$f_k \to f$ in $L^1(\mathbb{R})$ implies that its Fourier transform converges uniformly, i.e. $\hat{f}_k \to \hat{f}$ uniformly on $\mathbb{R}$

Question

Assumme $f_k \to f$ in $L^1(\mathbb{R})$ and show that its Fourier transform converges uniformly, i.e. $\hat{f}_k \to \hat{f}$ uniformly on $\mathbb{R}$. Using the definiton and triangle inequality for integral, we obtain \begin{align} |\hat{f}_k(\xi) - \hat{f}(\xi)| \leq \int_{\mathbb{R}}|e^{-ix\xi}||f_k(x)-f(x)|\ dx = \int_{\mathbb{R}}|f_k(x)-f(x)|\ dx. \end{align} If I can just interchange the limit and the integral sign, then I believe I'm done because our bound is independent of $\xi$. In order to be able to do to that, I need dominated convergence theorem. But I have no idea how to find an integrable function $g$ such that $|f_k(x)-f(x)| \leq |g(x)|$. Any hint would be appreciated.