Fourier transform is uniformly continuous 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. 
 A: 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.
A: 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 $|\eta|<\delta$,
$$
\begin{align}
|\hat{f}(\xi+\eta)-\hat{f}(\xi)|
&\le\int_{\mathbb{R}^n}|f(x)|\;|e^{-2\pi ix\cdot\eta}-1|\;\mathrm{d}x\\
&=\int_{|x|<R}|f(x)|\;|e^{-2\pi ix\cdot\eta}-1|\;\mathrm{d}x
+\int_{|x|\ge R}|f(x)|\;|e^{-2\pi ix\cdot\eta}-1|\;\mathrm{d}x\\
&\le\|f\|_{L^1}\frac{\epsilon}{2\|f\|_{L^1}}+\;2\frac{\epsilon}{4}\\
&=\epsilon
\end{align}
$$
A: 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:
$$
\int_{\mathbb{R}}|u_n-u| \to 0\ as \ n \to \infty $$
A: 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.
