I came across this exercise and I am stuck on how to solve it:

Consider a Fredholm Integral of the second kind: $u(x)=f(x)+\lambda\int_{0}^{L}K(x,t)u(t)dt$.

Define the 2nd iterate Kernel $K_2(x,y)=\int_{0}^{L}K(x,t)K(t,s)dt$. If $\lambda$ is an eigenvalue for $K_2$, then we must have that $\sqrt\lambda$ or $-\sqrt{\lambda}$ is an eigenvalue of the original kernel $K(x,t)$. I want to show this, and this seems like a relatively straightforward problem but I am struggling.

I am assuming the Kernel is defined on $L^2([0,L]^2)$.

  • 1
    $\begingroup$ Do you mean that if $\int_{0}^{L}K_2(x,t)u(t)dt=\lambda u(t)$, then $\int_{0}^{L}K(x,t)v(t)dt=\pm\sqrt{\lambda}v$ for some choice of $\pm$ and some $v$? $\endgroup$ – DisintegratingByParts Mar 4 '18 at 6:48
  • $\begingroup$ Yes, I suppose that is a better way to write it. $\endgroup$ – Kernel_Dirichlet Mar 5 '18 at 2:16

For any linear operator $K$ on a complex vector space, if $$ K^2 v = \lambda v, \;\; v \ne 0, $$ one has $(K-\sqrt{\lambda}I)(K+\sqrt{\lambda}I)v=0$. So either $(a) Kv=-\sqrt{\lambda}v$ or (b) $w=(K+\sqrt{\lambda}I)v \ne 0$ and $Kw=\sqrt{\lambda}w$.

  • $\begingroup$ Thanks, it was indeed much simpler than I thought! $\endgroup$ – Kernel_Dirichlet Mar 6 '18 at 2:54

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