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In the literature on inverse problems, the Riemann-Lebesgue lemma is often used to demonstrate the ill-posedness of integral equations with square-integrable kernels. For example, in Groetsch (1984),

A more serious concern arises from the Riemann-Lebesgue lemma which states that if $k(\cdot, \cdot)$ is any square integrable kernel, then

$$\int_0^\pi k(x,s) \sin(ns) \ ds \rightarrow 0 \quad \mathrm{as} \quad n \rightarrow \infty$$

Elsewhere on the internet, however, the same lemma often appears to be written for absolutely integrable kernels. For example, on ProofWiki:

Let $f \in L^1$. Then:

$$\lim_{n \rightarrow \infty} \int f(x) e^{inx} \ dx = 0$$

Are these two definitions consistent? Can the Riemann-Lebesgue lemma be used to show that integral equations are ill-posed for kernels in $L^1$, kernels in $L^2$, or both?

(I am not a mathematician - just an interested scientist.)

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1 Answer 1

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In the first scenario, you’re working on the finite interval $[0,\pi]$ so being in $L^2$ already implies you’re in $L^1$, so the Riemann-Lebesgue lemma (which indeed is stated for $L^1$) can already be applied.

The proof that $L^2\subset L^1$ on a finite interval is by Cauchy-Schwarz: \begin{align} \int_a^b|f(x)|\,dx=\int_a^b|f(x)|\cdot 1\,dx\leq \sqrt{\int_a^b|f(x)|^2\,dx}\sqrt{\int_a^b1^2\,dx}=\sqrt{b-a}\|f\|_{L^2([a,b])}<\infty. \end{align} More generally, you can run a similar argument using Holder’s inequality for arbitrary finite measure spaces.

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