Question about the Carleson's Theorm Recently, I've been trying to understand the proof of Carleson's Theorem on Fourier Analysis given that $f\in L^2$
But I still find it quite hard to understand the relationship of the boundedness of Carleson operator and pointwise continuous of Fourier inversion series.
That is, how
$$\|\mathcal{C}f\|_2 \leq M\|f\|_2$$
where $\mathcal{C}$ is the Carleson Operator, would indicate
$$\lim_{N\rightarrow \infty}\int_{-N}^N \widehat{f}(\xi)e^{2\pi i x\xi} \,\mathrm{d}\xi=f(x)$$
Thank you in advance!
 A: The Carleson operator isn't bounded $L^2\rightarrow L^2$ or at least the proofs I've seen show the weaker bound
$$|\{x\mid Cf(x)>\lambda\}|\leq \lambda^{-2}||f||_{L^2}$$
which says that the Carleson operator is bounded $L^2\rightarrow L^{2,w}$, the latter pronounced weak $L^2$. Note that by Markov's inequality this would be implied by the $L^2$ bound.
Anyway, let $Lf(x)=\limsup_{N\rightarrow\infty}|f(x)-\int_{-N}^N e^{2\pi i \xi x}\hat{f}(\xi)\,d\xi|$ and fix $\epsilon>0$. We will show that $E_\epsilon=\{x\mid Lf(x)>\epsilon\}\rightarrow 0$ as $\epsilon\rightarrow 0$, which shows that $f(x)=\int_{-\infty}^\infty e^{2\pi i \xi x}\hat{f}(\xi)\,d\xi$ for almost every $x$. Let $g\in C^\infty_{c}(\mathbb{R})$ satisfy $||f-g||_{L^2}<\epsilon^2$ and notice that
$$E_\epsilon=\{x\mid \limsup_{N\rightarrow\infty}|f(x)-g(x)-\int_{-N}^Ne^{2\pi i \xi x}\hat{f}(\xi)-\hat{g}(\xi)\,d\xi|>\epsilon/2\}$$ where we have used that $g(x)=\int_{-\infty}^\infty e^{2\pi i \xi x}\hat{g}(\xi)\,d\xi$ for all $x$ because $g\in C^\infty_c(\mathbb{R})$ (this is just Fourier inversion). Then
$$|E_\epsilon|\leq |\{x\mid C(f-g)(x)>\epsilon/2\}|+ |\{x\mid |f-g|(x)>\epsilon/2\}|\leq 2(\epsilon/2)^{-2}\epsilon^4=2\epsilon^2$$
where we have used Chebyeshev's inequality to estimate the second term.
