The a function $f = \dfrac{\sin x}{x} \in L_2(\mathbb{R}^n)$ is not in $L_1(\mathbb{R}^n)$, but $\hat{f} \in L_1(\mathbb{R}^n)$.
Under what circumstances can this happen?
2026-04-01 20:01:02.1775073662
Under what circumstances can this happen?
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Actually
The Fourier transformation is bijective only on $L_2(\mathbb{R}^n)$ not on $L_1(\mathbb{R}^n)$. So there is no contradiction to this.
Furthermore, because there is no inclusion between $L_1$ and $L_2$ these can happen without contradiction.
I hope it answers to your question.