# Why do the absolute values of functions in Hardy spaces tend to be non-oscillatory?

I'm trying to find a rigorous formulation of an impression/intuitive notion related to Hardy spaces. It seems to me that functions in Hardy spaces tend to have modulus functions which do not oscillate. This apparently generalizes a property of the exponential function. In particular: $$\exp(ikz) = \cos(kz) + i\sin(kz)$$ with $$\sin(kz)$$ the harmonic conjugate of $$\cos(kz)$$, and $$|\exp(ikz)|^2 = \cos^2(kz) + \sin^2(kz) = 1.$$

For many examples of functions $$F$$ in Hardy spaces, one has $$F(x) = u(x) + i v(x)$$ with $$u(x)^2 + v(x)^2$$ very apparently non-oscillatory. For certain specific examples, I am able to prove something. Bessel functions are a good example, I think.

The spherical Hankel function $$h_n(z)$$ of the first kind admits the representation
$$h_n(z) = \exp(iz) C_n \int_0^\infty \exp(-zt) P_n(1+it) dt$$ with $$C_n$$ a constant depending on $$n$$ and $$P_n$$ the Legendre function of the first kind of order $$n$$. This can can be verified for integers $$n$$ by direct substitution of the (finite) series expansions for each of these functions. This holds for noninteger orders as well, but proving it is a bit harder. It follows from the equivalent formula
$$h_n(z) = C_n \int_1^\infty \exp(izt) P_n(t) dt$$ that $$h_n$$ is in a Hardy space of functions analytic on the upper half of the complex plane (not $$H^p$$ for any $$p>0$$, though, since $$P_n(t)$$ belows up at infinity).

It can be shown by direct expansion of the relevant series that
$$|h_n(x)|^2 = 1+ \int_0^\infty \exp(-xt) \frac{d}{dt} P_n(1+t^2/2) dt.$$
The derivative of $$P_n(1+t^2/2)$$ is positive on $$(0,\infty)$$, so $$|h_n(x)|^2$$ is completely monotone on $$(0,\infty)$$. An equivalent formula is $$|h_n(x)|^2 = z \int_0^\infty \exp(-xt) P_n(1+t^2/2) dt.$$

Obviously, it is too much to ask for $$|F(z)|^2$$ to be completely monotone for all $$F$$ in a Hardy space. But is there a general principal here? It is not enough for $$F$$ to be in a Hardy space because there are obvious examples where $$|F|$$ is highly oscillatory. For example, take $$F(z) = \int_0^\infty \exp(izt) \phi(t) dt$$ with $$\phi(t)$$ a smooth function which is equal to $$1$$ on $$[1,100000]$$. The function $$|F(x)|^2$$ will oscillate rapidly on $$(0,\infty)$$.

I have the strong impression that this must related to a well-known, standard result, but I don't know where to look for it.

A function $$f\ge 0$$ on the circle is the absolute value $$f=|F|$$ of an $$F\in H^p$$, $$F\not\equiv 0$$, if and only if $$f\in L^p$$ and $$\log f\in L^1$$, and obviously you can make such functions as oscillatory as you desire them to be, so there's no general principle at work here.