Compactly support function with compactly supported Hilbert transform.

Question

I was reading a bit on the Hilbert transform. I was wondering if it is possible to have a compactly supported $L_p \ni f : \mathbb{R} \to \mathbb{R}$ such that its Hilbert transform is compactly supported as well.

Answer 1

No, apart from the function constantly equal to $0$, there is no such function.

Note that as $f$ is compactly supported and $L^p$, it is in fact $L^1$ (apply Hölder to $f$ times the characteristic function of its support). By a similar argument, also $x\mapsto x^kf(x)$ is $L^1$ for all $k\geq 0$, and so is $x\mapsto\frac{f(x)}{y-x}$ for $y$ large enough in absolute value so that it is outside the support of $f$. One can then infer that $$ Hf(y)=\int_\mathbb{R}\frac{f(x)}{y-x}dx $$ for almost every $y$ large enough in absolute value. But then note that $$ y^n\int_\mathbb{R}\frac{f(x)}{y-x}dx-\sum_{k=0}^{n-1}y^{n-1-k}\int_\mathbb{R} x^kf(x)dx= \int_\mathbb{R}\frac{x^nf(x)}{y-x}dx $$ for such $y$, which implies that $$ \lim_{|y|\to+\infty} y^n\int_\mathbb{R}\frac{f(x)}{y-x}dx-\sum_{k=0}^{n-1}y^{n-1-k}\int_\mathbb{R} x^kf(x)dx=0. $$ But then if we suppose that $Hf$ is of compact support, then $y\mapsto\int_\mathbb{R}\frac{f(x)}{y-x}dx$ is too, and hence by the above $$ \lim_{|y|\to+\infty} \sum_{k=0}^{n-1}y^{n-1-k}\int_\mathbb{R} x^kf(x)dx=0. $$ This implies that $$ \int_\mathbb{R}p(x)f(x)dx=0 $$ for all polynomials $p\in\mathbb{R}[x]$, which by density of the polynomials implies $f=0$.