In a proof of the Riemann-Lebesgue lemma in Hunter's Applies Analysis, he first proves the statement in the Schwartz space and then uses a density argument:
Here are my questions:
- What goes wrong if one only assumes $\varphi\in L^1$ in the red box?
- In the very last line, why use $\liminf$ instead of $\lim$ directly?
The argument establishing the inequality for $\varphi\in\mathcal S$ would work just as well if it had been assumed only that $\varphi\in L^1$, but the author wanted to focus on $\mathcal S$ for a reason. The space $\mathcal S$ is actually closed under the Fourier transform. So $\widehat\varphi$, the Fourier transform of $\varphi\in\mathcal S$, is continuous. Later he shows a sequence of such Fourier transforms that are in $\mathcal S$ converges uniformly. A sequence of continuous functions can converge in various senses to a discontinuous function, including pointwise convergence. But if the convergence is uniform, then the limit is continuous. And he shows that limit is the Fourier transform of the function in $L^1$ that he started with. Thus he concludes that the Fourier transform of a function in $L^1$ is continuous.