Asymptotic behaviour of Hilbert transform

Question

Let $f$ be a bounded function on $\mathbb{R}$ with compact support include in $[-K,K]$. Show that $$ H(f)(x)=\frac{a}{\pi x}+O(\frac{1}{x^2})$$ where $a=\int f(t)dt$ and $H$ denote the Hilbert transform, $H(f)(x)= \lim _{\epsilon \to 0}\pi ^{-1}\int _{|t|>\epsilon}f(x-t)\frac{dt}{t}$ (in the $L^2$ norm).

Answer 1

Technicality of the sort often passed over in silence: If a sequence converges in $L^2$ norm to one thing and pointwise to another the two limits are the same. Hence for $x>K$ we have simply $$Hf(x)=\frac1\pi\int_{x-K}^{x+K}f(x-t)\frac{dt}t.$$Hence$$Hf(x)=\frac1{\pi x}\int_{x-K}^{x+K}f(t-x)\,dt+\frac1\pi\int_{x-K}^{x+K}f(t-x)\left(\frac1t-\frac1x\right)\,dx$$The first term is exactly $a/\pi x$, while the absolute value of the second term is no larger than $$\frac1\pi\int_{x-K}^{x+K}|f(t-x)|\left|\frac{t-x}{tx}\right|\,dt.$$ Now $t\in[x-K,x+K]$ implies that $|t-x|\le K$.

Suppose now that in fact $x>2K$. Then for any $t\in[x-K,x+K]$ we have $|t|\ge|x|/2$ (in fact $t\ge x-K>x-x/2=x/2$). So the absolute value of the second term is no larger than $c/x^2$.

In fact all this works if just $f\in L^2(\mathbb R)$ and $f$ vanishes outside $[-K,K]$; note that this implies $\int|f|<\infty$.