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I'm trying to employ Mercer's theorem on the kernel $k(x,y)=\min(x,y)$. It is known (and easy to verify) that this is a nonnegative-definite kernel over $[0,T]$ for any $T>0$.

Fix $T>0$. Let's calculate the eigenfunctions of the transformation $\mathcal T_kf=\intop_{0}^T k(x,y)f(y)dy$:

$$ \lambda\psi(x)=\intop_{0}^T \min(x,y)\psi(y)dy=$$ $$ \intop_{0}^x \min(x,y)\psi(y)dy - \intop_{T}^x \min(x,y)\psi(y)dy=$$ $$ \intop_{0}^x y\psi(y)dy - x\intop_{T}^x \psi(y)dy\implies$$ $$ \lambda\psi'(x)= x\psi(x)-x\psi(x)-\intop_{T}^x \psi(y)dy\implies$$ $$ -\lambda\psi''(x)= \psi(x)\implies$$ $$\psi(x)=C_1\sin\frac x {\sqrt\lambda} + C_2\cos\frac x {\sqrt\lambda}$$

it seems like we're allowed to pick $C_1=1$ and $C_2=0$. So we pick $\psi_n(x)=\sin nx $ and $\lambda_n = n^{-2}$. Then Mercer's theorem actually says that we should get: $$ \min(x,y)=\sum_{n=1}^\infty n^{-2}\sin nx\sin ny$$

this all seem very nice, but when evaluating this numerically, it doesn't work. I tried also to normalize $\psi$ by dividing by its norm which is $\sqrt {\frac 1 {4n} (2nT-\sin2nT)}$, and it didn't help.

So what's wrong here?

EDIT: I also tried to substitute the original solution with $C_1,C_2$ in the original eigenvalue problem equation, and then to calculate $C_1,C_2$. It didn't work. It's a bit tedious, I did it with wxMaxima so I won't bring it here.

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You should note that all those implications are only forward implications. Have you checked when a general linear combination $\psi(x) = C_1 \sin \tfrac{x}{\sqrt{\lambda}} + C_2 \cos \tfrac{x}{\sqrt{\lambda}}$ satisfies the actual eigenvalue equation itself? –  Branimir Ćaćić Jan 8 '13 at 21:20
yes, it is satisfied in its general form, but I cannot find $C_1,C_2$ correctly. I verified it with wxMaxima, see my last edit –  Troy McClure Jan 8 '13 at 21:30

1 Answer 1

Problem solved here:


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