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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!

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    $\begingroup$ I guess you want to assume that $f,g$ are $\omega$-periodic in the first variable. Otherwise this operator will not be well-defined (the image will not be in $E\times E$). $\endgroup$ – Severin Schraven May 18 '19 at 17:05
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    $\begingroup$ The compactness follows from the Arzelá-Ascoli theorem. Take a bounded sequence in $E\times E$ and show that the image of the sequence is uniformly bounded (easy) and equicontinuous. You have to use that due to periodicity everything is defined on compacta and continuity will allow you to conclude. $\endgroup$ – Severin Schraven May 18 '19 at 17:17
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    $\begingroup$ In fact all you need is to convince yourself that if $(x_n,y_n)$ a bounded sequence in $E\times E$ then $$ \sup_{n} \Vert f(\cdot, x_n(\cdot), y_n(\cdot)) \Vert_{\infty}<\infty$$ Which follows from the fact that continuous functions are bounded on compact domains. $\endgroup$ – Severin Schraven May 18 '19 at 17:22
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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.

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  • $\begingroup$ Thank you for your comments and reply, and sorry for my late reply (very late). I have followed what you have done, it seems to me that it is very correct. I have one fear left to conclude: Suppose that we omit periodicity on the definition of $E$ and on $f,\:g$ and working on an interval $[a,b] $ instead of $\mathbb R$, are we going to have the same results? I think YES, what do you think? $\endgroup$ – Motaka Jul 20 '20 at 12:09
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    $\begingroup$ Indeed, if we trade periodicity for a compact interval, then essentially the same proof will work and tell us that the operator is compact. $\endgroup$ – Severin Schraven Jul 20 '20 at 12:19

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