# Scalar product on Hardy spaces on the strip $\mathcal{S} = \mathbb{R} + i(-1,1)$

I want to learn more about the Hardy space $$H^2(\mathcal{S})$$ on the complex strip $$\mathcal{S} = \{ x+iy \in \mathbb{C} ~|~ x,y \in \mathbb{R},~ |y|<1 \}$$. In particular, I am interested in it as a reproducing kernel Hilbert space.

I am following this publication by Bakan and Kaijser.

Usually, one defines a Hardy space on the disc as follows: $$H^2(\mathbb{D}) = \left\{ f \text{ holomorphic } ~\bigg|~ ||f||_{H^2(\mathbb{D})}^2 = \sup_{0 \leq r < 1} \frac{1}{2\pi} \int_{-\pi}^\pi |f(r e^{i\theta})|^2 d\theta < \infty \right\}.$$ The strip is isomorphic to the disc via the following isomorphism: $$\phi : \mathcal{S} \to \mathbb{D},~ \phi(z) = \tanh\frac{\pi}{4} z ~,~~~~ \phi^{-1} : \mathbb{D} \to \mathcal{S},~ \phi^{-1}(w) = \frac{4}{\pi} \text{artanh}(w)$$ One can define the a Hardy space on the strip as $$H^2(\mathcal{S}) = \{ f \text{ holomorphic } ~\Big|~ ||f||_{H^2(\mathcal{S})}^2 := ||f \circ \phi^{-1}||_{H(\mathbb{D})}^2 < \infty \}$$. But this definition is not very elegant because one has to integrate $$f(4/\pi~ \text{artanh}(w))$$.
However, $$H^2(\mathcal{S})$$ and its norm are equal to the following space and norm: $$H'^2(\mathcal{S}) = \left\{ f \text{ holomorphic } ~\bigg|~ ||f||_{H'^2(\mathcal{S})}^2 = \sup_{0\leq y < 1} \int_{\mathbb{R}} \frac{|f(t - iy)|^2 + |f(t + iy)|^2}{4 (\cosh \frac{\pi}{2} t + \cos\frac{\pi}{2} y)} dt < \infty \right\}$$ But these are not Hilbert spaces, they do are only Banach and do not have a scalar product. To obtain a Hilbert space from $$H^2(\mathbb{D})$$, one has to replace $$\sup_{0\leq r < 1} \to \lim_{r \to 1}$$ and define the scalar product: $$\langle f, g \rangle_{H^2(\mathbb{D})} = \lim_{r \to 1} \frac{1}{2\pi} \int_{-\pi}^\pi \overline{f(r e^{i\theta})} g(re^{i\theta}) d\theta.$$ My Question: What is the scalar product on $$H'^2(\mathcal{S})$$?

I suspect that one has to replacing $$\sup_{0\leq y < 1} \to \lim_{y\to 1}$$ in the norm and that the scalar product is given by replacing $$|f(t \mp iy)|^2$$ with $$\overline{f(t \mp iy)} g(t \mp iy)$$. But I do not know how to prove it because I do not understand the equality of $$H^2(\mathcal{S})$$ and $$H'^2(\mathcal{S})$$ (which is proven in the publication).

I am also grateful for any references on the topic.