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 [MSE-SOLVED] Subfunction is functorial!!!! JmsNxn Long Time Fellow Posts: 628 Threads: 102 Joined: Dec 2010 06/05/2021, 03:11 AM (This post was last modified: 06/05/2021, 07:31 AM by JmsNxn.) 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. MphLee Fellow Posts: 220 Threads: 22 Joined: May 2013 06/05/2021, 09:11 PM (06/05/2021, 03:11 AM)JmsNxn Wrote: This isn't exactly adjoints, Mphlee; but I believe it borders well enough. OK, I see what you are trying to do. My intuition seems to follow you but my brain explodes when I try to make it more precise. You're cool and everything and I understand that you know those constructions like the palms of your hands and do not need to specify from where to where those operations go... but now I need to reconstruct clearly the domains and the codomain of everything you're saying or my brain wont connect hahah Need time. MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ JmsNxn Long Time Fellow Posts: 628 Threads: 102 Joined: Dec 2010 06/06/2021, 01:24 AM (This post was last modified: 06/06/2021, 03:43 AM by JmsNxn.) It may help to remind you of some facts about hilbert spaces. Every functional $L : \mathcal{H} \to \mathbb{C}$ has a representation as, $ L f = (f,c)_{\mathcal{H}}\,\,\text{for some}\,\,c = c(f) \in \mathcal{H}\\$ With that in mind, $ L f = \uparrow F |_{z=z_0} = (f,c_{z_0})\,\,\text{for some}\,\,c_{z_0} \in \mathcal{H}\\$ So we can represent the operator, $ \uparrow F = (Hf, x^{-\overline{z}})_{\mathcal{H}}\\ \uparrow F = (f,c_{z})_{\mathcal{H}}\\$ Such that $c_z$ depends on $f$; so we should write $c_z = c_z(f)$ (in the linear case we get to throw away the dependence; c is constant) And we can create such a chain as $c_z^n(f)$ such that, $ \uparrow^n F = (f,c_z^n)_{\mathcal{H}}\\$ Because the uparrow operator sends the space to itself. I'm wondering if this will admit an obvious extension to $n \mapsto s \in \mathbb{C}$. As, this superfunction operation is well managed in this space; and always sends to this space, we're okay. The spaces in question, again, are $\mathbb{S}_\theta$ for $\theta < \pi/2$ and $F \in \mathbb{E}_{\theta}$ for $\theta < \pi/2$; with the additional requirement that, $ |F(z)| \le M\,\,\text{for}\,\,\Re(z) > 0\\ F : \mathb{R}^+ \to \mathbb{R}^+\,\,\,\text{and}\,\,F' : \mathbb{R}^+ \to \mathbb{R}^+\\ F : \mathbb{C}_{\Re(z) > 0} \to \mathbb{C}_{\Re(z) > 0}\\$ Where $\uparrow$ maps such $F$ to such $F$. Additionally, for any two $F_1,F_2$ for $c_1,c_2 > 0$ we get $c_1F_1 + c_2F_2 = F$ still belongs to this space. This means we have some linear span data. Honestly Mphlee, I'm thinking adjoints is the way to do this. This is making more and more sense. Especially as you describing the categorical perspective; it just looks like complicated hilbert spaces. MphLee Fellow Posts: 220 Threads: 22 Joined: May 2013 06/06/2021, 11:18 AM Ok... I must take a pause...my brain is exploding and I think I'm going crazy. I literally think about this all day every minute. This is too big for me atm especially after I wrote down clearly the proof of functoriality and now that you remind me the thing about representations of functionals. Under further investigation it turns out that the category I've defined, as long as we stick with groups, i.e. we keep invertibility in the time monoid, seems equivalent to a much simpler and fundamental one that has to do with "distance and outer composition in the group". That also links up heavily with the abstract nonsense pattern I found in the Jabotinsky/Compositional integral business of infinitesimal generators. I'm a bit surprised by this... I didn't expect to see the link with the subfunction operator this soon. Also yes! it has something to do with adjoints in Hilbert spaces... and also metric spaces.... and also as "it just looks like complicated hilbert spaces": in the analogy functionals are to hilbert spaces what presheaves are to categories. And dual space is to the original space what the cateogory of presheaves is. I see something big. But I don't know what is. I need to rework all of this from scratches. I need also to go back to basics of category theory. I need several weeks of stop to tidy up my brain. I'm sorry. MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ JmsNxn Long Time Fellow Posts: 628 Threads: 102 Joined: Dec 2010 06/06/2021, 11:16 PM (06/06/2021, 11:18 AM)MphLee Wrote: I'm sorry. No problem. When you said adjoint though; something clicked in me, and I was thinking it might have something to do with this. « Next Oldest | Next Newest »

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