Does square integrable fourier transform mean the original function square integrable?

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if $f\in L^1(R^n)$ and its Fourier transform $\hat{f}=\int f(x) e^{-ixt} \ dx \in L^2(R^n)$, then can we show $f \in L^2(R^n)$?

If yes, can you tell me why? If not, can you give me some counterexamples? I also see some obscure answers elsewhere with some high-order mathematcical concepts such as fourier transform in schwardz space, I can't understand it. If yes, can you prove it with the knowledge within the scope of Rudin's 'Real and Complex Analysis'?