# 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

$$\lim_{N\rightarrow \infty}\int_{-N}^N \widehat{f}(\xi)e^{2\pi i x\xi} \,\mathrm{d}\xi=f(x)$$

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.