# 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 http://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.

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Seems to be a bug in your numerics... Calculations look fine and my plots also... –  Dirk Jan 8 '13 at 21:42
wow that's amazing. you actually got $\min(x,y)=\sum_{n=1}^\infty n^{-2}\sin nx\sin ny$? –  Ohad Asor Jan 8 '13 at 21:50
here's my code: ## #include <iostream> #include <cmath> using namespace std; long double phi(double x, double n) { return sin(nx)/n; } int main(int, char*) { long double sum = 0, x=8,y=1; for (long double n = 1;n< 100000;n++) { sum += phi(x,n)*phi(y,n); if (((int)n)%1000) cout<<n<<' '<<sum<<endl; } return 0; } ## –  Ohad Asor Jan 8 '13 at 22:01
Seems like I did not look careful enough on my mesh plot... Sorry. –  Dirk Jan 8 '13 at 22:26

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$\psi_0(x)\psi_0(y) = \sin(\tfrac{\pi x}{2T})\sin(\tfrac{\pi y}{2T})$. - now it really works :) – Ohad Asor Jan 9 '13 at 10:22 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!