Elaborating Mercer's theorem (RKHS) on Cameron-Martin space $k(x,y)=\min(x,y)$ 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.
 A: If my chalkboard scribblings are correct, if $f(x) = \cos(\alpha x)$ and $g(x) = \sin(\alpha x)$ for $\alpha \neq 0$, then
$$ \mathscr{T}_k f(x) = \alpha^{-2}(f(x) + \alpha x\sin(\alpha T) - 1), \quad \mathscr{T}_k g(x) = \alpha^{-2}(g(x) - \alpha x \cos(\alpha T))$$ 
(up to some inconsequential signs), so that your orthonormal basis of eigenfunctions is
$$ \psi_k(x) = \sqrt{\frac{2}{T}} \sin\left(\frac{\pi(k+\tfrac{1}{2})}{T}x\right) $$
with corresponding eigenvalues
$$ \lambda_k = \frac{T^2}{\pi^2(k+\tfrac{1}{2})^2}. $$
Hence,
$$ k(x,y) = \sum_{k=0}^\infty \frac{2T}{\pi^2(k+\tfrac{1}{2})^2} \sin\left(\frac{\pi(k+\tfrac{1}{2})}{T}x\right) \sin\left(\frac{\pi(k+\tfrac{1}{2})}{T}y\right).$$
I hope this works!
A: 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 $\psi_0(x)\psi_0(y) = \sin(\tfrac{\pi x}{2T})\sin(\tfrac{\pi y}{2T})$.
