Let $k(t) \in \mathcal{L}(E; E)$, $t \geq 0$ be a family of compact operators on a Banach space $E$, such that $$ \int_0^\infty \lVert k(t)\rVert dt < \infty. $$ Let $r(t) \in \mathcal{L}(E; E)$ be the solution of integral equation $$\forall t \geq 0, \quad r(t) = k(t) + \int_0^t{ k(t-s) r(s) ds}. $$ The integral with respect to $s$ is well-defined as an Bochner integral and $k(t-s)r(s)$ is the composition of $k(t-s)$ and $r(s)$. For $\Re(z) \geq 0$, denote by $\hat{k}(z) = \int_0^\infty{e^{-zt} k(t) dt}$ the Laplace transform of $k$. Assume that $$\forall z \in \mathbb{C} \text{ with } \Re(z) \geq 0, \quad I - \hat{k}(z) \text{ is invertible}, $$ where $I$ stands for the identity operator on $\mathcal{L}(E; E)$. It is true that $\int_0^\infty{\lVert r(t)\rVert dt} < \infty$ ?

It is easy to check that the condition on the Laplace transform is required for $r \in L^1(\mathbb{R}_+; \mathcal{L}(E;E))$. Indeed, when $r \in L^1(\mathbb{R}_+; \mathcal{L}(E;E))$, we have for $\Re(z) \geq 0$: $(I+\hat{r}(z))(I-\hat{k}(z)) = (I-\hat{k}(z))(I+\hat{r}(z)) = I $, so $I-\hat{k}(z)$ is invertible. When $dim(E) < \infty$, it is known that this condition is sufficient by the Paley–Wiener theorem [1, Th. 4.1]. I am particularly interested by the case where for all $t \geq 0$, $k(t)$ is a Hilbert–Schmidt operator on a Hilbert space $E$. Also, I would be already be happy with the slightly weaker conclusion: $$ \forall \epsilon > 0, \quad \int_0^\infty{e^{-\epsilon t } \lVert r(t)\rVert dt } < \infty. $$

[1]: Gripenberg, G. ; Londen, S.-O. ; Staffans, O. Volterra integral and functional equations.

Let $k(t) \in \mathcal{L}(E; E)$, $t \geq 0$ be a family of compact operators on a Banach space $E$, such that $$ \int_0^\infty \lVert k(t)\rVert dt < \infty. $$ Let $r(t) \in \mathcal{L}(E; E)$ be the solution of integral equation $$\forall t \geq 0, \quad r(t) = k(t) + \int_0^t{ k(t-s) r(s) ds}. $$ The integral with respect to $s$ is well-defined as an Bochner integral and $k(t-s)r(s)$ is the composition of $k(t-s)$ and $r(s)$. For $\Re(z) \geq 0$, denote by $\hat{k}(z) = \int_0^\infty{e^{-zt} k(t) dt}$ the Laplace transform of $k$. Assume that $$\forall z \in \mathbb{C} \text{ with } \Re(z) \geq 0, \quad I - \hat{k}(z) \text{ is invertible}, $$ where $I$ stands for the identity operator on $\mathcal{L}(E; E)$. It is true that $\int_0^\infty{\lVert r(t)\rVert dt} < \infty$ ?

It is easy to check that the condition on the Laplace transform is required for $r \in L^1(\mathbb{R}_+; \mathcal{L}(E;E))$. Indeed, when $r \in L^1(\mathbb{R}_+; \mathcal{L}(E;E))$, we have for $\Re(z) \geq 0$: $(I+\hat{r}(z))(I-\hat{k}(z)) = (I-\hat{k}(z))(I+\hat{r}(z)) = I $, so $I-\hat{k}(z)$ is invertible. When $dim(E) < \infty$, it is known that this condition is sufficient by the Paley–Wiener theorem [1, Th. 4.1]. I am particularly interested by the case where for all $t \geq 0$, $k(t)$ is a Hilbert–Schmidt operator on a separable Hilbert space $E$. Also, I would be already be happy with the slightly weaker conclusion: $$ \forall \epsilon > 0, \quad \int_0^\infty{e^{-\epsilon t } \lVert r(t)\rVert dt } < \infty. $$

[1]: Gripenberg, G. ; Londen, S.-O. ; Staffans, O. Volterra integral and functional equations.

