^2$ is condensing: for all $x,y\in E$ and all $g,h \in G$ we have $\tilde A^2((x,g),\, (y,h)) = (0, 2K^2(g+h))$, so $\tilde A^2(S\_1,S\_2)$ is compact for all bounded sets $S\_1, S\_2 \subseteq \tilde E$.
* If $K^2$ is not $0$, then $\tilde A^2 \not= 0$. More generally, if $K$ is not nilpotent, then $A^n$ is non-zero for any $n \in \mathbb{N}$.


**Example 3.** If we want a still less trivial example, we can construct a space $E$ and a map $A$ with the required properties and such that $A^n(E,E)$ is dense in $E$ for every $n \in \mathbb{N}$. To this end, we redo the construction from Example 1, but with a few modifications.


Again, let $F$ be an infinite dimensional Banach space and set $E = F \times F$ (with the maximum norm). Let $K: F \to F$ be a compact linear operator with dense range (it is easy to construct such an operator if we choose $F$, for instance, to be a classical sequence space such as $\ell^p$ or $c\_0$).


Now we define an operator $T: E \to E$ as the operator matrix
$$
\begin{pmatrix}
K & I\_F \\
0 & K
\end{pmatrix}.
$$
Then we have
$$
T^n =
\begin{pmatrix}
K^n & nK^{n-1} \\
0 & K^n
\end{pmatrix}.
$$
for each integer $n \ge 1$. In particular, $T^n$ is compact for each $n \ge 2$.


Again, we define $A: E \times E \to E$ by $A(x,y) = T(x+y)$ for all $x,y \in E$. Then:


* $A$ is not condensing. To see this, let $D$ be the same set as in Example 1 and let $P: E \to E$ be the projection onto the first component. Then $P$ is contractive, so $\alpha(P(X)) \le \alpha(X)$ for each subset $X$ of $E$. Moreover, we have
$$
A(D,D) = T(D) + T(D) \supseteq T(D),
$$
so it follows that
$$
\alpha(A(D,D)) \ge \alpha(T(D)) \ge \alpha(P(T(D))) = \alpha(B \times \{0\}) = \alpha(D).
$$
* For each $n \in \mathbb{N}$ and all $x,y \in E$ we have $A^n(x,y) = nT^n(x+y)$, so $A^n(S\_1,S\_2)$ is relatively compact for all $n \ge 2$ and all bounded sets $S\_1, S\_2 \subseteq E$. This shows that $A^n$ is condensing whenever $n \ge 2$.
* We have $A^n(E,E) = T^n(E)$ for each $n \in \mathbb{N}$, so it remains to show that $T^n$ has dense range for each $n$, and to this end it suffices to show that $T$ has dense range. So let $(f,g) \in E = F \times F$. Since $K$ has dense range in $F$; there exists a sequence $(g\_k)$ in $F$ such that $Kg\_k \to g$. On the other hand, again since $K$ has dense range in $F$, we can find, for each index $k \ge 1$, a vector $f\_k \in F$ such that $\|Kf\_k + g\_k - f\| < 1/k$. Hence, $T(f\_k, g\_k) \to (f,g)$.


