Consider a function $f:\mathbb R^2\to \mathbb R$, and define its Fourier transform by $$\hat f(\xi_1,\xi_2)=\int_{\mathbb R\times\mathbb R}f(x_1,x_2)e^{-i(x_1\ \xi_2+x_2\ \xi_2)}\,dx_1\,dx_2, \qquad \xi_1,\xi_2\in\mathbb R.$$ For each $t>0$, define $$g(t,x)=\int_{\mathbb R}\hat f(\xi,\xi t) e^{ix\xi}\,d\xi,\qquad x\in\mathbb R.$$ Show that $\|g(t,\cdot)\|_{L^1(\mathbb R^1)}\lesssim \|f\|_{L^1(\mathbb R^2)}$.
This question comes from the formula $(3.1)$ in Lemma $3.1$ of this paper, where all the $\mathbb R$ above are replaced by $\mathbb R^3$. The authors said in the paper that "The inequality $(3.1)$ is clear", but I cannot figure the proof.
I can prove an $L^2$ variant as follows: \begin{align*} \int_{\mathbb R}|g(t,x)|^2\,dx&\approx \int_{\mathbb R}|\hat f(\xi,\xi t)|^2\,d\xi\\ &=\int_{\mathbb R}\langle (\xi,\xi t)\rangle^{-1-2\varepsilon}\left|\langle (\xi,\xi t)\rangle^{\frac12+\varepsilon}\hat f(\xi,\xi t)\right|^2\,d\xi\\ &\lesssim \|\langle \nabla\rangle^{1/2+\varepsilon} f\|_{L^2(\mathbb R^2)}^2\int_{\mathbb R}\langle (\xi,\xi t)\rangle^{-1-2\varepsilon}\,d\xi\\ &\lesssim_\varepsilon \langle t\rangle^{-1}\|\langle \nabla\rangle^{1/2+\varepsilon} f\|_{L^2(\mathbb R^2)}^2, \end{align*} for any $\varepsilon>0$.
However, as for the original problem, I do not know how to start.
Any help would be appreciated.
Many thanks to Jose27's comments, which motivated me to write down the proof here.
Define $$h(t,x)=\int_\mathbb R f(x-tv,v)\,dv,\qquad t>0, x\in\mathbb R.$$ Denoting the Fourier transform of $h$ with respect to $x$ as $\hat h(t,\xi)$, then $$\hat h(t,\xi)=\hat f(\xi, \xi t).$$ This implies that $h$ equals to $g$ multiplied by a constant that dosen't depend on $t$. A simply application of Fubini implies the estimate $$\|h(t,\cdot)\|_{L^1(\mathbb R^1)}\le \|f\|_{L^1(\mathbb R^2)},$$ which gives the desired $L^1$ estimate on $g$.
Remark: For any $\sigma\in\mathbb R$, define $$g_\sigma(t,x)=\int_\mathbb R \langle(\xi, \xi t)\rangle^\sigma \hat f(\xi,\xi t)e^{ix\xi}\,d\xi,\qquad t\in\mathbb R, x\in\mathbb R.$$ Using a similar argument, we can also prove $$\|g_\sigma(t,\cdot)\|_{L^1(\mathbb R^1)}\lesssim \|\langle\nabla\rangle^\sigma f\|_{L^1(\mathbb R^2)}.$$