Prove that this operator is compact Let $E$ be the real Banach space of all real and continuous $\omega $-periodic functions defined on $\mathbb{R}$ with the norm $$\max_{0\leq t\leq\omega}\left | x(t) \right |
\:,\:\forall x\in E$$
Let $\tau_1$ and $\tau_2$ are two strictly postive constants, and $f,g :\mathbb{R}\times I_1 \times I_2\rightarrow \mathbb{R}$, are continuous functions where $I_1$, $I_2$ are subintervals of $[0,+\infty)$, and $\omega $-periodic with respect to the first variable.
Let $T:E\times E\rightarrow E\times E $ defined by $$T(x,y)(t)=\Big( \int_{t-\tau_1}^{t} f(s, x(s),y(s))ds,\int_{t-\tau_2}^{t} g(s, x(s),y(s))ds\Big)\:,\:\:\forall(x,y)\in E\times E\,,\: \forall t\in \mathbb{R}
$$
I want to prove that this operator is compact!
 A: I will assume that $f,g$ are $\omega$-periodic in the first variable.
We want to show that $T$ is a compact operator. Hence, we want to show that for every bounded sequence $(x_n, y_n)_{n\in \mathbb{N}}\subseteq E\times E$ there exists a subsequence such that $(T(x_{n_l}, y_{n_l}))_{l\in \mathbb{N}}$ converges in $E\times E$. However, this follows from the Arzela-Ascoli theorem as soon as we verify that $(T(x_{n_l}, y_{n_l}))_{l\in \mathbb{N}}$ is bounded and equicontinuous.
First we define $M=\sup_{n\in \mathbb{N}} \max \{ \Vert x_n \Vert_\infty, \Vert y_n\Vert_\infty \}$. Then by the continuity of $f$ and $g$ we have
$$ \sup_{(t,x_0,y_0)\in [0,\omega]\times [-M,M] \times [-M,M]} \vert f(t,x_0,y_0) \vert < \infty$$
and
$$ \sup_{(t,x_0,y_0)\in [0,\omega]\times [-M,M] \times [-M,M]} \vert g(t,x_0,y_0) \vert < \infty.$$
However, as I assumed that $f$ and $g$ are $\omega$-periodic in the first variable, I actually obtain
$$ C_1:=\sup_{(t,x_0,y_0)\in \mathbb{R}\times [-M,M] \times [-M,M]} \vert f(t,x_0,y_0) \vert < \infty $$
and
$$ C_2:=\sup_{(t,x_0,y_0)\in \mathbb{R}\times [-M,M] \times [-M,M]} \vert g(t,x_0,y_0) \vert < \infty $$
From this we get
$$ \vert T(x_n,y_n)(t) \vert \leq \tau_1 \cdot C_1 + \tau_2 \cdot C_2.  $$
Hence, we get boundedness. The equicontinuity follows in a similar manner using
$$ \left\vert \int_{u-\tau_1}^{u} f(s,x_n(s), y_n(s)) ds - \int_{t-\tau_1}^{t} f(s,x_n(s), y_n(s)) ds \right\vert
\leq 2 C_1 \vert t-u \vert $$
and the analogous bound for $g$.
Added:
Why it seems to me, that we have to assume that $f$ and $g$ are $\omega$-periodic for $T(x,y)$ to be in $E\times E$. Indeed, we want to have
$$ \int_{t+\omega-\tau_1}^{t+\omega} f(s,x(s), y(s)) ds = \int_{t-\tau_1}^{t} f(s,x(s), y(s)) ds $$
as this needs to be an element of $E$. However, we have with the change of variables $u=s-\omega$ and the fact that $x,y$ are $\omega$-periodic
$$ \int_{t+\omega-\tau_1}^{t+\omega} f(s,x(s), y(s)) ds
= \int_{t-\tau_1}^{t} f(u+ \omega,x(u+\omega), y(u+\omega)) du
= \int_{t-\tau_1}^{t} f(u+ \omega,x(u), y(u)) du $$
This implies by taking constant functions that for all $x_0, y_0, t\in \mathbb{R}$ we have
$$ \int_{t-\tau_1}^t f(s,x_0,y_0) ds = \int_{t-\tau_1}^t f(s+\omega, x_0,y_0) ds $$
This seems to suggest that we get periodicity, however, I am not able to prove it.
