## Partial sum operator of fourier series and Hilbert transform

$\mathbf{Problem:}$ Let $D_N=\sum\limits_{-N}^{N}e^{inx}$, the partial sum operator of fourier series can be represented by

$\displaystyle S_Nf(x)=\int_{-\pi}^\pi f(t)D_N(x-t)dt=f\ast D_N$

actually $\displaystyle D_N=\frac{\sin(N+\frac{1}{2})x}{\sin\frac{x}{2}}$

$\displaystyle S_Nf(x)=\int_{-\pi}^\pi f(t)\frac{\sin(N+\frac{1}{2})(x-t)}{\sin\frac{x-t}{2}}$

$\displaystyle =\int_{-\pi}^\pi f(t)\frac{\cos(N+\frac{1}{2})x\sin(N+\frac{1}{2})t-\sin(N+\frac{1}{2})x\cos(N+\frac{1}{2})t}{\sin\frac{x-t}{2}}$

$\displaystyle =P.V.\cos(N+\frac{1}{2})x\int_{-\pi}^\pi\frac{f(t)\sin(N+\frac{1}{2})t}{\sin\frac{x-t}{2}}+P.V.\sin(N+\frac{1}{2})x\int_{-\pi}^\pi\frac{f(t)\cos(N+\frac{1}{2})t}{\sin\frac{x-t}{2}}$

$\displaystyle =P.V.\frac{1}{\sin\frac{t}{2}}\ast f_N^1+P.V.\frac{1}{\sin\frac{t}{2}}\ast f_N^2$

Since $\displaystyle |\frac{1}{\sin\frac{t}{2}}-\frac{1}{\frac{t}{2}}|\leq ct$, for $t\in (-\pi,\pi)$, then $S_Nf$ behavior much like Hilbert transform

$\displaystyle Hf(x)=P.V.\int_{-\pi}^\pi\frac{f(x)}{x-t}$

$\mathbf{Thm:}$

(a) $||Hf||_p\leq c_p||f||_p$, $1

(b) $\{x\in(-\pi,\pi)||Hf(x)|>\lambda\}\leq \frac{C||f||_1}{\lambda}$

By the fact in this post, $S_Nf\to f$ in $L^p(-\pi,\pi)$

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$