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Question on Mercer's theorem

  1. Feb 23, 2014 #1
    Hello,

    if we consider a [itex]n\times n[/itex] symmetric positive definite matrix, we can prove that it has n positive eigenvalues and n orthogonal eigenvectors, and that such matrix can be expressed as a linear combination [itex]\sum_{i=1}^n \lambda_i e_i\otimes e_i[/itex]

    Mercer's theorem extends this result to continuous symmetric positive definite functions [itex]K:[a,b]\times [a,b]\rightarrow \mathbb{R}[/itex] by stating that K(x,y) can be expressed as [itex]\sum_{i=1}^\infty \lambda_i e_i(x)e_i(y)[/itex] where e_i are eigenfunctions of the linear operator associated with K.

    My question is: can Mercer's theorem be generalized to square integrable functions K defined on the whole domain [itex]\mathbb{R}^2[/itex] instead of just [itex][a,b]\times[a,b][/itex]?
     
  2. jcsd
  3. Mar 7, 2014 #2
    I don't know mercer's theorem, but I quote Wikipedia:

    "A recent generalization replaces this conditions by that follows: the set X is a first-countable topological space endowed with a Borel (complete) measure μ. X is the support of μ and, for all x in X, there is an open set U containing x and having finite measure. Then essentially the same result holds:

    Theorem. Suppose K is a continuous symmetric non-negative definite kernel on X. If the function κ is L1μ(X), where κ(x)=K(x,x), for all x in X, then there is an orthonormal set {ei}i of L2μ(X) consisting of eigenfunctions of TKa such that corresponding sequence of eigenvalues {λi}i is nonnegative. The eigenfunctions corresponding to non-zero eigenvalues are continuous on X and K has the representation
    K(s,t) = \sum_{j=1}^\infty \lambda_j \, e_j(s) \, e_j(t)
    where the convergence is absolute and uniform on compact subsets of X."
     
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