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I'm looking at the eigenfunction expansion of Brownian motion on the interval [0,1]:

$W_t = \sqrt{2} \sum_{k=1}^\infty Z_k \frac{\sin((k - \frac{1}{2}) \pi t)}{(k - \frac{1}{2}) \pi}.$

One deduces this fact by computing the eigenfunctions of the integral operator

$f(s) \to \int_0^1 \min(s,t)f(t)dt.$

Mercer's theorem says the eigenfunctions of this operator should form an orthonormal basis of $L^2[0,1]$. However, eigenfunctions of the integral operator are $\sqrt{2}\sin((k - \frac{1}{2}) \pi t)$, which clearly don't form a basis since they all satisfy $f(0) = 0$. What am I missing here?

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    If I've understood the replies correctly, you;re saying that the functions $\{ \sqrt{2} \sin((k-.5)\pi t)\}$ do form an orthonormal basis?2011-07-25

2 Answers 2

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The eigenfunction expansion in this case is the "sine Fourier series". This corresponds to the ordinary (periodic) Fourier series of your function extended to be periodic with period 4 such that $f(-t) = -f(t)$ and $f(2-t) = f(t)$. The Fourier series converges in $L^2$, but not pointwise in general, especially not when the function is discontinuous. In particular if your original function has $\lim_{t \to 0+} f(t) \ne 0$, the periodic version will have a jump discontinuity at 0.

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What you're missing may be what an orthonormal basis of $L^2$ is. It should be a set of functions $f_n,\quad n = 1,2,3,\dots$ belonging to $L^2(0,1)$ such that $ \int_0^1 f_n f_m = 0\text{ if } n\ne m, \text{ and} $ $ \int_0^1 \|f_n\|^2 = 1 \text{ for all } n,\text{ and} $ such that every function $g$ in $L^2(0,1)$ can be arbitrarily closely approximated in $L^2$ norm by a finite linear combination of members of the basis. "Approximated in $L^2$ norm" does not mean uniformly or pointwise or the like, but rather $\int_0^1 |h -g|^2$ can be made as small as desired be letting $h$ be an appropriate finite linear combination of members of the basis.

Crucial point: the value of the integral is not at all affected by altering the value of any function at just one point, such as (in this case) $0$. So the value at $0$ is of no consequence. You're not trying to make the value of the linear combination at $0$ approximate the value of $g$ at $0$; you're just trying to make the integral arbitarily small.