$O(\lambda^{-1/k})$ estimate for $\int_a^b e^{i\lambda \phi(x)}\, dx$ given $|\phi^{(k)}(x)| \ge 1$ for fixed $k$

Question

In what follows, $\phi$ is a single-variable real-valued function. The excerpt is from Chapter 8, of Stein's Harmonic Analysis: Real-Variable Methods, Orthogonality, and Oscillatory Integrals.

Suppose we know only that $$\left|\frac{d^k\phi(x)}{dx^k} \right| \ge 1$$ for some fixed $k$, and we wish to obtain an estimate for $$\int_a^b e^{i\lambda \phi(x)}\, dx$$ that is independent of $a$ and $b$. The change of variable $x\mapsto \lambda^{-1/k} x'$ shows that the only possible estimate for the integral is $$O(\lambda^{-1/k})$$

How does the change of variable show that the only possible estimate for the integral is $O(\lambda^{-1/k})$? Putting $x = \lambda^{-1/k} t$, we get $$\int_a^b e^{i\lambda \phi(x)}\, dx = \lambda^{-1/k} \int_{\lambda^{1/k} a}^{\lambda^{1/k} b} e^{i\lambda \phi( \lambda^{-1/k} t)}\, dt$$ but I'm not sure how to go from here. Specifically, how do we use $\left|\frac{d^k\phi(x)}{dx^k} \right| \ge 1$? Perhaps there is some integration by parts involved. Thank you!

Answer 1

Consider the following claim.

Let $a < b$, and let $\phi \in C^k([a,b])$ be such that $|\phi^{(k)}(x)| \geq 1$ for all $x \in [a,b]$. Then for any $\lambda>0$, one has $\int_a^b e^{i\lambda \phi(x)}\ dx \lesssim_k \lambda^\beta$.

The hypotheses of this claim obey the following scale invariance: if $a,b,\lambda$, and $\phi: x \mapsto \phi(x)$ obey the above hypotheses, and $\lambda_0>0$ is a scaling parameter, then $\lambda_0^{1/k} a$, $\lambda_0^{1/k} b$, $\lambda/\lambda_0$, and $\phi_{\lambda_0}: x \mapsto \lambda_0 \phi(x/\lambda_0^{1/k}$) also obey the above hypotheses. Applying the conclusion of the above claim to this rescaled data and applying a change of variables, one concludes that $$ \displaystyle \int_a^b e^{i\lambda \phi(x)}\ dx \lesssim_k \lambda_0^{-1/k-\beta} \lambda^\beta $$ for any $\lambda_0>0$ and any $a,b,\phi,\lambda$ obeying the hypotheses of the claim.

Since one can easily construct $a,b,\phi,\lambda$ for which the left-hand side is non-zero, we arrive at a contradiction by sending $\lambda_0$ to either zero or infinity unless $\beta=-1/k$. Thus we conclude that $\beta = -1/k$ is a necessary condition for this claim (though this does not show that it is sufficient; indeed, the claim is false when $k=1$).

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One can run the scaling argument above with $\lambda_0=\lambda$ if desired (this amounts to “spending” the scale invariance to normalise $\lambda=1$). If one rescales $x =\lambda^{-1/k} x'$, then the estimate $$\int_a^b e^{i\lambda \phi(x)}\ dx \lesssim_k \lambda^\beta$$ becomes $$\int_{a'}^{b'} e^{i\phi'(x')}\ dx' \lesssim_k \lambda^{\beta+1/k}$$ where $a' := \lambda^{1/k} a, b' := \lambda^{1/k} b$, and $\phi'(x') := \lambda \phi(\lambda^{-1/k} x')$. The rescaled data $a'$,$b'$,$\phi'$ range in exactly the same class as the original data $a,b,\phi$. Fixing a single $a',b',\phi'$ in this class for which the integral $\int_{a'}^{b'} e^{i\phi'(x')}\ dx'$ does not vanish, and sending $\lambda$ to either zero or infinity, one recovers the scaling condition $\beta = -1/k$.