Estimate of an oscillatory integral in Stein's book In Stein's book "Harmonic Analysis, Real-variable methods, orthogonality and oscillatory integrals", the author claims on page 335 that if $\eta\in C^{\infty}_{c}(\mathbb{R})$ and $l\geq 0$, then
$$\left\vert\int_{-\infty}^{\infty}e^{i\lambda x^2}x^l\eta(x)dx\right\vert\leq A\lambda^{-\frac{l+1}{2}}$$
for every $\lambda>0$, where the constant $A$ does not depend on $\lambda$. To prove this estimate, the author considers a bump function $\alpha\in C^{\infty}(\mathbb{R})$ such that $\alpha(x)=1$ for $|x|\leq 1$ and $\alpha(x)=0$ for $|x|\geq 2$, and he writes
$$\int_{-\infty}^{\infty}e^{i\lambda x^2}x^l\eta(x)dx=\int_{-\infty}^{\infty}e^{i\lambda x^2}x^l\eta(x)\alpha\left(\frac{x}{\epsilon}\right)dx+\int_{-\infty}^{\infty}e^{i\lambda x^2}x^l\eta(x)\left[1-\alpha\left(\frac{x}{\epsilon}\right)\right]dx.$$
I am having trouble estimating the second integral in the rhs above. The author basically integrates by parts, considering the differential operator
$$D:=\frac{1}{2i\lambda x}\frac{d}{dx}$$
and its adjoint
$$D^{*}=-(i\lambda)^{-1}\frac{d}{dx}\left(\frac{\cdot}{2x}\right)$$
to write for every integer $N\geq 0$ that
\begin{align}
\int_{-\infty}^{\infty}e^{i\lambda x^2}x^l\eta(x)\left[1-\alpha\left(\frac{x}{\epsilon}\right)\right]dx&=\int_{-\infty}^{\infty}D^{N}\left(e^{i\lambda x^2}\right)x^l\eta(x)\left[1-\alpha\left(\frac{x}{\epsilon}\right)\right]dx\\
&=\int_{-\infty}^{\infty}e^{i\lambda x^2}\left(D^{*}\right)^{N}\left[x^l\eta(x)\left[1-\alpha\left(\frac{x}{\epsilon}\right)\right]\right]dx.
\end{align}
Next, the author claims that a "simple computation" shows that
$$\left|\left(D^{*}\right)^{N}\left[x^l\eta(x)\left[1-\alpha\left(\frac{x}{\epsilon}\right)\right]\right]\right|\leq C_{N}\lambda^{-N}\vert x\vert^{l-2N}$$
holds for every $N\geq 0$ and every $x\in\mathbb{R}$.
I am unable to prove this inequality, even when $N=1$. It is clear that the factor $\lambda^{-N}$ immediately comes from the operator $\left(D^{*}\right)^{N}$. However, I don't see how to obtain the factor $\vert x\vert^{l-2N}$ in this inequality. When $N=1$ we can compute that
\begin{align}-2\lambda iD^{*}\left(x^l\eta(x)\left[1-\alpha\left(\frac{x}{\epsilon}\right)\right]\right)&=(l-1)x^{l-2}\eta(x)\left[1-\alpha\left(\frac{x}{\epsilon}\right)\right]\\
&+x^{l-1}\eta'(x)\left[1-\alpha\left(\frac{x}{\epsilon}\right)\right]\\
&-\epsilon^{-1}x^{l-1}\eta(x)\alpha'\left(\frac{x}{\epsilon}\right).
\end{align}
Looking at the three terms in the rhs, I see that the first and the third satisfy the desired bound of $\vert x\vert ^{l-2}$. However, the second term does not seem to satisfy such an upper bound. How can I prove the estimate?
 A: I think there may be a few pieces you are missing. $\DeclareMathOperator{\supp}{supp}$ First note that the operator $D^*$ is local in the sense that $\supp(D^*f)\subset \supp(f)$ for any function on $\mathbb R$. Let $\phi(x)=1-\alpha(x/\epsilon)$. By the observation that $D^*$ is local, $(D^*)^N[x^l\eta\phi]$ is supported in $\supp\phi=\{|x|\ge \epsilon\}$, so we only need to prove estimates for this range of $x$.
Second, notice that Stein is asking for an estimate when $l\ge 0$ is given, and $l-2N<-1$. Therefore, it suffices to prove the estimate for $N$ sufficiently large, and at one point in the sketch I give below, I will assume $N > 10l$. Optimizing the lower bound on $N$ is not important, but it may be possible to do the calculation without assuming $N$ is quite this large.
Here are some details to help you digest the estimate. Let $I(\lambda) = \int e^{i\lambda x^2}(D^*)^N[x^l\eta\phi]\,dx$.

