Fourier transform is uniformly continuous

Solution 1:

I don't know if your questions has been answered in full. For completeness, we apply DCT for the reasons you mentioned in your post. The punchline of the story is:

$$\begin{align} \left|\widehat{f}(\xi + h) - \widehat{f}(\xi)\right| &= \left| \int f(x) \left(e^{-2 \pi i x \cdot (\xi + h)} - e^{-2 \pi i \xi \cdot x} \right)dx \right| \\ &\leq \int |f(x)| \left|e^{2 \pi i x \cdot h} - 1 \right| dx \end{align}$$

which tends to zero as $h \to 0$, and this is enough to show uniform continuity.

Solution 2:

I like Olivier's comment suggesting the use of the Riemann-Lebesgue Lemma, but here is a different approach. $$ \begin{align} \hat{f}(\xi+\eta)-\hat{f}(\xi) &=\int_{\mathbb{R}^n}f(x)\left(e^{-2\pi ix\cdot(\xi+\eta)}-e^{-2\pi ix\cdot\xi}\right)\mathrm{d}x\\ &=\int_{\mathbb{R}^n}f(x)\left(e^{-2\pi ix\cdot\eta}-1\right)e^{-2\pi ix\cdot\xi}\;\mathrm{d}x\tag{1} \end{align} $$ For any $f\in L^1$ and $\epsilon>0$, by Dominated Convergence, we can find an $R>0$ so that $$ \int_{|x|>R}|f(x)|\mathrm{d}x<\frac{\epsilon}{4}\tag{2} $$ Let $\delta=\frac{\epsilon}{4\pi R\|f\|_{L^1}}$. For $|x|\le R$ and $|\eta|<\delta$, $$ \left|e^{-2\pi ix\cdot\eta}-1\right|\le\frac{\epsilon}{2\|f\|_{L^1}}\tag{3} $$ whereas for all $x$, $$ \left|e^{-2\pi ix\cdot\eta}-1\right|\le2\tag{4} $$ Then, for $|\eta|<\delta$, $$ \begin{align} |\hat{f}(\xi+\eta)-\hat{f}(\xi)| &\le\int_{\mathbb{R}^n}|f(x)|\;|e^{-2\pi ix\cdot\eta}-1|\;\mathrm{d}x\\ &=\int_{|x|<R}|f(x)|\;|e^{-2\pi ix\cdot\eta}-1|\;\mathrm{d}x +\int_{|x|\ge R}|f(x)|\;|e^{-2\pi ix\cdot\eta}-1|\;\mathrm{d}x\\ &\le\|f\|_{L^1}\frac{\epsilon}{2\|f\|_{L^1}}+\;2\frac{\epsilon}{4}\\ &=\epsilon \end{align} $$