Let $f\in L^1(\mathbb{R}^n)$. Then the Fourier transform $\hat{f}$ is given by
$$\hat{f}(\xi):=\int_{\mathbb{R}^n} e^{-2\pi i x\cdot \xi} f(x) dx.$$
The dominated convergence theorem confirms that the Fourier transform is a continuous function. Indeed, if the sequence $\xi_n\rightarrow \xi_0$ then
$\hat{f}(\xi_n)\rightarrow\hat{f}(\xi_0)$ since, for every $n\geq 1$, $|e^{-2\pi i x\cdot \xi_n} f(x)|\leq |f(x)|\in L^1(\mathbb{R}^n)$.
I want to prove that $\hat{f}$ is continuous using the $\epsilon$ –$\delta$ definition.
My attempt:
Fix $\xi_0$ in $\mathbb{R}^n$ and let $\epsilon>0$. We need to prove that there exists $\delta>0$ such that
$$|\xi-\xi_0|<\delta \implies |\hat{f}(\xi)-\hat{f}(\xi_0)|<\epsilon.$$
We have
$$\hat{f}(\xi)-\hat{f}(\xi_0)=\int_{\mathbb{R}^n} (e^{-2\pi i x\cdot \xi}-e^{-2\pi i x\cdot \xi_0}) f(x) dx.$$
Define
$$g(t):=e^{-2\pi i x\cdot ((1-t)\xi+t\xi_0)}.$$
Then $g$ is a smooth function. By the mean-value theorem
$$e^{-2\pi i x\cdot \xi}-e^{-2\pi i x\cdot \xi_0}=g(1)-g(0)= g^{\prime}(s)=
2\pi i e^{-2\pi i x\cdot ((1-s)\xi+s\xi_0)}x\cdot (\xi-\xi_0)$$
for some $s\in (0,1)$.
Therefore
$$\hat{f}(\xi)-\hat{f}(\xi_0)=2\pi i (\xi-\xi_0)\cdot\int_{\mathbb{R}^n} e^{-2\pi i x\cdot ((1-s)\xi+s\xi_0)}x f(x) dx.$$
Now, if $f$ has compact support, or decay faster than $1/|x|$ as $|x|\rightarrow \infty$ we are done since, in that case,
$$|\hat{f}(\xi)-\hat{f}(\xi_0)|\leq C |\xi-\xi_0|$$
with $C=2\pi \int_{\mathbb{R}^n} |x| |f(x)| dx$.
If $f$ is just an $L^1$ function we must use the cancellation due to the oscillatory exponential factor (may be using Riemann-Lebesgue lemma).
Best Answer
You need some properties of $L^{1}$ functions (which may themselves involve DCT).
You cannot prove this directly for an arbitrary $L^{1}$ function without using some such property.
You can show that given $\epsilon >0$ there exist $M$ such that $\int_{\{x: |x|>M\}} |f| <\epsilon$. Note that $\int_{|x|>M} |e^{-2\pi ix.\xi}-e^{-2\pi ix.\xi_o}||f(x)|dx <2\epsilon$. For the integral over $\{x: |x| \leq M\}$ you can use your argument.