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I've been reading up on Hardy spaces and (sub)harmonic functions over the open unit disk $\mathbb{D}\subset\mathbb{C}$, and I've found myself working with atypical objects in mostly-typical situations. Whenever I look up Hardy spaces (in texts, or on the internet), they're virtually always referred to as “spaces of holomorphic/analytic functions”. However, the space-defining norm works for non-holomorphic functions, as well. The prototypical case I'm presently dealing with is as follows: let $f:\mathbb{D}\rightarrow\mathbb{C}$ be holomorphic, and let: $$g\left(z\right)\overset{\textrm{def}}{=}\left(1-\left|z\right|\right)^{1/2}f\left(z\right)$$ Then, the $H^{2}\left(\mathbb{D}\right)$-norm of $g\left(z\right)$ is given by: $$\left\Vert g\right\Vert _{H^{2}\left(\mathbb{D}\right)}=\sqrt{\sup_{r\in\left(0,1\right)}\int_{0}^{1}\left|g\left(re^{2\pi it}\right)\right|^{2}dt}=\sqrt{\sup_{r\in\left(0,1\right)}\left(1-r\right)\int_{0}^{1}\left|f\left(re^{2\pi it}\right)\right|^{2}dt}$$ Letting $f$ admit the power series expansion: $$f\left(z\right)=\sum_{n=0}^{\infty}a_{n}z^{n}$$ the Parseval-Gutzmer Theorem then allows us to write: $$\left\Vert g\right\Vert _{H^{2}\left(\mathbb{D}\right)}=\sqrt{\sup_{r\in\left(0,1\right)}\left(1-r\right)\sum_{n=0}^{\infty}\left|a_{n}\right|^{2}r^{2n}}$$ If, for example, the $a_{n}$s satisfy: $$\lim_{N\rightarrow\infty}\frac{1}{N}\sum_{n=0}^{N-1}\left|a_{n}\right|^{2}=\alpha\in\left(0,\infty\right)$$ then: $$\lim_{r\uparrow1}\left(1-r\right)\sum_{n=0}^{\infty}\left|a_{n}\right|^{2}r^{2n}=\alpha$$ and hence, necessarily the supremum over $r\in\left(0,1\right)$ is finite, which then guarantees the same for $\left\Vert g\right\Vert _{H^{2}\left(\mathbb{D}\right)}$. Despite all this, however, $g\left(z\right)$ is most definitely not holomorphic on $\mathbb{D}$, even when $f$'s coefficients satisfy the above asymptotic conditions.

In light of this, I then ask: to what extent, if any, does the 'classical' Hardy space theory apply to these functions? More generally, if we create an “extended” version of $H^{2}\left(\mathbb{D}\right)$ by considering the space of all, say, continuous functions $g:\mathbb{D}\rightarrow\mathbb{C}$ with finite $H^{2}\left(\mathbb{D}\right)$-norm, do we end up with a Banach space, and—if so—does it have any noteworthy properties? References to any books or papers on this sort of thing would be much appreciated.

Additionally, supposing that $f$ satisfies the conditions in the preivous pragraphs, of particular interest to me is the behavior of: $$\int_{0}^{1}\left|g\left(re^{2\pi it}\right)\right|^{2}dt=\left(1-r\right)\int_{0}^{1}\left|f\left(re^{2\pi it}\right)\right|^{2}dt$$ as a real-valued function of $r\in\left[0,1\right)$. I know that, for any subharmonic function $v:\mathbb{D}\rightarrow\left[-\infty,\infty\right]$, the quantity: $$\int_{0}^{1}\left|v\left(re^{2\pi it}\right)\right|^{2}dt$$ is strictly increasing as a function of $r$, and from this, one can express notions of harmonic majorants, and the like. To that end, it is easy to show that $\left|g\left(z\right)\right|^{2}$ is subharmonic on the annulus $A=\left\{ z\in\mathbb{D}:\frac{1}{2}\leq\left|z\right|<1\right\}$. Consequently, the mean value inequality property: $$\left|g\left(z\right)\right|^{2}\leq\int_{0}^{1}\left|g\left(z+re^{2\pi it}\right)\right|^{2}dt$$ holds for all z in the interior of $A$ whenever $r$ is sufficiently small. What happens, though, if we integrate $\left|g\left(z\right)\right|^{2}$ along a circle centered at $0$ of radius $r\in\left(\frac{1}{2},1\right)$?

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    $\begingroup$ Would you mind editing your title? Hardy spaces on the disk have established definitions, and if you want to consider larger spaces then you really really really really should not talk about "atypical elements of Hardy spaces" $\endgroup$
    – Yemon Choi
    Commented Dec 20, 2020 at 0:48
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    $\begingroup$ In particular, "the space-defining norm" is not the characteristic feature of Hardy spaces; the holomorphic structure (and the ensuing boundary behaviour as $r\nearrow 1$) is the crucial factor. $\endgroup$
    – Yemon Choi
    Commented Dec 20, 2020 at 0:50
  • $\begingroup$ Thank you for this comment. I have changed the title, as per your suggestion. Your comment is in part exactly why I asked this question. Now I know that what I'm working with doesn't fall under the provenance of classical Hardy Space theory. $\endgroup$
    – MCS
    Commented Dec 20, 2020 at 1:25
  • $\begingroup$ Thanks. Upon re-reading, perhaps I was a little hasty with my second comment; there is a notion of "harmonic Hardy space" which is by definition a space of real-valued harmonic functions on the disk satisfying the growth condition that you refer to as the space-defining norm. I think what is going on here is that "(1-|z|) times holomorphic" is still much more special than just continuous $\endgroup$
    – Yemon Choi
    Commented Dec 20, 2020 at 1:40
  • $\begingroup$ I think you can tie this up with mainstream theory by viewing $g$ as a function $g: (0,1)\to L^2(S)$, and now of course $g(r)$ is the restriction of your $g$ to the circle $re^{i\varphi}$. So I think the usual notation for your space would be $L^{\infty}((0,1); L^2(S))$ (and yes, this is a Banach space). $\endgroup$ Commented Dec 20, 2020 at 23:49

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