The operator $T_F(x)$ depending on the function $F\in L^1\cap L^2$ and the real number $x$ is formally defined as follows: $$ T_F(x)\psi(y)=\int_0^{\infty}\psi(t)F(x+t+y)dt $$
Now my question is: Firstly, is this operator well defined as an operator $L^1\to L^1$, $L^1\to L^2$ and $L^2\to L^2$ respectively? Why? Secondly, what is the operator norm corresponding to the three cases if the operator is well defined? Thank you!
Fix $x \geq 0$. Let's first suppose that $\psi, F \geq 0$, $\psi \in L^1$. Then, by Jensen's inequality and Tonelli's theorem,
$$\begin{align*} \int_{\mathbb{R}} |T_F(x)\psi(y)|^p \, dy &= \|\psi\|_1^p \cdot \int_{\mathbb{R}} \left| \int_0^{\infty} F(t+x+y) \cdot \frac{\psi(t)}{\|\psi\|_1} \, dt \right|^p \, dy \\ &\leq \|\psi\|_1^p \cdot \int_{(0,\infty)} \underbrace{\int_{\mathbb{R}} |F(x+t+y)|^p \, dy}_{=\|F(x+\cdot)\|_p^p} \frac{\psi(t)}{\|\psi\|_1} \, dt = \|F\|_p^p \cdot \|\psi\|_1^p \end{align*}$$
for any $p \geq 1$. This shows that $T_F(x) \psi < \infty$ almost surely and $\|T_F(x)\| \leq \|F(x+\cdot)\|_p$, i.e. $T_F(x):L^1 \to L^1$ and $T_F(x): L^1 \to L^2$ are bounded operators. Splitting up $\psi = \psi^+-\psi^-$ and $F= F^+-F^-$ proves this for not necessarily non-negative $\psi$, $F$.
In order to find the operator norm, we first consider the case $F \in C_c$ (i.e. $F$ has bounded support and is (uniformly) continuous). We choose a sequence $\psi_k \in L^1$ such that the following conditions are satisfied:
$$\psi_k \geq 0 \qquad \int \psi_k = 1 \qquad \{x; \psi_k(x) \neq 0\} \subseteq [0,k^{-1}]$$
Then
$$\begin{align*} \left| \int_0^{\infty} \psi_k(t) \cdot F(t+x+y) \, dt - F(x+y) \right| &= \left| \int_0^{\infty} (F(x+y+t)-F(x+y)) \cdot \psi_k(t) \, dt \right| \\ &\leq \sup_{t \leq k^{-1}} |F(x+y+t)-F(x+y)| \cdot \underbrace{\int_0^{k^{-1}} \psi_k(t) \, dt}_{1} \end{align*}$$
Since $F$ is uniformly continuous, this shows that $$T_F(x)\psi_k \to F(x+\cdot)$$ uniformly, hence in particular $\|T_F \psi_k\|_p \to \|F(x+\cdot)\|_p$. Thus $\|T_F(x)\| \geq \|F(x+\cdot)\|_p$. Since $C_c$ is dense in $L^1 \cap L^2$, this inequality holds for any $F \in L^1 \cap L^2$. Consequently, $$\|T_F(x)\| = \|F(x+\cdot)\|_p$$
As far as I can see, a similar reasoning applies if $\psi \in L^2$ (basiscally, we can interchange the roles of $F$ and $\psi$ in the first calculation).
