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I am studying singular integrals and I have understood that if we have an operator $T$ which is bounded in $L^2$, then we can extend this operator for functions in $L^{\infty}$ to BMO space. This is the case of the Hilbert transform.

$$Hf(x)=\frac{1}{\pi}\text{p.v}\int_{\mathbb{R}}\frac{f(y)}{x-y}dy$$

My question is, what is $H1?$ It is clear that the function $f(x)=1$ is bounded, so $H1$ would be a BMO function, but which function?

Maybe the key is to work with distributions instead of functions, or using the Fourier definition of the Hilbert tranform:

$$\widehat{(Hf)}(\xi)=-i\ \text{sgn}(\xi) \widehat{f}(\xi)$$

This is:

$$\widehat{(H1)}(\xi)=-i\ \text{sgn}(\xi) \delta(\xi)$$

but we know its inverse Fourier transform?

Thank you very much

pitàgores
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  • If $h$ is an odd function on $\Bbb R$ then $\int_{\Bbb R}h=\dots$ – David C. Ullrich Jun 18 '21 at 12:28
  • I see, then it is 0. Wow, how could I not see it? Thank you. And this Concordes with the Fourier definition? It would be in terms of distributions that $0=-I \ \text{sign}(\xi)\delta(\xi)$? Has this sense? – pitàgores Jun 18 '21 at 12:44
  • @micabua You can also observe that the supports of $\mathrm{sgn}$ and $\delta$ are disjoint. – Nicolas Jun 18 '21 at 12:46
  • @Nicolas The support of a distribution is by definition a closed set; in particular the support of $sgn$ is all of $\Bbb R$. Not just a technicality; this is needed to deduce that the product of two distributions with disjoint support is zero. – David C. Ullrich Jun 18 '21 at 12:52
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    @micabua the fact that sgn is discontinuous at the origin means one needs to be very careful about that. If $u$ is a distribution and $\psi$ is not smooth it's not clear in general what $\psi u$ even means... – David C. Ullrich Jun 18 '21 at 12:54
  • @DavidC.Ullrich Sometimes $\mathrm{sgn}(0):=0$. I do agree with your last comment anyway, we have to take care of wat happens at 0. – Nicolas Jun 18 '21 at 12:56
  • @Nicolas I don't see what $sgn(0)=0$ has to do with anything I said. It doesn't change the fact that the support is all of $\Bbb R$, for example... – David C. Ullrich Jun 18 '21 at 12:58
  • @DavidC.Ullrich For a (mere) function $f$, $f(x)=0\ \implies\ x\notin\mathrm{Supp\ }(f)$, so the definition for the sign function matters (for the question of the support). – Nicolas Jun 18 '21 at 13:02
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    @Nicolas Sorry, but that's simply not true. The support of the function $f$ is defined to be the closure of the set where $f\ne0$. (If you use the definition you gave it's not true that disjoint support implies the product vanishes!) – David C. Ullrich Jun 18 '21 at 13:04
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    @Nicolas For example: If $u=\delta'$ (the distribution derivative) and $\psi(x)=x$ then $\psi u=\delta\ne0$. So if the support were defined as you say than it's not true that disjoint supports imply the product is zero. – David C. Ullrich Jun 18 '21 at 13:13
  • @DavidC.Ullrich The common definition of the support for topological spaces is indeed with closure, but you can define it as I did for general cases. Of course, when we use distribution theory, topology matters and closure is indeed necessary for the disjoint support property to hold. My first comment was not an answer, rather something to quickly understand where things are interesting (at $0$); your first comment needs too some care as you do not deal with functions but distributions. – Nicolas Jun 18 '21 at 15:29

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