Let $k(t) \in \mathcal{L}(E; E)$, $t \geq 0$ be a family of compact operators on a Banach space $E$, such that $$ \int_0^\infty \lVert k(t)\rVert dt < \infty. $$ Let $r(t) \in \mathcal{L}(E; E)$ be the solution of integral equation $$\forall t \geq 0, \quad r(t) = k(t) + \int_0^t{ k(t-s) r(s) ds}. $$ The integral with respect to $s$ is well-defined as an Bochner integral and $k(t-s)r(s)$ is the composition of $k(t-s)$ and $r(s)$. For $\Re(z) \geq 0$, denote by $\hat{k}(z) = \int_0^\infty{e^{-zt} k(t) dt}$ the Laplace transform of $k$. Assume that $$\forall z \in \mathbb{C} \text{ with } \Re(z) \geq 0, \quad I - \hat{k}(z) \text{ is invertible}, $$ where $I$ stands for the identity operator on $\mathcal{L}(E; E)$. It is true that $\int_0^\infty{\lVert r(t)\rVert dt} < \infty$ ?

It is easy to check that the condition on the Laplace transform is required for $r \in L^1(\mathbb{R}_+; \mathcal{L}(E;E))$. Indeed, when $r \in L^1(\mathbb{R}_+; \mathcal{L}(E;E))$, we have for $\Re(z) \geq 0$: $(I+\hat{r}(z))(I-\hat{k}(z)) = (I-\hat{k}(z))(I+\hat{r}(z)) = I $, so $I-\hat{k}(z)$ is invertible. When $dim(E) < \infty$, it is known that this condition is sufficient by the Paley–Wiener theorem [1, Th. 4.1]. I am particularly interested by the case where for all $t \geq 0$, $k(t)$ is a Hilbert–Schmidt operator on a Hilbert space $E$. Also, I would be already be happy with the slightly weaker conclusion: $$ \forall \epsilon > 0, \quad \int_0^\infty{e^{\epsilon t } \lVert r(t)\rVert dt } < \infty. $$

[1]: Gripenberg, G. ; Londen, S.-O. ; Staffans, O. Volterra integral and functional equations.

Let $k(t) \in \mathcal{L}(E; E)$, $t \geq 0$ be a family of compact operators on a Banach space $E$, such that $$ \int_0^\infty \lVert k(t)\rVert dt < \infty. $$ Let $r(t) \in \mathcal{L}(E; E)$ be the solution of integral equation $$\forall t \geq 0, \quad r(t) = k(t) + \int_0^t{ k(t-s) r(s) ds}. $$ The integral with respect to $s$ is well-defined as an Bochner integral and $k(t-s)r(s)$ is the composition of $k(t-s)$ and $r(s)$. For $\Re(z) \geq 0$, denote by $\hat{k}(z) = \int_0^\infty{e^{-zt} k(t) dt}$ the Laplace transform of $k$. Assume that $$\forall z \in \mathbb{C} \text{ with } \Re(z) \geq 0, \quad I - \hat{k}(z) \text{ is invertible}, $$ where $I$ stands for the identity operator on $\mathcal{L}(E; E)$. It is true that $\int_0^\infty{\lVert r(t)\rVert dt} < \infty$ ?

It is easy to check that the condition on the Laplace transform is required for $r \in L^1(\mathbb{R}_+; \mathcal{L}(E;E))$. Indeed, when $r \in L^1(\mathbb{R}_+; \mathcal{L}(E;E))$, we have for $\Re(z) \geq 0$: $(I+\hat{r}(z))(I-\hat{k}(z)) = (I-\hat{k}(z))(I+\hat{r}(z)) = I $, so $I-\hat{k}(z)$ is invertible. When $dim(E) < \infty$, it is known that this condition is sufficient by the Paley–Wiener theorem [1, Th. 4.1]. I am particularly interested by the case where for all $t \geq 0$, $k(t)$ is a Hilbert–Schmidt operator on a Hilbert space $E$. Also, I would be already be happy with the slightly weaker conclusion: $$ \forall \epsilon > 0, \quad \int_0^\infty{e^{-\epsilon t } \lVert r(t)\rVert dt } < \infty. $$

[1]: Gripenberg, G. ; Londen, S.-O. ; Staffans, O. Volterra integral and functional equations.

