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 Hyper operator space JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 08/12/2013, 10:17 PM 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. « Next Oldest | Next Newest »

 Messages In This Thread Hyper operator space - by JmsNxn - 08/12/2013, 10:17 PM

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