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.

  • 5
    $\begingroup$ wow! big spender. (i have done the same. what's the point of getting some rep if you never spend it.) $\endgroup$ Jul 8, 2019 at 22:42
  • $\begingroup$ 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. $\endgroup$
    – Andrew
    Jul 11, 2019 at 8:39
  • $\begingroup$ @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? $\endgroup$
    – PPR
    Jul 12, 2019 at 1:45
  • $\begingroup$ @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. $\endgroup$
    – Andrew
    Jul 12, 2019 at 7:55
  • $\begingroup$ @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 $f$? $\endgroup$ Jul 12, 2019 at 8:41

1 Answer 1


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 $p$.

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

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