Bounding an Oscillatory Integral Operator involving the phase $|x-y|$

Question

I am currently studying Chapter 9 of Stein's "Harmonic Analysis", in particular Section 2.2.2. Consider the oscillatory integral operator $$ T_\lambda f(x) = \int a(x,y) e^{2 \pi i \lambda |x - y|} f(y)\; dy $$ The standard oscillatory integral operator techniques assume the phase function is $C^\infty$, so it is natural to replace this operator by the operator $$ \tilde{T_\lambda} f(x) = \int \tilde{a}(x,y) e^{2 \pi i \lambda |x - y|} f(y)\; dy $$ where $\tilde{a}(x,y) = (1 - \psi(x - y)) a(x,y)$, $\psi(z) = 1$ for $|z| \leq 1$, and $\psi(z) = 0$ for $|z| \geq 2$. Under these assumptions it is possible to justify estimates of the form $$ \| \tilde{T}_\lambda f \|_{L^q(\mathbf{R}^n)} \lesssim \lambda^{-n/q} \| f \|_{L^p(\mathbf{R}^n)}. $$ What I am confused about is the precise technique of using this estimate to infer an estimate of the form $$ \| T_\lambda f \|_{L^q(\mathbf{R}^n)} \lesssim \lambda^{-n/q} \| f \|_{L^p(\mathbf{R}^n)}. $$ Certainly to do so, we must study the operator $$ R_\lambda f(x) = \int \psi(x-y) a(x,y) e^{2 \pi i \lambda |x - y|} f(y)\; dy, $$ but to do so seems to require essentially the same techniques as understanding the operator $T_\lambda$ in the first place. What is the correct way to proceed?

Answer 1

The approach is not quite as you say: fix $\epsilon$ and let $$T_{\lambda}^{\epsilon}=\int{\psi_{\epsilon}(x,y)a(x,y)e^{2\pi i\lambda|x-y|}f(y)\,dy}$$ where $\psi_{\epsilon}(x,y)=1$ as long as $|x-y|>2\epsilon$ and $\psi_{\epsilon}(x,y)=0$ as long as $|x-y|<\epsilon$. Then one takes $\epsilon\to0$, showing:

1. $$\|T_{\lambda}^{\epsilon}f\|_q\lesssim\lambda_{\epsilon}^{-\frac{n}{q}}\|f\|_p$$ with $\sup_{\epsilon}{\lambda_{\epsilon}}<\infty$.
2. As long as $a$ is bounded in a neighborhood of the diagonal (this is too strong, but works in 2.2.2; I leave a sharp version to you), the integral defining $T_{\lambda}^{\epsilon}$ converges to the one defining $T_{\lambda}$ by the dominated convergence theorem.