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<p<\infty

(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:}

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