The Hardy spaces $H^p$ of holomorphic functions on the unit disk are Banach spaces.

Question: Are they also Banach lattices?

If yes, why is it less common to consider the Hardy spaces as Banach lattices than it is to consider $L^p$ as Banach lattices?

Motivation for question: the function spaces $L^p$ are often supplied with an order (pointwise order) to become Banach lattices. We can then derive properties like the Fatou property or order continuity of the norm. However, I cannot find any place where the Hardy space is given such an order to become a Banach lattice.

  • $\begingroup$ It depends on your definition of Banach lattice.# $\endgroup$
    – daw
    Oct 18, 2017 at 18:07

1 Answer 1


the function spaces $L^p$ are often supplied with an order

More precisely, the real function spaces $L^p$ have natural order $f\le g$ pointwise. There is no natural order to put on Banach spaces of complex-valued functions. Indeed, if there was one, it would induce an order on complex numbers, which can be identified with constant functions. And there is no useful order on complex numbers that respects the linear structure (one can think of ordering by magnitude, $|a|\le |b|$, but that fails translation invariance).

There are things called "complex Riesz spaces", which are defined as the complexification of a Riesz space. But the Hardy space of holomorphic functions is not the complexification of a real function space; indeed it doesn't contain any nonconstant real-valued functions.

Also, even if we were talking about real Hardy spaces, there would still be no lattice structure: the membership of a function $f$ in a real Hardy space may depend on some cancellation between its positive and negative parts, which means $|f|$ is not necessarily in the same space.


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