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+--- Thread: Hyper operator space (/showthread.php?tid=807)
Hyper operator space - JmsNxn - 08/12/2013
Well I've been muddling this idea around for a while. I have been trying to create a hyper operator space and I recently realized the form of this. I'll start as follows:
If \( [n] \) is a hyper operator then, \( [n] [m] \) is a hyper operator created by forming left composition. I.e:
\( [n] = x [n] y \) for all \( x,y \in \mathbb{N} \) then
\( [n][m] = x[n] (x[m] y) \)
Associate to every function that is a finite product a number as follows:
\( [e_1] [e_2] ... [e_n] = (\ p_1^{e_1} \cdot p_2^{e_2} \cdot ... \cdot p_n^{e_n}) \)
Where \( e_n \in \mathbb{N} \) and p_n is the nth prime.
Now hyper operator space is the following:
\( \frac{1}{(n)} \in \mathbb{H} \)
\( f,g \in \mathbb{H} \,\, \alpha,\beta \in \mathbb{C} \)
\( \alpha f + \alpha g \in \mathbb{H} \)
Now define the inner product as follows:
\( (f,g) = \sum_{x=1}^\infty \sum_{y=1}^{\infty} f(x,y) \bar{g(x,y)} \)
Where quite clearly (f,f) converges for all elements since the terms decay to zero across x and y faster or just as fast as \( \frac{1}{(x+y)^2} \)
We say all the functions \( \{ 1/(1), 1/(2), 1/(3),...,\} \) are dense in \( \mathbb{H} \)
Orthonormalize them to get \( \Delta_n \) such that:
\( (\Delta_i, \Delta_j) = \delta_{ij} \)
\( f = \sum_{i=0}^\infty (f, \Delta_i) \Delta_i \)
Now we have the advantage of being in a Hilbert space and having an orthonormal basis.
The first operator we have is the transfer operator:
\( T f = f(x,y+1) \)
Since \( [n-1][n] = T [n] \) this operator is well defined for any element of \( \mathbb{H} \) where \( T [a_1][a_2]...[a_n] = [a_1][a_2]...[(a_n) - 1][a_n] \)
Suppose:
\( [s] \) exists such that \( [s-1][s] = T [s] \) for all values that [s] returns natural numbers at, this is our solution to hyper operators.
I think the key is to invesetigate the inner product.