Hardy space inclusion in the right-half plane

Question

I'm looking for an example of a function $u \in H_2$ such that $u \notin H_\infty$, where $H_p$ is the Hardy space on the right-half plane. Since this notation is perhaps not standard, here is a complete definition.

Let $\mathbb{C}^+ := \{s\in \mathbb{C} \mid \text{Re}(s) > 0 \}$ and $\bar{\mathbb{C}}^+ := \{s\in \mathbb{C} \mid \text{Re}(s) \geq 0 \}$ denotes the open and closed right-half plane, respectively.

Let $u:\bar{\mathbb{C}}^+\to \mathbb{C}$ be a function. Consider the following four conditions:

1. $u$ is analytic in the open right-half plane $\mathbb{C}^+$.
2. For almost all $\omega\in\mathbb{R}$, we have $\displaystyle \lim_{\sigma \to 0^+} u(\sigma+i\omega) = u(i\omega)$
3. $\displaystyle \sup_{\sigma \geq 0} \int_{-\infty}^\infty |u(\sigma+j\omega)|^2\,\mathrm{d}\omega < \infty$
4. $\displaystyle \text{ess}\sup_{s \in \bar{\mathbb{C}}^+} |u(s)| < \infty\;$ (this is the essential supremum, so ignoring any sets of measure zero)

We say that $u\in H_2$ if conditions 1,2,3 hold. We say that $u\in H_\infty$ if conditions 1,2,4 hold. Note that when conditions 1 and 2 hold, the suprema in conditions 3 and 4 will occur on the imaginary axis.

Answer 1

Here is a suggestion (but I have not worked through the details). What follows is given for the upper-half plane (UHP for short) but of course a trivial rotation will convert the example to one on the right-half plane as you requested.

There is a weighted composition operator giving an isometry of Hilbert spaces from $H^2$ of the UHP to $H^2$ of the open unit disc. One can find an explicit formula on the Wikipedia page for Hardy spaces. If $M: H^2({\rm UHP})\to H^2({\bf D})$ is the particular isometry described at that link, then inspecting the formula there shows that for each $f\in (H^2\cap H^\infty)({\rm UHP})$, the function $M(f)$ belongs to $H^2({\bf D})$ and is essentially bounded as $z\to -1$.

It therefore suffices to pick any $g\in H^2({\bf D})$ which is not essentially bounded as $z\to -1$, and then $M^{-1}(g)$ will be a function in $H^2({\rm UHP})$ that is not essentially bounded. One such $g$ is given by $$ \displaystyle g(z)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}z^n=\log(1+z). $$

Answer 2

Function $$\log\frac{z+1}{z-1},$$ where $\log$ is the principal branch ($\log(1)=0$), is evidently in $H^2$ but not in $H^\infty$ in the upper half-plane.