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Riemann-Lebesgue Lemma — Definition, Formula & Examples

The Riemann-Lebesgue Lemma states that the Fourier coefficients of an integrable function decay to zero as the frequency grows without bound. In other words, if you decompose a function into sinusoidal components, the high-frequency components must have vanishingly small amplitudes.

If f∈L1(R)f \in L^1(\mathbb{R}), then f^(ξ)=∫−∞∞f(x) e−2πiξx dx→0\hat{f}(\xi) = \int_{-\infty}^{\infty} f(x)\, e^{-2\pi i \xi x}\, dx \to 0 as ∣ξ∣→∞|\xi| \to \infty. Equivalently, on a finite interval, if f∈L1([a,b])f \in L^1([a,b]), then ∫abf(x) einx dx→0\int_a^b f(x)\, e^{in x}\, dx \to 0 as ∣n∣→∞|n| \to \infty.

Key Formula

lim⁡∣ξ∣→∞∫−∞∞f(x) e−2πiξx dx=0\lim_{|\xi|\to\infty} \int_{-\infty}^{\infty} f(x)\, e^{-2\pi i \xi x}\, dx = 0
Where:
  • ff = An integrable function, i.e., $f \in L^1(\mathbb{R})$
  • ξ\xi = Frequency variable in the Fourier transform

How It Works

The lemma works because rapid oscillation of einxe^{i n x} causes positive and negative contributions of the integral to cancel more and more completely as ∣n∣|n| increases. For step functions this cancellation is straightforward to verify, and since every L1L^1 function can be approximated by step functions, the result extends by a density argument. In practice, you apply it to conclude that Fourier coefficients f^(n)→0\hat{f}(n) \to 0 without computing them explicitly, or to show that certain oscillatory integrals vanish in a limit.

Worked Example

Problem: Let f(x)=e−∣x∣f(x) = e^{-|x|} on R\mathbb{R}. Verify the Riemann-Lebesgue Lemma by computing f^(ξ)\hat{f}(\xi) and showing it tends to 00 as ∣ξ∣→∞|\xi| \to \infty.
Step 1: Confirm f∈L1(R)f \in L^1(\mathbb{R}). We have ∫−∞∞e−∣x∣ dx=2\int_{-\infty}^{\infty} e^{-|x|}\, dx = 2, so ff is integrable.
∫−∞∞e−∣x∣ dx=2\int_{-\infty}^{\infty} e^{-|x|}\, dx = 2
Step 2: Compute the Fourier transform. Using the standard formula:
f^(ξ)=∫−∞∞e−∣x∣e−2πiξx dx=21+4π2ξ2\hat{f}(\xi) = \int_{-\infty}^{\infty} e^{-|x|} e^{-2\pi i \xi x}\, dx = \frac{2}{1 + 4\pi^2 \xi^2}
Step 3: Observe the decay: as ∣ξ∣→∞|\xi| \to \infty, the denominator 1+4π2ξ2→∞1 + 4\pi^2 \xi^2 \to \infty, so f^(ξ)→0\hat{f}(\xi) \to 0, exactly as the Riemann-Lebesgue Lemma guarantees.
lim⁡∣ξ∣→∞21+4π2ξ2=0\lim_{|\xi|\to\infty} \frac{2}{1 + 4\pi^2 \xi^2} = 0
Answer: f^(ξ)=21+4π2ξ2→0\hat{f}(\xi) = \dfrac{2}{1+4\pi^2\xi^2} \to 0 as ∣ξ∣→∞|\xi| \to \infty, confirming the lemma.

Why It Matters

The Riemann-Lebesgue Lemma is essential in proving convergence theorems for Fourier series, including pointwise convergence under Dirichlet conditions. It also underpins results in signal processing: any finite-energy signal must have high-frequency content that dies off, which is why truncating a Fourier series is a reasonable approximation strategy.

Common Mistakes

Mistake: Assuming the lemma guarantees a specific rate of decay (e.g., 1/n1/n) for Fourier coefficients.
Correction: The lemma only guarantees f^(ξ)→0\hat{f}(\xi) \to 0. The rate of decay depends on the smoothness of ff; smoother functions have faster-decaying coefficients, but this requires additional theorems to quantify.