Generalization of van der Corput's estimate on oscillatory integrals

Question

Question: Given exponents $0<\alpha<\beta$ and an interval $[a,b]\subset(0,\infty)$ are there constants $C,d>0$ such that for any $\lambda_1,\lambda_2\in\mathbb{R}$, $$\left|\int_a^be(\lambda_1x^\alpha+\lambda_2x^\beta)dx\right|\leq C\max\left(\frac1{|\lambda_1|^d},\frac1{|\lambda_2|^d}\right)?$$ (where I'm using $e(x):=e^{2\pi ix}$.)

The reason I hope the answer may be positive is a lemma of van der Corput which implies that for any $[a,b]\subset\mathbb{R}$ and any smooth $f:[a,b]\to\mathbb{R}$ with $|f''(x)|\geq1$ on $[a,b]$, $$\left|\int_a^be\big(\lambda f(x)\big)dx\right|\leq 4|\lambda|^{-1/2}.$$

I would (perhaps naively) expect a generalization for several smooth functions $f_1,\dots,f_k:[a,b]\to\mathbb{R}$ satisfying some conditions; namely that $$\left|\int_a^be\big(\lambda_1 f_1(x)+\cdots+\lambda_kf_k(x)\big)dx\right|\leq C\max|\lambda_i|^{-d},$$ where $C$ and $d$ depend on the $f_i$ and $[a,b]$ but not on the $\lambda_i$. It's not clear to me which conditions to impose, but it seems reasonable to ask that such an extension holds for $f_i(x)=x^{\alpha_i}$ for distinct positive $\alpha_i$.

I would also appreciate any references treating this kind of extension of the van der Corput's lemma.

Answer 1

For convenience of notation write $\Phi(x) = \lambda_1 x^\alpha + \lambda_2 x^\beta$. The general argument is based on the integration by parts argument $$ \int_a^b e\circ\Phi = \int_a^b \frac{1}{2\pi i \Phi'} \frac{d}{dx} e\circ \Phi = \frac{1}{2\pi i \Phi'(b)} e(\Phi(b)) - \frac{1}{2\pi i \Phi'(a)} e(\Phi(a)) + \int_a^b \frac{\Phi''}{2\pi i (\Phi')^2} e\circ\Phi $$

If $\lambda_1 = \lambda_2 = 0$, then there is nothing to prove, since the integral is exactly $b-a$.

Case 1: $\lambda_1 \lambda_2 \geq 0$ and not both zero.

In this case the key observation is that $\Phi'(x) = \alpha \lambda_1 x^{\alpha - 1} + \beta \lambda_2 x^{\beta - 1}$ is signed, and has the lower bound $$ |\Phi'(x)| \geq |\lambda_1| \alpha \min(a^{\alpha - 1}, b^{\alpha - 1}) + |\lambda_2| \beta \min(a^{\beta-1}, b^{\beta -1}). $$ Additionally, $\Phi''(x)$ has the bound $$ |\Phi''(x)| \leq |\lambda_1 \alpha (\alpha - 1) \max(a^{\alpha - 2}, b^{\alpha - 2})| + |\lambda_2 \beta (\beta - 2) \max(a^{\beta - 2}, b^{\beta - 2})| $$ So there is some constant $C$ depending on $a, b, \alpha, \beta$ such that $$ |\int_a^b e\circ \Phi| \leq C ( \max( |\lambda_1|, |\lambda_2| )^{-1} $$

(Note that it is the inverse of the max, not the max of the inverses.)

Case 2: $\lambda_1 \lambda_2 < 0$

At issue here is whether $\Phi'$ has a root. Its unique positive root is at the point $$ c = \left( \frac{\alpha |\lambda_1|}{\beta |\lambda_2|} \right)^{\frac{1}{\beta - \alpha}} $$ Note that for $|\lambda_1| \gg |\lambda_2|$ or $|\lambda_2| \gg |\lambda_1|$, necessarily $c \not\in [a,b]$, and in these cases we have $|\Phi'(x)| > \tilde{C} \max(|\lambda_1|, |\lambda_2|)$ again, and the analyses as in Case 1 will show the same decay rate.

So the only issue is when $|\lambda_1| \approx |\lambda_2|$.

