Though… it occurs to me that one could construct something that acted like a “square root of a delta”, which would then make an orthonormal basis (though still not part of the hilbert space).
I’m not sure what you’re trying to construct, but note that one can only multiply distributions under rather restrictive conditions. There are some even more abstract classes of distributions which permit an associative multiplication (Colombeau algebras, generalized Gevrey classes of ultradistributions, and so on) but they’re neither terribly common nor fun to work with.
Ah, nevermind then. I was thinking something like let b(x,k) = 1/sqrt(2k) when |x| < k and 0 otherwise
then define integral B(x)f(x) dx as the limit as k->0+ of integral b(x,k)f(x) dx
I was thinking that then integral (B(x))^2 f(x) dx would be like integral delta(x)f(x) dx.
Now that I think about it more carefully, especially in light of your comment, perhaps that was naive and that wouldn’t actually work. (Yeah, I can see now my reasoning wasn’t actually valid there. Whoops.)
I’m not sure what you’re trying to construct, but note that one can only multiply distributions under rather restrictive conditions. There are some even more abstract classes of distributions which permit an associative multiplication (Colombeau algebras, generalized Gevrey classes of ultradistributions, and so on) but they’re neither terribly common nor fun to work with.
Ah, nevermind then. I was thinking something like let b(x,k) = 1/sqrt(2k) when |x| < k and 0 otherwise
then define integral B(x)f(x) dx as the limit as k->0+ of integral b(x,k)f(x) dx
I was thinking that then integral (B(x))^2 f(x) dx would be like integral delta(x)f(x) dx.
Now that I think about it more carefully, especially in light of your comment, perhaps that was naive and that wouldn’t actually work. (Yeah, I can see now my reasoning wasn’t actually valid there. Whoops.)
Ah well. thank you for correcting me then. :)