# Gronwall estimate with a Fourier transform

Suppose I have the following equality $$\hat{u}_\epsilon(t,k) = \alpha(t,k) + \int_0^t\int_{\mathbb{R}^n} e^{ik\cdot(x +\epsilon \phi(s,x))}u_\epsilon(s,x)dxds$$ Where $$\alpha(t,k) \geq 0$$ and $$\alpha(t,\cdot)\in L^1( \mathbb{R}^n)$$. Moreover $$\phi \in C^\infty\cap L^\infty(\mathbb{R}\times\mathbb{R}^n)$$ and $$\epsilon \in (0,1]$$. Furthermore, $$\hat{u}_\epsilon$$ denotes the Fourier transform of $$u_\epsilon$$. I would like to deduce a uniform estimate for $$\|\hat{u}_\epsilon(t,\cdot)\|_{L^1}$$ with respect to $$\epsilon$$ (as $$\epsilon\rightarrow 0^+$$) using Gronwall's inequality. I am wondering if it's possible in general due to the non-linear exponent $$e^{ik\cdot(x+\epsilon\phi(s,x))}$$. The relevant theorem is the Beurling-Helson theorem which makes me think I cannot get such an estimate unless $$\phi$$ is linear in $$x$$. But the fact that the map $$x\mapsto x + \epsilon\phi(t,x)$$ is nearly the identity makes me think otherwise. Any ideas would be appreciated.

I will more comfortable with the notation $$v_\epsilon=\hat{u_\epsilon}$$; you have then $$v_\epsilon(t,x)=\alpha(t,x)+\int_0^t\int e^{2πix(\xi+\epsilon\phi(s,\xi))} \hat{v_\epsilon}(s,\xi) d\xi ds=\alpha(t,x)+\int_0^t \bigl(\textrm{Op}(e^{2πix\epsilon\phi(s,\xi)}) v(s,\cdot)\bigr) (x) ds$$ where $$\textrm{Op}(e^{2πix\epsilon\phi(s,\xi)})=A_{\epsilon, s}$$ is the operator with symbol $$e^{2πix\epsilon\phi(s,\xi)}$$. Let me now assume that $$A_{\epsilon, s}$$ is bounded on $$L^2(\mathbb R^n)$$ and that $$\alpha$$ is in $$L^2$$: you get with $$L^2$$ norms (the triple norm is the operator-norm) $$\Vert v_\epsilon(t)\Vert\le\Vert\alpha(t)\Vert+\int_0^t\vert\!\Vert A_{\epsilon, s}\Vert\!\vert \Vert v_\epsilon(s)\Vert ds,$$ and if $$\vert\!\Vert A_{\epsilon, s}\Vert\!\vert$$ as a function of $$s$$ is in $$L^1$$, you can use Gronwall. The real problem is to get an estimate for the triple norm (by the way $$\phi$$ is certainly real-valued): that operator is likely to be a Fourier Integral Operator and not a pseudo-differential operator, so to get the sought bound you should consider $$A_{\epsilon, s}^* A_{\epsilon, s},$$ which will be a pseudo-differential operator under some assumptions on $$\phi$$.