Is harmonic Hardy space $h^p (D)$ complete for $0<p<1$?

Question

Let $D$ be the unit disk and $0<p<1$. Let $h^p(D)$ be the harmonic Hardy space on the unit disk (i.e. $u\in h^p(D)$ if and only if $u$ is harmonic on the disk and $\sup_{0<r<1}\int_{-\pi}^\pi|u(re^{it})|^p\frac{\operatorname{d}t}{2\pi}<+\infty$), equipped with its natural metric, i.e. $$d(u,v):=\sup_{0<r<1}\int_{-\pi}^\pi|u(re^{it})-v(re^{it})|^p\frac{\operatorname{d}t}{2\pi}$$

I've the following question: is $(h^p(D),d)$ complete? The failure of Young's convolution inequality prevents me to prove the result following the same path that works for harmonic Hardy spaces with $p\ge1$ (i.e.: using the Poisson reproducing formula for harmonic functions and then use Young's convolution inequality), to get uniform convergence on compact subsets of $D$ (does it even hold true here? Or at least the point evaluation functionals are continuous?), so I don't know if the limit function defined circle by circle (using the fact that $L^p$ is a complete metric space) is actually continuous, even less harmonic.

Answer 1

If I didn't make any silly mistake $h_p$ indeed is complete. We will first prove that for every $z\in D$ $u\to u(z)$ is continuous. To do so we will need the following estimate (see Fefferman and Stein "$H^p$ spaces of several variables", 9; Lemma 2, page 172):

Theorem 1. There exists constant $c_p > 0$ such that for every harmonic function $u$ in $D$ we have \begin{equation}\label{H-L} |u(0)|^p \le \frac{c_p}{|D|}\int_D |u(z)|^pdz. \end{equation}

(Note that we are integrating over the whole disk, not over spheres!)

From this theorem it is easy to deduce that $z\to u(z)$ is indeed continuous and we also have uniform control of the constant on the compact subsets of $D$ (consider a small ball centred at $z$ and estimate rhs in Theorem 1 by constant times $h_p$-norm of $u$). Now we are ready to prove the desired result:

Theorem 2. $h_p$ space is complete.

Proof

Let $u_n\in h_p$ be a Cauchy sequence. From above reasoning we can easily see that for every $z\in D$ $u_n(z)$ is a Cauchy sequence and moreover convergence is uniform on compact subsets of $D$. From this it follows that we have $u(z) := \lim_{n\to \infty} u_n(z)$ is a harmonic function on $D$. It remains to prove that $u_n\to u$ in $h_p$ (note that $u\in h_p$ obviously by Fatou lemma).

For $n \ge N$ we have that $d(u_n, u_N) \le \varepsilon$. Then $u_n - u_N$ pointwise converges to $u - u_N$ and again by Fatou $d(u, u_N) \le \liminf d(u_n, u_N) \le \varepsilon$. Thus since $\varepsilon$ is arbitrary $u_N \to u$ in $h_p$.

Answer 2

Not an answer to the question, but a discussion of related issues arising in the comments.

The closest I have to an answer is that I'd be very surprised if $h^p$ was not complete, but I have no idea how to prove it's complete.

General Comment: Reading Duren is a Good Thing. If you like Duren you'll love Garnett Bounded Analytic Functions. But you should be aware that these days "complex methods" are deprecated, because they don't extend to other contexts where people want to do analogous things. And from that point of view Blaschke products are positively evil, being totally and irredeemably "complex". You've seen an example of this already - Blashcke products are obviously useless for the current question. Otoh the argument by subharmonicity could have worked, it just doesn't.

What argument?

The completeness of $H^p$ for $0<p<1$ is immediate from the fact that $f$ holomorphic implies that $|f|^p$ is subharmonic.

If $K\subset\Bbb D$ is compact there exists $c_K$ such that if $u$ is subharmonic in $\Bbb D$ and continuous on $\overline{\Bbb D}$ then $$\sup_{z\in K}u(z)\le c_K\frac1{2\pi}\int_0^{2\pi}u(e^{it})\,dt.$$

Say $(f_n)$ is Cauchy in $H^p$. Since $|f_n-f_m|^p$ is subharmonic it follows that $|f_n-f_m|^p\to0$ uniformly on compact sets, hence $f_n\to f$ uniformly on compact sets. And this implies by Fatou's Lemma that $||f_n-f||_p\to0$, where $||.||_p$ is the sup over $r$ of the $L^p$ norm on circles of radius $r$ about the origin. (If $||f_n-f_m||_p<\epsilon$ for all $n,m>N$ and $n>N$ then FL says $$\int_0^{2\pi}|f(re^{it})-f_n(re^{it})|^p\,dt\le\liminf_{m\to\infty}\int_0^{2\pi}|f_m-f_n|^p\le\epsilon$$for every $r\in(0,1)$.)

