Polynomial delta function

The function {k_1(x,y) = (3xy+1)/2} has a curious property: for any linear function {\ell}, and any point {y\in \mathbb R}, the integral {\int_{-1}^1 \ell(x)k_1(x,y)\,dx} evaluates to {\ell(y)}. This is easy to check using the fact that odd powers of {x} integrate to zero:

\displaystyle \frac12 \int_{-1}^1 (ax+b)(3xy+1)\,dx = \frac12 \int_{-1}^1 (3ax^2y+b)\,dx = \frac12(2ay+2b) = ay+b

More generally, for any integer {n\ge 0} there exists a unique symmetric polynomial {k_n(x,y)} that has degree {n} in {x} and {y} separately and satisfies {\int_{-1}^1 p(x)k_n(x,y)\,dx = p(y)} for all polynomials {p} of degree at most {n}. For example, {k_0(x,y)=1/2} (obviously) and

\displaystyle k_2(x,y)=\frac98+\frac32xy+\frac{15}{8}(x^2+y^2)+\frac{45}{8}x^2y^2

The formula is not really intuitive, and a 3d plot would not help the matter much. To visualize {k_n}, I plotted {k_n(x,-3/4)}, {k_n(x,0)}, and {k_n(x,1/2)} below (green, red, blue respectively).

Degree 1
Degree 1
Degree 2
Degree 2
Degree 4
Degree 4

For {y\in [-1,1]} and large {n}, the function {k_n(\cdot, y)} approaches the Dirac delta at {y}, although the convergence is slow, especially when {|y|} is close to {1}. I don’t think there is anything good to be said about the case {|y|>1}.

Degree 10
Degree 10
Degree 20
Degree 20

The existence and uniqueness of {k_n} are a consequence of the Riesz representation of linear functionals on an inner product space. Indeed, polynomials of degree at most {n} form such a space {\mathbb P_n} with inner product {\langle p,q\rangle = \int_{-1}^1p(x)q(x)\,dx}, and the functional {p\mapsto p(y)} is linear for any fixed {y\in\mathbb R}. Hence, this functional can be written as {p\mapsto \langle p, k_y\rangle } for some {k_y}. The function {(x,y) \mapsto k_x(y)} is a reproducing kernel for this space. Its symmetry is not immediately obvious.

The Legendre polynomials {P_0,\dots,P_n} are an orthogonal basis of {\mathbb P_n}; more precisely, {\widetilde{P}_j = \sqrt{j+1/2}P_j} form an orthonormal basis. It’s a general fact about reproducing kernels that

\displaystyle k(x,y) = \sum_j \widetilde{P}_j(x)\widetilde{P}_j(y)

(which, incidentally, proves the symmetry {k(y,x)=k(x,y)}). Indeed, taking this sum as the definition of {k} and writing {p = \sum_{j=0}^n c_j \widetilde{P}_j}, we find

\displaystyle \langle p, k(\cdot, y)\rangle = \sum_j \widetilde{P}_j(y) \langle p, \widetilde{P}_j\rangle = \sum_j \widetilde{P}_j(y) c_j = p(y)

This is the Sage code used for the above plots.

n = 20
k = sum([(j+1/2)*legendre_P(j,x)*legendre_P(j,y) for j in range(0,n+1)])
plot(k(x,y=-3/4),(x,-1,1),color='green') + plot(k(x,y=0),(x,-1,1),color='red') +  plot(k(x,y=1/2),(x,-1,1),color='blue')

Higher degrees cause some numerical issues…

Degree 22
Degree 22

Post motivated by Math.SE question

2 thoughts on “Polynomial delta function”

    1. You are right; I corrected this. It was a typo; the code with (j+1/2)*legendre_P(j,x)*legendre_P(j,y) was based on sqrt(j+1/2) being the factor, which becomes (j+1/2) when applied to both polynomials.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s