Hi,


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 $ \mathscr 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.


I also tried to substitute the original solution with $C\_1,C\_2$ in the original eigenvalue problem equation, and then to calculate $C\_1,C\_2$, but they turned out to depend on $x$, which is of cource unacceptable.


So what's wrong here?


I also asked the question on <https://math.stackexchange.com/questions/272857/elaborating-mercers-theorem-rkhs-on-cameron-martin-space-kx-y-minx-y>, but no answers there. I'll post in either forum immediately if an answer will come.



Elaborating Mercer's theorem (RKHS) on Cameron-Martin space $k(x,y)=\min(x,y)$
What Branimir Ćaćić writes is correct. Another way to see that your $\lambda$'s where not right is as follows: 


From $$\lambda\psi(x) = \int\_0^x y\psi(y) dy + x\int\_x^T \psi(y)dy$$ you get that $
$$\psi(0)=0.$$


Similarly, from $$\lambda\psi'(x) = \int\_x^T \psi(y)dy$$ you get $$\psi'(T)=0.$$


Hence, you have two boundary conditions for the differential equation $\lambda\psi''(x) = -\psi(x)$. The first forces $C\_2=0$, the second gives $$\lambda = \frac{T^2}{\pi^2(n+\tfrac{1}{2})^2}$$
($T/\sqrt{\lambda}$ has to be a root of $\cos$) and no condition on $C\_1$. Since you want an orthonormal basis, you have to normalize the functions in $L^2([0,T])$ which gives $C\_1=\sqrt{2/T}$.


What you are missing in your numerics is that the series starts with $n=0$ and hence, your result differs from $\min(x,y)$ by $\psi\_0(x)\psi\_0(y) = \sin(\tfrac{\pi x}{2T})\sin(\tfrac{\pi y}{2T})$.


