Decomposition conjecture for normalized gauge-norm Hardy spaces

A weighted-gauge counterexample to the residue-class decomposition

Statement

Let $m$ be normalized Lebesgue measure on the unit circle $\mathbb T$. For a continuous, normalized, $\lVert\cdot\rVert _1$-dominating gauge norm $\alpha$, let $H^\alpha$ be the $\alpha$-closure of $H^\infty$, and let

$$M_\alpha(z^n)=[H^\infty(z^n)]_\alpha.$$

Conjecture 4.4 in the source asks whether

$$H^\alpha=M_\alpha(z^n)\oplus zM_\alpha(z^n)\oplus\cdots \oplus z^{n-1}M_\alpha(z^n) \tag{1}$$

for every such norm. We disprove (1) already for $n=2$.

The gauge norm

Put

$$a=b=\frac23, \qquad c_b=\left(\int_{\mathbb T}|1-\zeta|^{-b}\,dm(\zeta)\right)^{-1}, \qquad w(\zeta)=c_b|1-\zeta|^{-b}.$$

The integral defining $c_b$ is finite because $b<1$, and $\int_{\mathbb T}w\,dm=1$. For $h\in L^\infty(\mathbb T)$, define

$$\alpha(h)=\max\left\{ \int_{\mathbb T}|h|\,dm, \int_{\mathbb T}|h|w\,dm \right\}. \tag{2}$$

This is the maximum of two norms. Moreover,

• $\alpha(1)=1$;
• $\alpha(h)=\alpha(|h|)$;
• $\alpha(h)\geq\lVert h\rVert _1$; and
• if $m(A)\to0$, then $\alpha(\mathbf1_A)=\max\{m(A),\int_Aw\,dm\}\to0$, by absolute continuity of the integral of $w\in L^1$.

Thus (2) is a continuous $\lVert\cdot\rVert _1$-dominating normalized gauge norm. Its extension to measurable functions is the same maximum of the two displayed integrals. The norm is deliberately not rotationally symmetric: the weight has its singularity at $1$.

A function in $H^\alpha$

On the disk, take the analytic branch

$$f(z)=(1+z)^{-a}. \tag{3}$$

Its boundary modulus has an integrable singularity of order $a<1$ at $-1$. The weight $w$ is bounded near $-1$, while $f$ is bounded near the only singularity $1$ of $w$. Consequently

$$\int_{\mathbb T}|f|\,dm<\infty, \qquad \int_{\mathbb T}|f|w\,dm<\infty. \tag{4}$$

For completeness, membership in $H^\alpha$ follows directly from bounded analytic approximants. Let $f_r(z)=(1+rz)^{-a}$, $0<r<1$. For $r\geq1/2$ and $\zeta=e^{i\theta}$,

$$|1+r e^{i\theta}|^2 =(1-r)^2+4r\cos^2(\theta/2),$$

so both the unweighted and weighted differences $f_r-f$ are dominated by integrable multiples of

$$|\cos(\theta/2)|^{-a} \quad\text{and}\quad |\cos(\theta/2)|^{-a}|\sin(\theta/2)|^{-b},$$

respectively. These are integrable because $a,b<1$. Dominated convergence therefore gives $\alpha(f_r-f)\to0$. Since every $f_r\in H^\infty$, equation (3) indeed defines an element of $H^\alpha$.

Its even part is not in $H^\alpha$

The even Fourier-residue component of $f$ is

$$E_0f(z)=\frac{f(z)+f(-z)}2 =\frac12\left((1+z)^{-a}+(1-z)^{-a}\right). \tag{5}$$

As $\zeta\to1$, the first summand in (5) stays bounded and the second has modulus $|1-\zeta|^{-a}\to\infty$. Hence, on a sufficiently small punctured arc about $1$,

$$|E_0f(\zeta)|\geq\frac14|1-\zeta|^{-a}.$$

It follows that

$$\int_{\mathbb T}|E_0f|w\,dm \geq C\int_0^\delta \theta^{-(a+b)}\,d\theta =\infty, \tag{6}$$

because $a+b=4/3>1$. Thus $E_0f\notin L^\alpha$, and in particular $E_0f\notin H^\alpha$.

Contradiction to the proposed decomposition

Every $g\in M_\alpha(z^2)$ is even. Indeed, choose $g_j\in H^\infty(z^2)$ with $\alpha(g_j-g)\to0$. Since $\alpha\geq\lVert\cdot\rVert _1$, this convergence also holds in $L^1$. Every odd Fourier coefficient of every $g_j$ is zero, and Fourier coefficients are continuous on $L^1$; hence every odd coefficient of $g$ is zero. Therefore $g(-z)=g(z)$ in the disk.

If (1) held for $n=2$, the function (3) would have a representation

$$f=g_0+zg_1, \qquad g_0,g_1\in M_\alpha(z^2).$$

Both $g_0$ and $g_1$ are even, so replacing $z$ by $-z$ and adding gives

$$g_0(z)=\frac{f(z)+f(-z)}2=E_0f(z).$$

But $g_0\in M_\alpha(z^2)\subset H^\alpha\subset L^\alpha$, whereas (6) shows that $E_0f\notin L^\alpha$. This contradiction disproves the conjecture.