Let \( \mathcal{H} \) and \( \mathbb{H} \) be two Hilbert spaces with,
\(
(f,g)_{\mathcal{H}} = \int_0^\infty f(x)\overline{g(x)}\,dx\\
(F,G)_{\mathbb{H}} = \int_{1/2 - i\infty}^{1/2+i\infty}|\Gamma(z)|^2F(1-z)\overline{G(1-z)}\,dz\\
\)
The function,
\(
\Gamma(1-z)F(z) = (f,x^{-\overline{z}})_{\mathcal{H}}\\
\Gamma(1-z)G(z) = (g,x^{-\overline{z}})_{\mathcal{H}}\\
\)
We can create an operator,
\(
f = \sum_{n=0}^\infty F(n+1) \frac{w^n}{n!}\\
Hf = \sum_{n=0}^\infty F^{\circ n+1}(1) \frac{w^n}{n!}\\
\)
Such that,
\(
(H^n f, x^{-\overline{z}})_{\mathcal{H}} = \Gamma(1-z)\uparrow^n F\\
\)
This is the Hilbert space interpretation of the last paper. We want to use this to find a function \( g_{s}^z(x) \) such that,
\(
(f,g_s^z) = \Gamma(1-z) \uparrow^s F\\
\)
Note that \( g \) is not really an adjoint; but it plays the part well enough. Also; in this hilbert space; these operators are always solvable. These are really just functional's in disguise; and functional's always have this "adjoint" kind of flavour to them.
This isn't exactly adjoints, Mphlee; but I believe it borders well enough.
\(
(f,g)_{\mathcal{H}} = \int_0^\infty f(x)\overline{g(x)}\,dx\\
(F,G)_{\mathbb{H}} = \int_{1/2 - i\infty}^{1/2+i\infty}|\Gamma(z)|^2F(1-z)\overline{G(1-z)}\,dz\\
\)
The function,
\(
\Gamma(1-z)F(z) = (f,x^{-\overline{z}})_{\mathcal{H}}\\
\Gamma(1-z)G(z) = (g,x^{-\overline{z}})_{\mathcal{H}}\\
\)
We can create an operator,
\(
f = \sum_{n=0}^\infty F(n+1) \frac{w^n}{n!}\\
Hf = \sum_{n=0}^\infty F^{\circ n+1}(1) \frac{w^n}{n!}\\
\)
Such that,
\(
(H^n f, x^{-\overline{z}})_{\mathcal{H}} = \Gamma(1-z)\uparrow^n F\\
\)
This is the Hilbert space interpretation of the last paper. We want to use this to find a function \( g_{s}^z(x) \) such that,
\(
(f,g_s^z) = \Gamma(1-z) \uparrow^s F\\
\)
Note that \( g \) is not really an adjoint; but it plays the part well enough. Also; in this hilbert space; these operators are always solvable. These are really just functional's in disguise; and functional's always have this "adjoint" kind of flavour to them.
This isn't exactly adjoints, Mphlee; but I believe it borders well enough.