Let 
and 
be two Hilbert spaces with,
_{\mathcal{H}} = \int_0^\infty f(x)\overline{g(x)}\,dx\\<br />
(F,G)_{\mathbb{H}} = \int_{1/2 - i\infty}^{1/2+i\infty}|\Gamma(z)|^2F(1-z)\overline{G(1-z)}\,dz\\<br />
)
The function,
F(z) = (f,x^{-\overline{z}})_{\mathcal{H}}\\<br />
\Gamma(1-z)G(z) = (g,x^{-\overline{z}})_{\mathcal{H}}\\<br />
)
We can create an operator,
 \frac{w^n}{n!}\\<br />
Hf = \sum_{n=0}^\infty F^{\circ n+1}(1) \frac{w^n}{n!}\\<br />
)
Such that,
_{\mathcal{H}} = \Gamma(1-z)\uparrow^n F\\<br />
)
This is the Hilbert space interpretation of the last paper. We want to use this to find a function)
such that,
 = \Gamma(1-z) \uparrow^s F\\<br />
)
Note that
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.
The function,
We can create an operator,
Such that,
This is the Hilbert space interpretation of the last paper. We want to use this to find a function
Note that
This isn't exactly adjoints, Mphlee; but I believe it borders well enough.