Computing $$ \Phi''(c) = c^{\alpha - 2} \lambda_1 \alpha (\alpha - \beta) $$ Let $\delta$ be a number of size $1/\sqrt{\lambda_1}$. Split the integral into $$ \int_a^{c-\delta} + \int_{c-\delta}^{c+\delta} + \int_{c+\delta}^b $$

On the interval $[a,c-\delta]$, it is not too hard to check that the minimum of $|\Phi'(x)|$ occurs at one of the end points $a$ or $c-\delta$, at $a$ one sees $\Phi'(x)$ is of size $\lambda_1$ (recall that $|\lambda_1| \approx |\lambda_2|$), and at $c-\delta$ we have $\Phi'(x)$ is of size $\Phi''(c) \delta \approx \sqrt{|\lambda_1|}$.

We could almost apply the method of Case 1: however, we still have $\Phi''$ is size $\lambda_1$ and now $(\Phi')^2$ may only be lower bounded by $\lambda_1$ also, which is not good enough. So we need a trick. Looking at $\Phi''$ we see that on $[a,b]$ it has at most one zero. Which means we can decompose $[a, c-\delta]$ and $[c+\delta,b]$ into finitely many disjoint intervals on which $\Phi''$ is signed. On such an interval, we have $$ \left| \int_a^b \frac{\Phi''}{2\pi i(\Phi')^2} e\circ \Phi \right| \leq \int_a^b \frac{|\Phi''|}{2\pi (\Phi')^2} = \left| \int_a^b \frac{|\Phi''|}{2\pi(\Phi')^2} \right| = \left| \frac{1}{2\pi \Phi'(b)} - \frac{1}{2\pi \Phi'(a)} \right| $$ Using this trick in addition to the analyses of Case 1, we see that the integrals over $[a,c-\delta]$ and $[c+\delta,b]$ contribute terms of size $\frac{1}{\sqrt{|\lambda_1|}}$ asymptotically.

The integral over $[c-\delta, c+\delta]$ is bounded in absolute values by $2\delta$ which is also size $\frac{1}{\sqrt{|\lambda_1|}}$.

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Summary

Skipping some details tidying up the analysis, you find, in the end, that there exists a constant $C$ depending on $a, b, \alpha, \beta$, such that for any $\lambda_1, \lambda_2\in \mathbb{R}$, you have

$$ \left| \int_a^b e\circ \Phi \right| \leq C \min( 1, |\alpha \lambda_1 + \beta \lambda_2|^{-1}, |\lambda_1|^{-1/2} ) $$

Remark

Case 1 directly generalizes to multiple sums of the form $\lambda_1 f_1 + \cdots + \lambda_n f_n$, when all the $\lambda_i$ have the same sign and all of the $f_i$ have first derivatives bounded below by some positive number, and second derivatives bounded.

Case 2 is more delicate in multiple sums. In the analysis here we used that for $\Phi$ of the form specified, $\Phi^{(k)}(x)$ has at most one positive root for any derivative. When the sum is of three or more terms, one has to worry about not just roots but "approximate roots" of $\Phi'$. But what you would generally expect is that the "decay rate" is directional (similar to how in Case 2 above, when $\lambda_1 \approx - \lambda_2$ you can only get $|\lambda_1|^{-\frac12}$ decay, but you get better decay outside of this zone).

I should note that in many cases such asymptotics have been computed. For constant coefficient PDEs, you have a representation formula for solutions via the Fourier transform. And these kinds of computations are exactly used to establish decay rates of the solution in different directions. You can find some examples in Rauch's PDE textbook, for Stein's Functional Analysis textbook (in the Princeton Analysis series).

Answer 2

(This is not a full answer, but rather a comment that got too long)

I think https://people.kth.se/~parissis/PhD_thesis/thesis.pdf has some useful insights for your question. In particular, when $\alpha,\beta$ are integers, it works well.

I did not do the computations, but I would expect that, with some modifications, it could work for your original question about monomials. The key insight is that, if one denotes the phase $\lambda_1 x^{\alpha} + \lambda_1 x^{\beta} = \phi_{\lambda}(x),$ then we just have to find a good bound on the measure of the set $$ \{ x \in (a,b) : |\phi_{\lambda}(x)| < \alpha \}, $$ and prove that the complement has 'boundedly many' connected components. Then one uses 'regular van der Corput' on the complement, and bounds the integral by $1$ in the set above.

For the aforementioned bound on the measure of the set, maybe Lemma 2.10 in the link I sent can be useful. For the bound on the amount of components of the complement, one basically needs to count roots of a polynomial equation in the integer case, and I would assume this would not be too hard to generalize.

Notice that these bounds would yield a $C$ which, in principle, depends on the interval under consideration. It seems that this is not relevant in the statement of your question by what I explained in the previous sentence, but in case one needs $C$ independent - or the optimal dependency on the interval - then one might need to dive deeper into the link I sent...

In any case, I hope this helps :)