From this source, on page 36 (bottom) there is a conjecture stated and it was said that the case of $d = 1$ corresponds to Carleson's theorem. A picture included here :
But when I look at wiki page on Carleson's theorem and at this expression of Carleson operator 
there is no $\frac{1}{y}$ inside the integral. So I'd like to know that how does the case of $d=1$ corresponds to Carleson's theorem.
Let's first prove that the Carleson operator is bounded on $L^p$ if Conjecture 9.11 holds for $d=1$. $$ \int_{-N}^{N} \hat f(y) e^{ixy} \, dy $$ is the inner product of $y \mapsto e^{i x y} \hat f(y)$ and $y \mapsto I_{|y|< N}$. By Parseval's Theorem this is equal to the inner product of the inverse Fourier transforms of these two functions, namely $z \mapsto f(x-z)$ and $z \mapsto \frac{\sin(N z )}{z}$. This can be shown to be bounded if the following two quantities can be shown bounded: $$ \int f(x-z) \frac{e^{\pm iN z }}{z} \, dz .$$ And $z \mapsto \pm Nz$ is a polynomial of degree 1.
Now suppose that the Carleson operator is bounded on $L^p$. If $f \in L^p$, then so are $(I\pm iH)f$, where $H$ is the Hilbert transform. The supports of their Fourier transforms are on $[0,\infty)$ and $(-\infty,0]$ respectively. Hence we have that both $$ \sup_N \left|\int_{0}^{N} \hat f(y) e^{ixy} \, dy \right| $$ and $$ \sup_N \left|\int_{-N}^{0} \hat f(y) e^{ixy} \, dy \right| $$ are in $L^p$. Now using Parsevl's identity, we obtain $$ \sup_N \left|\int f(x-z) \frac{e^{\pm iN z }-1}{z} \, dz \right| $$ are in $L^p$. Since $x \mapsto \int \frac{f(x-z)}{z} \, dz$ is the Hilbert transform of $f$ (perhaps times a constant), this is bounded in $L^p$. From here, Conjecture 9.11 for $d=1$ follows easily.