Why is the inverse fourier transform of the fourier transform of this function equal the original function?

Question

I know that for function in the Schwarz space I have the realtion $\mathcal{F}^{-1}\mathcal{F}(g)=g$, i.e. the inverse Fourier transform of the Fourier transform of a function is the function. But here suppose $f$ is a function in $L^p(\mathbb{R})$ where $1<p<\infty$. I know the Hilbert transform is bounded for $1<p<\infty$ but I was told that $\mathcal{F}^{-1}\mathcal{F}(\frac{ie^{-iNx}}{\pi}H(e^{iNx}f))=\frac{ie^{-iNx}}{\pi}H(e^{iNx}f)$. I dont understand this since $\frac{ie^{-iNx}}{\pi}H(e^{iNx}f) \notin S(\mathbb{R})$ no?