Hilbert transform of Schwartz function is in $L^1$ iff $\int_{\mathbb R} f=0$

Question

Suppose $f:\mathbb R \to \mathbb R$ be a Schwartz function. Then the Hilbert transform $Hf\in L^1(\mathbb R)$ iff $\int_{\mathbb R} f(x) \mathrm dx=0$.

I could prove $Hf\in L^1(\mathbb R)$ implies $\int_{\mathbb R} f(x) \mathrm dx=0$ using the fact $\lim_{|x|\to \infty}xHf(x)=\frac{1}{\pi}\int_{\mathbb R}f(x)\mathrm dx$.

But I couldn't prove the converse, please help.

Context (from comments): I have seen Integrability of the Hilbert transform of a Schwartz function but I couldn't undestand how it is done there. I had seen that before posting the question

Does this answer your question? Integrability of the Hilbert transform of a Schwartz function — Calvin Khor, Mar 26 '21 at 09:50
No that arises more questions. I couldn't undestand how it is done there. I had seen that before posting the question. @CalvinKhor — Noob mathematician, Mar 26 '21 at 12:25
If you've seen it and didn't want other people to tell you about it, its a good idea to say so in the post — Calvin Khor, Mar 27 '21 at 02:44

reuns · Answer 1 · 2021-03-27T06:25:50.680

2

$Hf$ is the inverse Fourier transform of $\widehat{Hf}=-i\ sign(\omega) \hat{f}$.

Assume only that $\hat{f},\hat{f}'$ are $L^2$. Thus, $\widehat{Hf}$ is $L^2$ and $\hat{f}$ is continuous.

If $\hat{f}(0)=0$ then $$\widehat{Hf}'=-i\ sign(\omega)\hat{f}'$$ is $L^2$ so that $(1+|x|)Hf \in L^2$ and $Hf\in L^1$ (Cauchy-Schwarz).
If $\hat{f}(0)\ne 0$ then $\widehat{Hf}$ is not continuous whence $Hf$ is not $L^1$.

edited Mar 27 '21 at 06:25

answered Mar 27 '21 at 06:20

reuns

77,999

I didn't get why $(1+|x|)Hf\in L^2$ in the second point . @reuns – Noob mathematician Mar 27 '21 at 07:08
@Noobmathematician (i'm not reuns but) the previous line shows that $\widehat {x Hf } = C\partial_{\xi} \widehat{Hf} \in L^2$ – Calvin Khor Mar 27 '21 at 07:15
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Oh yes. My bad .. I forgot this good old identity . Thanks @CalvinKhor . – Noob mathematician Mar 27 '21 at 07:31
@reuns: What is $\hat{f}'$? – eggplant Oct 15 '22 at 20:55
@eggplant The derivative of the Fourier transform of $f$ – reuns Oct 15 '22 at 22:17

Calvin Khor · Accepted Answer · 2021-03-27T07:08:39.037

Mild elaboration of linked post: Let $A\lesssim B$ mean that $|A|\le C B$. For $x<-1$,

$$g(x):=\int_{-\infty}^xf(t)dt = \int_{-\infty}^x \langle t\rangle ^{-N} \langle t\rangle ^Nf(t) \ dt\ dx \lesssim_{N} \int_{-\infty}^x\langle t\rangle ^{-N} dt \lesssim_N \langle x\rangle^{-N+1} $$ The condition $\int_{-\infty}^\infty f(t)dt = 0$ lets you prove this also for $x>1$ (and of course there is no issue near $x=0$.) So $g$ is Schwartz. Then

$$Hf(x) =\int_{|t|<1} \frac{f(x-t)-f(x)}tdt + \int_{|t|>1} \frac{f(x-t)}t dt =: I_1+I_2,$$ with $$ I_2 =\int_{|t|>1} \frac{f(x-t)}t dt = -\int_{|t|>1} \frac{d}{dt}(g(x-t))\frac{dt}t=g(x-1)-g(x+1) -\int_{|t|>1} \frac{g(x-t)}{t^2} dt$$ so that $$ \|I_2\|_{L^1} \le 2\|g\|_{L^1} + \int_{|t|>1} \frac{\|g\|_{L^1}}{t^2} dt<\infty.$$ For $I_1$, let $I(x,t)$ denote whichever of $[x-t,x]$ or $[x,x-t]$ that is not empty, then $$ |I_1| \le \int_{|t|<1}\frac1t\int_{x-t}^x |f'(r)|dr \le \int_{|t|<1} \max_{r\in I(x,t)}|f'(r)| dt \lesssim \max_{r\in [x-1,x+1]}|f'(r)| $$ and $$f'(r) \lesssim_N \langle r\rangle^{-N}$$ giving $\|I_1\|_{L^1}<\infty$ too; thus $\|Hf\|_{L^1}<\infty$.

@eggplant hello; $\langle t \rangle = \sqrt{1+t^2}$ is the so-called "japanese bracket". — Calvin Khor, Oct 29 '22 at 06:01

Hilbert transform of Schwartz function is in $L^1$ iff $\int_{\mathbb R} f=0$

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