Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In classical functional analysis, one can construct a reproducing kernel Hilbert space by starting with a positive definite kernel, say $K: [0,1]\times [0,1] \rightarrow \mathbb{R}$. One then creates linear combinations of the form $f(x) = \sum^n a_i k(x_i,x)$, together with an inner product

$\langle f,g \rangle = \sum \sum a_i b_j k(x_i,x_j),$

and completes the space as usual.

As far as I can see, the 'essential' properties of a reproducing kernel Hilbert space are the ability to represent linear operators as integral kernels, and the Riesz representation theorem.

I'm just wondering if there is a similar construction to the one outlined above in the framework of the max-plus algebra. Both of the properties I mentioned above have max-plus analogues (see this introduction). If such a construction exists, how far is it possibe to take it? Is there an idempotent version of Mercer's theorem, for example?

share|improve this question

1 Answer 1

up vote 8 down vote accepted

See

  1. G.L. Litvinov, V.P. Maslov and G.B. Shpiz. Idempotent functional analysis. An algebraic approach // Mathematical Notes, v. 69, # 5, 2001, p. 696-729. E-print math.FA/0009128 (http://ArXiv.org).

  2. G.L. Litvinov and G.B. Shpiz. Kernel theorems and nuclearity in idempotent mathematics. An algebraic approach, Journal of Mathematical Sciences, v. 141, #4, 2007, p. 1417-1428. See also E-print math.FA/0609033 (http://arXiv.org), 2006.

  3. G.L. Litvinov. Tropical mathematics, idempotent analysis, classical mechanics and geometry. - in: Spectral Theory and Geometric Analysis M.Braverman et al., Eds., AMS Contemporary Mathematics, vol. 535, 2011, p. 159-186. See also E-print arXiv: arXiv: 1005.1247 (http://arXiv.org).

share|improve this answer
    
I took the liberty to add hyperlinks to the arXiv references; hope that's ok :-) –  Suvrit Jul 13 '12 at 8:03
    
Excellent, thanks! –  Simon Lyons Jul 13 '12 at 9:42

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.