/sci/ - Science & Math

I love this problem. The solution is pretty straightforward — if you've had a graduate course in functional analysis. I realize there must be a more elementary solution, but since I deal with integral operators so often these days, this is what came to me first:

Let <span class="math">\mathcal{K}:L^2([0,1])\to L^2([0,1])[/spoiler] be the compact, self-adjoint, positive-definite operator <span class="math">\mathcal{K}(u) =\int_0^1 u(y)K(x,y)\,dy[/spoiler] (positive-definiteness follows easily from the fact that <span class="math">K[/spoiler] is strictly positive and continuous on its compact domain, and therefore bounded away from 0).

A well-known result from functional analysis:
If <span class="math">\mathcal{A}[/spoiler] is a compact, self-adjoint, positive definite operator on <span class="math">L^2(\Omega)[/spoiler], then
(a) the spectral radius <span class="math">\rho_mathcal{A}[/spoiler] is a simple eigenvalue of <span class="math">\mathcal{A}[/spoiler] (i.e., the corresponding eigenspace has dimension one),
(b) <span class="math">\rho_\mathcal{A}[/spoiler] has a corresponding eigenfunction that is positive a.e. <span class="math">(\Omega)[/spoiler],
(c) <span class="math">\rho_\mathcal{A}[/spoiler] is the _only_ eigenvalue that corresponds to a positive (a.e.) eigenfunction (this is a consequence of <span class="math">\mathcal{A}[/spoiler] being self-adjoint, and therefore having orthogonal eigenfunctions: two positive functions cannot possibly be orthogonal).


Note that <span class="math">\mathcal{K}^2[/spoiler] is compact with <span class="math">\mathcal{K}^2(f)=f[/spoiler] and <span class="math">\mathcal{K}^2(g)=g[/spoiler]. Since <span class="math">f[/spoiler] and <span class="math">g[/spoiler] are positive, by (c) it follows that 1 must be the spectral radius of <span class="math">\mathcal{K}^2[/spoiler]. By (a), the corresponding eigenspace must have dimension one. Hence, <span class="math">f[/spoiler] and <span class="math">g[/spoiler] must be a multiples of each other: <span class="math">f= \alpha g[/spoiler] for some <span class="math">\alpha>0[/spoiler].

Applying <span class="math">\mathcal{K}[/spoiler] to both sides yields <span class="math">g = \alpha f[/spoiler]. Consequently, <span class="math">\alpha^2=1[/spoiler]. Since <span class="math">\alpha[/spoiler] must be positive, we conclude <span class="math">\alpha=1[/spoiler].

>>4333163
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