*

*Prove that for $N\ge 0$, there is a constant $C_N>0$ such that
$$
|(D^*)^N(f)|\le C_N\lambda^{-N}\sum_{j=0}^N\frac{1}{|x|^{2N-j}}|\partial^jf(x)|.
$$

*For each $j\ge 0$, and $0\le p,q,r\le j$ such that $p+q+r=j$, there are constants $C^j_{pqr}$ such that
\begin{align*}
|\partial^j[x^l\eta\phi]|&=|\sum_{p+q+r=j}C^j_{pqr}\partial^px^l\,\partial^q\eta\,\partial^r\phi|.
\end{align*}
Integrating and using (1),
\begin{align*}
|I(\lambda)|&\le C_N\lambda^{-N}\int_{|x|\ge \epsilon}\sum_{j=0}^N|x|^{j-2N}|\partial^j[x^l\eta\phi]|\,dx\\
&\lesssim C_N\lambda^{-N}\bigg[\underbrace{\int_{|x|\ge\epsilon}|x|^{l-2N}\,dx}_{\equiv A_0} + \sum_{j=1}^N\underbrace{\int_{|x|\ge\epsilon}|x|^{j-2N}|\partial^j[x^l\eta\phi]|\,dx}_{\equiv A_j}\bigg]
\end{align*}
I claim $A_0$ is the dominant term, so it suffices to show each $A_j$, $1\le j\le N$ has the same size as $A_0$. Let's expand $A_j$ for each $j\in\{1,\dots,N\}$:
\begin{align*}
A_j &\equiv \int_{\epsilon\le|x|\le 2\epsilon}|x|^{j-2N}|\partial^j[x^l\eta\phi]|\,dx+\int_{|x|>2\epsilon}|x|^{j-2N}|\partial^j[x^l\eta\phi]|\,dx.
\end{align*}
For $|x|>2\epsilon$, we note $\phi \equiv 1$ there. We have:
\begin{align*}
\int_{|x|>2\epsilon}|x|^{j-2N}|\partial^j[x^l\eta\phi]|\,dx&=\int_{|x|>2\epsilon}|x|^{j-2N}|\partial^j[x^l\eta]|\,dx\\
&\lesssim \sum_{p=0}^j\int_{|x|>2\epsilon}|x|^{j-2N+l-p}\,dx\\
&\sim \epsilon^{l-2N+1}\sum_{p=0}^j\epsilon^{j-p}\lesssim \epsilon^{l-2N+1}.
\end{align*}
(For the $\sim$ in the above, it is sufficient to assume $N>10l$ so the overall exponent is $<-1$.) Since
$$
\epsilon^{l-2N+1}\sim\int_{|x|\ge \epsilon}|x|^{l-2N}\,dx \equiv A_0,
$$
this part has the right size. Now for $\epsilon\le|x|\le2\epsilon$:
\begin{align*}
\int_{\epsilon\le|x|\le 2\epsilon}|x|^{j-2N}|\partial^j[x^l\eta\phi]|\,dx&\sim\epsilon^{j-2N}\int_{\epsilon\le|x|\le 2\epsilon}|\partial^j[x^l\eta\phi]|\,dx \\
&\lesssim \epsilon^{j-2N}\int_{\epsilon\le|x|\le 2\epsilon}\sum_{p+q+r=j}|x|^{l-p}\,|\partial^q\eta|\,|\partial^r\phi|\,dx\\
&\lesssim \epsilon^{j-2N}\int_{|x|\sim\epsilon}\sum_{p+q+r=j}\epsilon^{l-p-r}\,dx\\
&\lesssim \epsilon^{j-2N}\epsilon^1\sum_{q=0}^j\epsilon^{l-j+q}\lesssim \epsilon^{l-2N+1},
\end{align*}
which we already noted is the right size.

These steps show for $N>10l$,
$$
|I(\lambda)|\lesssim C_N\lambda^{-N}\int_{|x|\ge \epsilon}|x|^{l-2N}\,dx,
$$
which is essentially the claim that Stein made at this point in the book, except our lower bound on $N$ is a little worse than the $(l+1)/2$ that Stein had.
