Let $\lambda$ denote the Lebesgue measure on $\mathbb R$, $d\in\mathbb N$ and $$\hat f(\omega):=\int e^{-{\rm i}2\pi\langle x,\;\omega\rangle}f(x)\:\lambda^{\otimes d}({\rm d}x)\;\;\;\text{for }\omega\in\mathbb R^d$$ denote the Fourier transform of $f\in\mathcal L^1(\lambda^{\otimes d};\mathbb C)$.
Let $p\in[1,\infty]$ and $(f_n)_{n\in\mathbb N}\subseteq\mathcal L^1(\lambda^{\otimes d};\mathbb C)\cap\mathcal L^p(\lambda^{\otimes d};\mathbb C)$ with $$\left\|f_n-f\right\|_{L^p(\lambda^{\otimes d};\:\mathbb C)}\xrightarrow{n\to\infty}0\tag1$$ for some $f\in\mathcal L^1(\lambda^{\otimes d};\mathbb C)\cap\mathcal L^p(\lambda^{\otimes d};\mathbb C)$. Is this enough to conclude $$\hat f_n(\omega)\xrightarrow{n\to\infty}\hat f(\omega)\tag2$$ for all $\omega\in\Omega$?
Let $\omega\in\mathbb R^d$. The function $\mathbb R^d\ni x\mapsto e^{-{\rm i}2\pi\langle x,\;\omega\rangle}$ is not in $\mathcal L^p(\lambda^{\otimes d};\mathbb C)$, unless $p=\infty$. So, am I missing something, or do we really need $p=1$ in order for $(2)$ to hold?
Here is an example that shows that bounds in $L_p$, $p\neq1$, don't imply pointwise convergence of an $L_p$ convergent sequence of Fourier series.
Let $f(x)=\frac12\mathbb{1}_{[-1,1]}(x)$ and define for $n\in\mathbb{N}$ $f_n(x)=f^{*n}(x)$ (convolution of $f$ with itself $n$ times). Then $$\widehat{f_n}(t)=\Big(\frac{\sin 2\pi t}{2\pi t}\Big)^n$$ It is easy to check that $\|f_n\|_1=1$ for all $n$, while $\|f_n\|_2=\|\widehat{f_n}\|_2\xrightarrow{n\rightarrow\infty}0$. The sequence $\widehat{f_n}$ converges to $0$ almost surely pointwise, however, it fails to converge to $0$ at the point $t=0$.