Suppose that $f_n$ and $f$ are $L^1(\mathbb R^n)$ functions with $f_n \to f$ in $L^1$ sense. Then is it true that their Fourier transforms defined as
$$ \hat f(\xi) := \int_{\mathbb R^n} e^{-ix\cdot\xi}f(x)dx $$
converge uniformly? I think it is true though not sure. The following is my argument.
$$ 0 \leq \lim_{n \to \infty} \sup_\xi \left| \hat f_n(\xi) - \hat f(\xi) \right| = \lim_{n \to \infty} \sup_\xi \left| \int_{\mathbb R^n} e^{-ix\cdot\xi}f_n(x)dx - \int_{\mathbb R^n} e^{-ix\cdot\xi}f(x)dx \right| \\ = \lim_{n \to \infty} \sup_\xi \left| \int_{\mathbb R^n} e^{-ix\cdot\xi} [f_n(x) - f(x)] dx \right| \leq \lim_{n \to \infty} \sup_\xi \int_{\mathbb R^n} \left| e^{-ix\cdot\xi} [f_n(x) - f(x)] \right| dx \\ \leq \lim_{n \to \infty} \sup_\xi \int_{\mathbb R^n} \left| f_n(x) - f(x) \right| dx = \sup_\xi \lim_{n \to \infty} \int_{\mathbb R^n} \left| f_n(x) - f(x) \right| dx = 0. $$
Therefore, the result is proved. The third inequality sign used the fact that $|e^{-ix\cdot\xi}| \leq 1$ and the last equality used the fact that the sup was taken over $\xi$ instead of $n$. Is the above argument correct, please? Thank you!
Yes, the argument is correct. It rests on the fact that for an integrable function $g$, we have $$|\widehat g(\xi)|\leqslant \lVert g\rVert_1$$ due to the triangle inequality. Hence $$\sup_\xi|\widehat{f_n}(\xi)-\widehat f(\xi)|\leqslant \lVert f_n-f\rVert_1$$ and we conclude using convergence in $\mathbb L^1$.