So the completeness of $H^p$ really is immediate from the fact that $|f|^p$ is subharmonic. But if $u$ is harmonic then $|u|^p$ need not be subharmonic for $0<p<1$.

(For example: Say $u=P[\mu]$, where $\mu$ is a measure on the circle. If $|u|^{1/2}$ is subharmonic then $\mu$ is absolutely continuous.)

A proof of that last statement leads to stuff about the F&M Riesz theorem. We need to recall a fact about $1<p\le\infty$, in order to be able to talk about what goes wrong for $p=1$:

Proposition 0. If $1<p\le\infty$ and $u\in h^p$ then $u=P[f]$ for some $f\in L^p(\Bbb T)$.

Proof: For $0<r<1$ define $u_r:\Bbb D\to\Bbb C$ and $f_r:\Bbb T\to\Bbb C$ by $$u_r(z)=u(rz), f_r(e^{it})=u(re^{it}).$$Then $$u_r=P[f_r].$$Since $||f_r||_p$ is bounded and $p>1$ there is a sequence $r_j\to 1$ with $f_{f_j}\to f$ weakly in $L^p$. (Not that the sequential part really matters, but if $X$ is a separable Banach space then the unit ball of $X^*$ is sequentially compact in the weak* topology.) Hence $P[f_{r_j}]\to P[f]$ pointwise, hence $u=P[f]$.

Of course if $\mu$ is a singular measure on $\Bbb T$ then $u=P[\mu]$ shows that Proposition 0 fails for $p=1$. And it's clear what goes wrong with the proof: The unit ball of $L^1$ is not compact in any relevant topology. For $p=1$ the proposition becomes

Proposition 0.1. If $u\in h^1$ there exists a (regular complex) measure $\mu$ on $\Bbb T$ such that $u=P[\mu]$.

The reason I mention all that is to show how surprising the following is:

F&M Riesz Theorem. If $f\in H^1$ then there exists $F\in L^1(\Bbb T)$ with $f=P[F]$.

(If that's not the statement you think of as the F&M Riesz theorem it's in any case the hard part; various versions of the theorem are immediate from the above. I'm going to call it F&M Riesz below.)

In fact F&M Riesz follows easily from the fact that $|f|^{1/2}$ is subharmonic. This is actually analogous to the proof you know, by Blaschke products: Let's define another space - say $sh^p$ is the space of subharmonic functions with bounded $p$-th means on circles. The proof of F&M Riesz in Duren uses the fact that $f$ is the product of two $H^2$ functions. The proof below uses the analogous but much more trivial fact that $|f|$ is the product of two functions in $sh^2$ (it's obvious that if $f\in H^1$ then $|f|^{1/2}\in sh^2$.)

For the proof:

Lemma. If $1<p\le\infty$ and $u\in sh^p$ then there exists $f\in L^p(\Bbb T)$ with $u\le P[f]$.

Proof: Same as the proof of Proposition 0, with $u_r\le P[f_r]$ in place of $u_r=P[f_r]$.

Oops In that proof we need to assume that $u$ is continuous, to get $u_r\to u$ pointwise. That case is sufficient for our application below. In fact the Lemma holds for a general (usc) subharmonic function, by essentially the same argument - use the solution to the Dirichlet problem on the disk $D(0,r)$ instead of using those dilations to transfer to $\Bbb D$.

Now if $\phi:\Bbb D\to\Bbb C$ let $M\phi$ be the radial maximal function: $$M\phi(e^{it})=\sup_{0<r<1}|\phi(re^{it})|.$$

Corollary If $1<p\le\infty$ and $u\in sh^p$ then $Mu\in L^p(\Bbb T)$.

Now, just to emphasize that we're really using nothing but the fact that $|f|^{1/2}$ is subharmonic, we point out that F&M Riesz is an immediate corollary of this:

Theorem. Suppose $\mu$ is a (regular complex) measure on $\Bbb T$ and $u=P[\mu]$. If $|u|^{1/2}$ is subharmonic then $\mu$ iis absolutely continuous.

Proof: Since $u\in h^1$ it is clear that $|u|^{1/2}\in sh^2$. So $$M|u|^{1/2}\in L^2,$$hence $$Mu=(M|u|^{1/2})^2\in L^1(\Bbb T).$$

And now we're done: We know that there exists $F\in L^1$, namely the absolutely continuous part of $\mu$, such that $u(re^{it})\to F(e^{it})$ almost everywhere. The fact that $Mu\in L^1$ gives precisely the domination we need to conclude by DCT that $$\int_0^{2\pi}|u(re^{it})-F(e^{it})|\,dt\to0.$$