Riemann Lebesgue Lemma for polynomial?

Question

I was asked to prove that

$$\lim_{n\to\infty} \int_{0}^{1} \exp(i\cdot n\cdot p(x))\;dx =0 $$

for nonconstant real polynomial $p(x)$.

if $p(x)$ is of degree $1$... It reduces to Riemann-Lebesgue lemma.

I think similar motivation will work... but not able to

You may want to proofread your question whether the limit symbol is omitted or not. — Sangchul Lee, Oct 02 '12 at 16:35

Micah · Accepted Answer · 2022-07-17T03:52:15.600

1

Suppose first that $p'$ is nonzero on $(0,1)$, so that $p$ has an inverse $q$ on $[0,1]$ which is differentiable everywhere except possibly at the endpoints. Then making the substitution $u=p(x)$ yields $$\int_0^1 e^{i \, n \, p(x)} \, dx=\int_{p(0)}^{p(1)} e^{i \, n \, u} \, \frac{1}{p'(q(u))}\, du \, .$$ The function $f(u)=\dfrac{1}{p'(q(u))}$ is continuous on the interior of the interval of integration. At $p(0)$ and $p(1)$, $f(u)$ is either continuous or grows like $u^{-(n-1)/n}$ for some $n>1$. So $f(u)$ is $L^1$, which means the ordinary Riemann-Lebesgue lemma applies.

In general, $p'$ may not be nonzero on the entire interval. But since $p$ is a polynomial, we can split $[0,1]$ up into some finite collection of intervals over which $p'$ is nonzero, and make the above substitution separately on each of them...

edited Jul 17 '22 at 03:52

answered Oct 02 '12 at 18:29

Micah

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I never thought change of variable will work.. Thank you! – Detectives Oct 03 '12 at 16:03
Shouldn't you have the derivative dividing after the change of variable? $du = p'(x) dx$, then $du = \frac{1}{p'( p^{-1} (u) )} du.$ – Gonzalo Benavides Jul 16 '22 at 06:07
The good thing is that the function $\frac{1}{p'(p^{-1}(u))}$ is indeed $L^1$. – Gonzalo Benavides Jul 16 '22 at 06:27
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@Gonzalo Benavides: Yes, 10-years-ago-me should have known better than that! – Micah Jul 16 '22 at 19:13

Riemann Lebesgue Lemma for polynomial?

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