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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.

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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 $\phi$.

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