If $f\in L^1$, why the Fourier transform $\hat f$ is uniformly continuous?

Question

Let $f\in L^1(\mathbb R)$. So, I know that the Fourier transform is continuous. Indeed, $$|\hat f(\alpha +h)-\hat f(\alpha )|\leq \int_{\mathbb R}|f(x)||e^{-2i\pi xh}-1|dx,$$ and since $|e^{-2i\pi xh}-1|\leq 2$, we can apply dominated convergence. But I don't get why is it uniformly continuous. If $\hat f$ would be supported on a compact set, then it's obvious, but here, how can I conclude ?

Answer 1

Starting with your inequality $$|\hat f(\alpha +h)-\hat f(\alpha )|\leq \int_{\mathbb R}|f(x)||e^{-2i\pi xh}-1|dx,$$ observe that the r.h.s does not depend on $\alpha$, hence $$\sup_{\alpha\in\mathbb{R}}|\hat f(\alpha +h)-\hat f(\alpha )|\leq \int_{\mathbb R}|f(x)||e^{-2i\pi xh}-1|dx,$$ Since $f\in L^1(\mathbb{R})$, the r.h.s tends to zero as $h\to 0$, (by the dominated convergence theorem), hence $$\lim_{h\to 0}\sup_{\alpha\in\mathbb{R}}|\hat f(\alpha +h)-\hat f(\alpha )|=0,$$ and this is equivalent to saying that $\hat{f}$ is uniformly continuous.

Answer 2

The Riemann Lebesgue Lemma tells you that $\hat {f}(x) \to 0$ as $x \to \pm \infty$. Any continuous function with this property is uniformly continuous.