Does the Riemann-Lebesgue Lemma Apply to $L^1$ or $L^2$ Space?

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.)

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.