Okay previously my lecturer showed that this is so by proving in the following way:
Proof by contradiction. Suppose the transform is in $L^{1}$. Then as $f \in L^1$, we may use Fourier Inversion Formula thereby getting $ f(x) = \int_{\mathbb{R}} \widehat f(t) e^{2\pi i xt} dt. $ Therefore, $f$ is bounded almost everywhere and therefore globally.
Is this proof correct? I think according to my previous post, it seems not. If this proof is incorrect, can I modify? Or change it entirely and if then how?