Let $k(t) \in \mathcal{L}(E; E)$, $t \geq 0$ be a family of compact operators on a Banach space $E$, such that $$ \int_0^\infty ||k(t)|| dt < \infty. $$ Let $r(t) \in \mathcal{L}(E; E)$ be the solution of integral equation $$\forall t \geq 0, \quad r(t) = k(t) + \int_0^t{ k(t-s) r(s) ds}. $$ The integral with respect to $s$ is well-defined as an Bochner integral and $k(t-s)r(s)$ is the composition of $k(t-s)$ and $r(s)$. For $\Re(z) \geq 0$, denote by $\hat{k}(z) = \int_0^\infty{e^{-zt} k(t) dt}$ the Laplace transform of $k$. Assume that $$\forall z \in \mathbb{C} \text{ with } \Re(z) \geq 0, \quad I - \hat{k}(z) \text{ is invertible}, $$ where $I$ stands for the identity operator on $\mathcal{L}(E; E)$. It is true that $\int_0^\infty{||r(t)|| dt} < \infty$ ?

It is easy to check that the condition on the Laplace transform is required for $r \in L^1(\mathbb{R}_+; \mathcal{L}(E;E))$. Indeed, when $r \in L^1(\mathbb{R}_+; \mathcal{L}(E;E))$, we have for $\Re(z) \geq 0$: $(I+\hat{r}(z))(I-\hat{k}(z)) = (I-\hat{k}(z))(I+\hat{r}(z)) = I $, so $I-\hat{k}(z)$ is invertible. When $dim(E) < \infty$, it is known that this condition is sufficient by the Palay-Wiener thereom [1, Th. 4.1]. I am particularly interested by the case where for all $t \geq 0$, $k(t)$ is an Hilbert-Schmidt operator on an Hilbert space $E$. Also, I would be already be happy by the slightly weaker conclusion: $$ \forall \epsilon > 0, \quad \int_0^\infty{e^{\epsilon t } ||r(t)||dt } < \infty. $$ Thank you for any help!

[1]: Gripenberg, G. ; Londen, S.-O. ; Staffans, O. Volterra integral and functional equations.

Let $k(t) \in \mathcal{L}(E; E)$, $t \geq 0$ be a family of compact operators on a Banach space $E$, such that $$ \int_0^\infty \lVert k(t)\rVert dt < \infty. $$ Let $r(t) \in \mathcal{L}(E; E)$ be the solution of integral equation $$\forall t \geq 0, \quad r(t) = k(t) + \int_0^t{ k(t-s) r(s) ds}. $$ The integral with respect to $s$ is well-defined as an Bochner integral and $k(t-s)r(s)$ is the composition of $k(t-s)$ and $r(s)$. For $\Re(z) \geq 0$, denote by $\hat{k}(z) = \int_0^\infty{e^{-zt} k(t) dt}$ the Laplace transform of $k$. Assume that $$\forall z \in \mathbb{C} \text{ with } \Re(z) \geq 0, \quad I - \hat{k}(z) \text{ is invertible}, $$ where $I$ stands for the identity operator on $\mathcal{L}(E; E)$. It is true that $\int_0^\infty{\lVert r(t)\rVert dt} < \infty$ ?

It is easy to check that the condition on the Laplace transform is required for $r \in L^1(\mathbb{R}_+; \mathcal{L}(E;E))$. Indeed, when $r \in L^1(\mathbb{R}_+; \mathcal{L}(E;E))$, we have for $\Re(z) \geq 0$: $(I+\hat{r}(z))(I-\hat{k}(z)) = (I-\hat{k}(z))(I+\hat{r}(z)) = I $, so $I-\hat{k}(z)$ is invertible. When $dim(E) < \infty$, it is known that this condition is sufficient by the Paley–Wiener theorem [1, Th. 4.1]. I am particularly interested by the case where for all $t \geq 0$, $k(t)$ is a Hilbert–Schmidt operator on a Hilbert space $E$. Also, I would be already be happy with the slightly weaker conclusion: $$ \forall \epsilon > 0, \quad \int_0^\infty{e^{\epsilon t } \lVert r(t)\rVert dt } < \infty. $$

[1]: Gripenberg, G. ; Londen, S.-O. ; Staffans, O. Volterra integral and functional equations.