In the theory of Hardy spaces of the unit disc, a fact that is implicitely used quite often is that if $f\in H^p, 1<p<\infty$, 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 1


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 :-)

  • $\begingroup$ 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. $\endgroup$ Sep 19, 2020 at 9:12
  • $\begingroup$ @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 $C$. $\endgroup$
    – fedja
    Sep 19, 2020 at 14:16
  • $\begingroup$ So that, combined with the observation about the Herglotz transform above is a quite roundabout proof that the Hilbert transform is unbounded on $L^1(\mathbb{T})$ ! $\endgroup$ Sep 19, 2020 at 14:28

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