Let $R > 0 $ and set $h = \frac{1}{R}$. Let $G \in C^\infty(\mathbb{R})\cap L^\infty(\mathbb{R})$. Further restrictions on $G$ are allowed.
Consider the (R-dependent) integral operator $K_R: L^2(\mathbb{R}) \rightarrow \ell^2(h\mathbb{Z})$ with integral kernel $k_R(z,x)$ given as
$$ k_R(z,x) = (G(x) - G(z)) \int_{\mathbb{R}_k}e^{-i(xk - zR\arctan(k/R))} \frac{1}{\sqrt{1 + k^2/R^2}} dk. $$
I want to show for the corresponding operator-norm, that $\|K_R\|_{L^2(\mathbb{R}) \rightarrow \ell^2(h \mathbb{Z})} \longrightarrow 0$ as $R \rightarrow \infty$.
One approach to this is via the Schur test. In order for this to work one would have to show the following decay:
$$ \lim\limits_{R \rightarrow \infty} \sup\limits_{z \in h\mathbb{Z}}\left\{ \int_{\mathbb{R}_x}\left|(G(x) - G(z)) \int_{\mathbb{R}_k}e^{-i(xk - zR\arctan(k/R))} \frac{1}{\sqrt{1 + k^2/R^2}}dk\right|dx \right\} \longrightarrow 0. $$
How can this best be established?
Partial observations:
For fixed $z$ and any $0<R<\infty$ we have $k_R(x,z) \in L^1(\mathbb{R}_x)$.
We have the naive limit
$$ e^{-i(xk - zR\arctan(k/R))} \frac{1}{\sqrt{1 + k^2/R^2}} \rightarrow e^{-ik(x-z)}. $$ Performing the $k$ integration yields a delta distribution.
- Replacing the exponential factor carrying the $\arctan$ with its naive pointwise limit yields
$$ \int_{\mathbb{R}_k}e^{-i(xk - zk)} \frac{1}{\sqrt{1 + k^2/R^2}}dk = R\sqrt{\frac{2}{\pi}} K_0(R|x|). $$
Here $K_0$ is the zeroth modified Bessel function of the second kind (which is in $L^1$). The family $\{RK_0(R\cdot)\}_{R>0}$ is a nascent delta function.
- For $\psi \in L^2(\mathbb{R})$ with $\hat{\psi} \in L^1(\mathbb{R})$, we have
$$ \int_{\mathbb{R}_x}\overline{\psi}(x) \left[\int_{\mathbb{R}_k}e^{-i(xk - zR\arctan(k/R))} \frac{1}{\sqrt{1 + k^2/R^2}}dk\right] dx = \int_{\mathbb{R}_k}\overline{\hat{\psi}}(k)e^{-izR\arctan(k/R)} \frac{1}{\sqrt{1 + k^2/R^2}}dk. $$ By dominated convergence, this converges to $\overline{\psi}(z)$.
- Since $z \in h\mathbb{Z}$, we have $Rz \in \mathbb{Z}$. Thus we also have $$ e^{i(zR\arctan(k/R))} = \left(\frac{1+ \frac{izk}{zR}}{\sqrt{1 + \frac{k^2}{R^2}}} \right)^{zR}. $$ This reproduces essentially the well known approximation $(1+x/n)^n \rightarrow e^x$. In principle one might plug in the above representation into the integral at hand, expand the n-fold power into a sum and integrate term by term. This will yield an unwieldy sum of derivatives of Bessel functions (with the number of summands dependent on $n = zR$), that I however do not know how to tame.
