Transcendental-free Riemann-Lebesgue lemma

Calculus books tend to introduce transcendental functions (trigonometric, exponential, logarithm) early. Analysis textbooks such as Principles of Mathematical Analysis by Rudin tend to introduce them later, because of how long it takes to develop enough of the theory of power series.

The Riemann-Lebesgue lemma involves either trigonometric or exponential functions. But the following version works with the “late transcendentals” approach.

Transcendental-free Riemann-Lebesgue Lemma

TFRLL. Suppose that ${f\colon [a, b]\to \mathbb R}$ and ${g\colon \mathbb R\to \mathbb R}$ are continuously differentiable functions, and ${g}$ is bounded. Then ${\int_a^b f(x)g'(nx)\,dx \to 0}$ as ${n\to\infty}$ .

The familiar form of the lemma is recovered by letting ${g(x) = \sin x}$ or ${g(x) = \cos x}$ .

Proof. By the chain rule, ${g'(nx)}$ is the derivative of ${g(nx)/n}$ . Integrate by parts:

$\int_a^b f(x)g'(nx)\,dx = \frac{f(b)g(nb)}{n} - \frac{f(a)g(na)}{n} - \int_a^b f'(x)\frac{g(nx)}{n}\,dx$

By assumption, there exists a constant ${M}$ such that ${|g|\le M}$ everywhere. Hence $\left| \frac{f(b)g(nb)}{n}\right| \le \frac{|f(b)| M}{n}$ , $\left|\frac{f(a)g(na)}{n}\right| \le \frac{|f(a)| M}{n}$ , and $\left|\int_a^b f'(x)\frac{g(nx)}{n}\,dx\right| \le \frac{M}{n} \int_a^b |f'(x)|\,dx$ . By the triangle inequality,

$\left|\int_a^b f(x)g'(nx)\,dx \right| \le \frac{M}{n}\left(|f(b)|+|f(a)| + \int_a^b |f'(x)|\,dx \right) \to 0$

completing the proof.

As a non-standard example, TFRLL applies to, say, ${g(x) = \sin (x^2) }$ for which ${g'(x) = 2x\cos (x^2)}$ . The conclusion is that $\int_a^b f(x) nx \cos (n^2 x^2) \,dx \to 0$ , that is, $\int_a^b xf(x) \cos (n^2 x^2) \,dx = o(1/n)$ which seems somewhat interesting. When ${0\notin [a, b]}$ , the factor of ${x}$ can be removed by applying the result to ${f(x)/x}$ , leading to $\int_a^b f(x) \cos (n^2 x^2) \,dx = o(1/n)$ .

What if we tried less smoothness?

Later in Rudin’s book we encounter the Weierstrass theorem: every continuous function on ${[a, b]}$ is a uniform limit of polynomials. Normally, this would be used to make the Riemann-Lebesgue lemma work for any continuous function ${f}$ . But the general form given above, with an unspecified ${g}$ , presents a difficulty.

Indeed, suppose ${f}$ is continuous on ${[a, b]}$ . Given ${\epsilon > 0 }$ , choose a polynomial ${p}$ such that ${|p-f|\le \epsilon}$ on ${[a, b]}$ . Since ${p}$ has continuous derivative, it follows that ${\int_a^b p(x)g'(nx)\,dx \to 0}$ . It remains to show that ${\int_a^b p(x)g'(nx)\,dx}$ is close to ${\int_a^b f(x)g'(nx)\,dx}$ . By the triangle inequality,

$\left| \int_a^b (p(x) - f(x))g'(nx)\,dx \right| \le \epsilon \int_a^b |g'(nx)|\,dx$

which is bounded by … um. Unlike for ${\sin }$ and ${\cos}$ , we do not have a uniform bound for ${|g'|}$ or for its integral. Indeed, with ${g(x) = \sin x^2}$ the integrals $\int_0^1 |g'(nx)| \,dx = \int_0^1 2nx |\cos (n^2x^2)| \,dx$ grow linearly with ${n}$ . And this behavior would be even worse with ${g(x) = \sin e^x}$ , for example.

At present I do not see a way to prove TFRLL for continuous ${f}$ , let alone for integrable ${f}$ . But I do not have a counterexample either.

2 thoughts on “Transcendental-free Riemann-Lebesgue lemma”

Yemon Choi says:

2020-03-07 at 11:52

Typo in the paragraph beginning ‘As a non-standard example, TFRLL applies…” — I think f'(x) should be g'(x)

1. Calculus7 says:
  
  2020-03-07 at 12:44
  
  Thanks, fixed.

Transcendental-free Riemann-Lebesgue lemma