Lower Bound on Oscillatory Integral

Let $$p,y\in\mathbb{R}^d\setminus\{0\},\beta>0$$ be given and fixed and define for all $$\alpha>0$$, $$I(\alpha) := \int_{x\in\mathbb{R}^d}\exp(\mathrm{i}\alpha p\cdot x-\alpha\beta \|x-y\|^2)f(x)\mathrm{d}x$$ where $$f:\mathbb{R}^d\to[0,1]$$ is some bump function (smooth, non-negative, of compact support).

I am interested in estimating (or obtaining a lower bound) on $$|I(\alpha)|$$ for very large values of $$\alpha$$. In particular, to see that $$|I(\alpha)|\geq g(\alpha)$$ as $$\alpha\to\infty$$, for some simple $$g$$ which vanishes at infinity (say, Gaussian).

$$I(\alpha)$$ is essentially the Fourier transform of a bump function and a Gaussian evaluated very far away from the origin. So I tried to use the convolution theorem and the fact that the Fourier transform of a Gaussian is a Gaussian, but it's still not clear to me how this helps, because I don't know quantitative estimates on the Fourier transform of a bump function.

Then I came across: this website which claims that "When controlling an oscillatory integral, bump functions and bounded phase corrections are not very important". So I replaced $$f$$ with another Gaussian:

If we take as a model for $$f$$ the function $$f(x) = \chi_{[-1,1]}(\|x\|)\exp\left(1-\frac{1}{1-\|x\|^2}\right)$$

then we replace it with some Gaussian $$\tilde{f}(x) := \exp\left(-\delta \|x\|^2\right)$$ where $$\delta>0$$ is some parameter to be adjusted later (say, $$\delta=2$$).

We then obtain \begin{align}I(\alpha)&=\int_{x\in\mathbb{R}^d}\exp(\mathrm{i}\alpha p\cdot x-\alpha\beta \|x-y\|^2-\delta \|x\|^2)\mathrm{d}x\\&=\exp(-\frac{\alpha\beta\delta}{\alpha\beta+\delta}\|y\|^2)\int_{x\in\mathbb{R}^d}\exp(\mathrm{i}\alpha p\cdot x-(\alpha\beta+\delta) \|x-\frac{\alpha\beta}{\alpha\beta+\delta}y\|^2)\mathrm{d}x\\&=\exp(-\frac{\alpha\beta\delta}{\alpha\beta+\delta}\|y\|^2)(\frac{\pi}{\alpha\beta+\delta})^{\frac{d}{2}}\exp\left(-\frac{\alpha^2}{4(\alpha\beta+\delta)}\|p\|^2+\mathrm{i}\alpha p \cdot y\right)\,.\end{align}

However, how do you estimate the error of replacing $$f$$ by a Gaussian?

We can do a similar exercise replacing $$f$$ by a Taylor approximation to its second degree, e.g..

Most of the texts I read about estimating oscillatory integrals deal with the case that the phase is rather complicated. However here it is just the Fourier transform, whose gradient never vanishes.

• wow! big spender. (i have done the same. what's the point of getting some rep if you never spend it.) Jul 8, 2019 at 22:42
• It seems you want a function $g(\alpha)$ positive for large enough values of $\alpha$. Numerical evaluation suggests that $I(\alpha)$ can have infinite number of zeroes (tending to infinity of course). If it is true, such a function $g$ does not exists. Jul 11, 2019 at 8:39
• @Andrew, so if $f$ is a Gaussian $g$ is also a nice Gaussian, and deforming $f$ into a smooth function with compact support suddenly makes $I(\alpha)$ have infinitely many zeros infinitely far away?
– PPR
Jul 12, 2019 at 1:45
• @PPR sorry, I missed factor $\alpha$ at the norm. So it's not just a Fourier transform. And the said numerical calculations corresponds therefore to the case $\beta=0$. For $\beta>0$ calculations of the same example don't show zeroes. Jul 12, 2019 at 7:55
• @PPR can you be more precise about what you want $I(\alpha)$ to depend on? Do you want it to depend on $p,y,\beta$, or $f$? Jul 12, 2019 at 8:41

Here's an example, not using bump functions admittedly, where $$I(\alpha) \equiv 0$$.

Taking $$\beta = 1$$ for simplicity, a slight modification of the Gaussian example included in the question shows that if we define $$f_{z}(x) = e^{-\| x - z \|^{2}}$$ for $$z \in \mathbb{R}^{d}$$, then \begin{align} I_{z}(\alpha) &= \int_{\mathbb{R}^{d}} \exp(i \alpha p \cdot x - \alpha \| x - y \|^{2}) f_{z}(x) \, dx \\ &= \left( \frac{\pi}{\alpha + 1} \right)^{d/2} \exp \left( -\frac{\alpha^{2}}{4(\alpha + 1)} \| p \|^{2} -\frac{\alpha}{\alpha + 1} \| y - z \|^{2} \right) e^{i \alpha p \cdot y}. \end{align}

The key property that I want to focus on here is that if $$z_1$$ and $$z_2$$ are such that $$\| y - z_1 \| = \| y - z_2 \|$$, then $$I_{z_1} \equiv I_{z_2}$$.

So if we define $$f(x) = f_{z_1}(x) - f_{z_2}(x)$$ for such $$z_1$$ and $$z_2$$, the resulting oscillatory integral is identically zero.

This isn't a representative example, of course, but I would suspect that bounding $$I(\alpha)$$ away from $$0$$ as $$\alpha \to \infty$$ would be pretty hard in general.

The form of the integral is very similar to that of the FBI (Fourier–Bros–Iagolnitzer) transform, so the behavior of $$I(\alpha)$$ might depend on local properties of $$f$$ at $$y$$, and in the direction of $$p$$.

• Hi Jason, thanks for this. However, shouldn't $f$ be positive?
– PPR
Jul 18, 2020 at 8:34