# A domination property for the Hardy space $H^1$

In the theory of Hardy spaces of the unit disc, a fact that is implicitely used quite often is that if $$f\in H^p, 1, then there exists a function $$F\in H^p$$ such that $$|f(z)| \leq |F(z)|, \,\, \forall z \in \mathbb{D}$$, $$Re F \geq 0$$ and $$\Vert F \Vert_p \leq c_p \Vert f \Vert_p$$.

To see why this is the case, given $$f\in H^p$$ define $$\begin{equation*} F(z) = \int_{\mathbb{T}} |f(\zeta)| \frac{1+\overline{\zeta}z}{1-\overline{\zeta}z} |d\zeta|. \end{equation*}$$

This is sometimes called Herglotz transform of $$|f|$$, but the point is that is a bounded linear operator from $$L^p(\mathbb{T})$$ into $$H^p$$, as a corollary of the M. Riesz Theorem. Hence $$F$$ defined like this has the required properties.

I was wondering if the existence of such an $$F$$ could be also true in the case $$p=1$$. Although the construction should be completely different because of the Failure of the M. Riesz Theorem for $$p=1$$.

## 1 Answer

In general it is impossible. We can take the square root, map the circle conformally to the half-plane, and arrive at the following problem: given any nonnegative $$f\in L^2(\mu)$$ with finite logarithmic integral, can we find $$g\in H^2(\mu)$$ with $$\Re g\ge f$$ and $$|\Im g|\le \Re g$$ where $$d\mu(x)=\frac{dx}{1+x^2}$$? Now, if that is possible for all $$f$$, it is also possible with the $$H^2(\mu)$$ norm of $$g$$ bounded by $$C\|f\|_{L^2(\mu)}$$.

Let's do the usual blow-up now taking some $$f\in L^2(dx)$$ and considering $$f(nx)$$ instead of $$f$$. Then, when we scale back, we'll get majorants $$g_n\in H^2(\mu_n)$$ such that $$\|g_n\|_{H^2(\mu_n)}\le C\|f\|_{L^2(\mu_n)}\le C\|f\|_{L^2(dx)}$$ with $$d\mu_n(x)=\frac{dx}{1+(x/n)^2}$$. We can pass to a subsequence and assume that $$g_n$$ converge to some $$g$$ weakly in $$L^2(dx)$$ on every subinterval of $$\mathbb R$$. Then we shall have $$g\in H^2(dx)$$ and we still have $$\Re g\ge f, |\Im g|\le \Re g$$(the vanishing of the Fourier transform on the negative semi-axis and the comparisons with non-negative functions can be tested by integrating against appropriately chosen $$L^2(dx)$$ functions, and the norm can only drop).

But for $$H^2(dx)$$ functions we have $$\int_{\mathbb R}|\Re g|^2dx=\int_{\mathbb R}|\Im g|^2$$, so we are forced to have $$|\Im g|=\Re g$$ almost everywhere on the line. But then $$g^2\in H^1(dx)$$ and $$\Re[g^2]=0$$ on $$\mathbb R$$, which is impossible.

Thus, a sufficiently strong bump at one point will give you a counterexample. To figure out exactly how strong is "sufficiently strong", one needs to make all that weak limit nonsense quantitative, which I leave to someone else :-)

• Thanks a lot, this is very nice ! Doing the calculations I don't see whe $\Re g$ is positive, I get instead $| \Im g| \leq |\Re g |$ and $\sqrt{2} |\Re g| \geq f$. Probably I am missing something. – an_ordinary_mathematician Sep 19 '20 at 9:12
• @an_ordinary_mathematician The square root of a function with non-negative real part has non-negative real part, doesn't it? $|\Im g|\le|\Re g|$ is insufficient because it isn't preserved under weak limits. $\sqrt 2$ is there, of course, but it is just absorbed into $C$. – fedja Sep 19 '20 at 14:16
• So that, combined with the observation about the Herglotz transform above is a quite roundabout proof that the Hilbert transform is unbounded on $L^1(\mathbb{T})$ ! – an_ordinary_mathematician Sep 19 '20 at 14:28