I am a Ph.D. student in statistic, and I wanted to know if there is an easy way to show that the Fourier inversion converge to $f$ in $L^p(\mathbb R)$ for $1<p\leq 2$. In other word, that $$\lim_{N\to \infty }\left\|\int_{|\alpha |\leq N}\hat f(\alpha )e^{2i\pi x\alpha }d\alpha-f(x)\right\|_{L^p(\mathbb R^n)}=0.$$ If the proof is to long could you give me at least the most important steps ?
Just in case, I denoted $\hat f$ as the Fourier transformation of $f$.
Thank you,
There is no "obvious proof". The proof is essentially based on the fact that Hilbert transform is $L^p-$bounded (i.e. that $$\|Hf\|_{L^p(\mathbb R)}\leq C\|f\|_{L^p(\mathbb R)}$$ for a suitable constant $C$. The proof of this is not that easy. But this boundness will allow you to show that $$T_Nf(x):=\int_{|\alpha |\leq N}\hat f(\alpha )e^{2i\pi x\alpha }\mathrm d \alpha $$ converge uniformly in $N$ provided $f\in \mathcal C^\infty _0(\mathbb R)$ (i.e. compact supported). Still, this will not be obvious to prove. A way to do it is to show that $$\widehat{Hf}(\alpha )=-i\pi \text{sign}(\alpha )\hat f(\alpha ),$$ remark that $$\chi_{[-N,N]}(\alpha )=(\text{sign}(N-\alpha )+1)(\text{sign}(N+\alpha )+1)$$ and use the fact that $$\mathfrak F\left(\frac{ie^{-iNx}}{\pi}H(e^{-Nx}f)(\alpha )\right)=\text{sign}(\alpha -N)\hat f(\alpha ).$$ I also denote $\mathfrak F(f)$ the Fourier transform of $f$. Finally, using all those informations, it will be (finally) easy to prove your result using then fact that Schwarz space with function compactly supported is dense in $L^p(\mathbb R)$ and that the Fourier transformation is a bijection $\mathcal S(\mathbb R)\longrightarrow \mathcal S(\mathbb R)$ (where $\mathcal S(\mathbb R)$ is the Schwarz space).
P.S.: All informations are in the Muscalu and Schlag (Vol I).