I am trying to prove the following statement:
If $f \in L^1$, then $\hat f$ is uniformly continuous.
The argument given is as follows :
$$|\hat f (\xi +h )-\hat f (\xi)| = \left| \int f(x) (e^{-2 \pi i x \cdot (\xi+h)}- e^{-2 \pi i x \cdot (\xi)})\mathrm dx \right| \leq 2 \|f\|_{L^1}$$
Now I suppose we have to use the Dominated Convergence Theorem, but I am unable to see to what sequence of functions we apply the theorem to.
Any help is greatly appreciated.
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.
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 $|\eta|<\delta$, $$ \begin{align} |\hat{f}(\xi+\eta)-\hat{f}(\xi)| &\le\int_{\mathbb{R}^n}|f(x)|\;|e^{-2\pi ix\cdot\eta}-1|\;\mathrm{d}x\\ &=\int_{|x|<R}|f(x)|\;|e^{-2\pi ix\cdot\eta}-1|\;\mathrm{d}x +\int_{|x|\ge R}|f(x)|\;|e^{-2\pi ix\cdot\eta}-1|\;\mathrm{d}x\\ &\le\|f\|_{L^1}\frac{\epsilon}{2\|f\|_{L^1}}+\;2\frac{\epsilon}{4}\\ &=\epsilon \end{align} $$
I) The proof for $L_1$ is simpler actually; here is an outline:
1) Prove that a linear map $f : E\to E'$ is continuous (even uniformly) iff it is continuous at zero (0); i.e.
$$\begin{align} (\exists c\ \epsilon \ \mathbb{R}) \ |f(x)|\leq c|x| \end{align}$$
2) Fourier transform is a linear functional defined on $L_1$. So, by (1) you only need to prove it is continuous at 0: We have:
$$\begin{align} F(f) = \int_{\mathbb{R}} f(x)e^{-j\omega x}dx \end{align}$$ Where F is the Fourier operator defined on L1. $$\begin{align} |F(f)| = \left|\int_{\mathbb{R}} f(x)e^{-j\omega x}dx\right| \leq \int_{\mathbb{R}} |f(x)e^{-j\omega x}|dx \leq \int_{\mathbb{R}} |f(x)|dx = \left \| f \right \|_{L^1} < {\infty} \end{align}$$ Thus $$\begin{align} \left | F(f) \right | \leq 1 \left \| f \right \|_{L^1}, \end{align}$$ This completes the proof (set c = 1)
II) Yes, you could use DCT, here's how: Take any sequence $$x_n \to 0$$ Set $$ u_n(x) = f(x)e^{-i\omega (x\pm x_n)} \ \\ \ u(x) = f(x)e^{-i\omega x} $$ Clearly, $$ u_n \to u $$
Now, $$\ u\ is\ L^1,\ so\ is\ each\ u_n$$
We also have $$\left |u_n \right | \leq \left | f \right |\ and\ f\ is\ L^1$$ Now use DCT to get: $$ \int_{\mathbb{R}}|u_n-u| \to 0\ as \ n \to \infty $$
Observe that for real $x,y \in \mathbb R$, $$|e^{ix}-e^{iy}|\leq |x-y| \wedge 2.$$ Then $$|\hat f (\xi +h )-\hat f (\xi)| \leq \int |f(x)| (|2\pi x h|\wedge 2) \mathrm dx, $$ so we have a bound not depending on $\xi$, which goes to zero as $|h| \to \infty$ by the dominated convergence theorem.
