Extreme values of a reproducing kernel for polynomials

For every nonnegative integer {n} there exists a (unique) polynomial {K_n(x, y)} of degree {n} in {x} and {y} separately with the following reproducing property:

{\displaystyle p(x) = \int_{-1}^1 K_n(x, y)p(y)\,dy}

for every polynomial {p} of degree at most {n}, and for every {x}. For example, {K_1(x, y)= (3xy+1)/2}; other examples are found in the post Polynomial delta function.

This fact gives an explicit pointwise bound on a polynomial in terms of its integral on an interval:

{\displaystyle |p(x)| \le M_n(x) \int_{-1}^1 |p(y)|\,dy}

where {M_n(x) = \sup\{|K(x, y)| \colon y\in [-1, 1]\}}. For example, {M_1(x) = (3|x|+1)/2}.

Although in principle {x} could be any real or complex number, it makes sense to restrict attention to {x\in [-1, 1]}, where integration takes place. This leads to the search for extreme values of {K} on the square {Q=[-1, 1]\times [-1, 1]}. Here is how this function looks for {n=1, 2, 3}:

K1
Degree 1
K2
Degree 2
K3
Degree 3

The symmetries {K(x, y)=K(-x, -y) = K(y, x)} are evident here.

Explicitly,

{\displaystyle K_n(x, y) = \sum_{k=0}^n \frac{2k+1}{2} P_k(x)P_k(y)}

where {P_k} is the Legendre polynomial of degree {k} and the factor {(2k+1)/2} is included to make the polynomials an orthonormal set in {L^2(-1, 1)}. Since {P_k} oscillates between {-1} and {1}, it follows that

{\displaystyle |K_n(x, y)|\le \sum_{k=0}^n \frac{2k+1}{2} = \frac{(n+1)^2}{2}}

and this bound is attained at {K(1, 1)=K(-1,-1)=(n+1)^2/2} because {P_k(1)=1} and {P_k(-1)=(-1)^k}.

Is

{\displaystyle K_n(-1, 1) =\sum_{k=0}^n (-1)^k\frac{2k+1}{2} = (-1)^n \frac{n+1}{2}}

the minimum value of {K} on the square {Q}? Certainly not for even {n}. Indeed, differentiating the sum

{\displaystyle S_n(x) = K_n(x, 1) = \sum_{k=0}^n \frac{2k+1}{2} P_k(x)}

with respect to {x} and using {P_k'(-1) =(-1)^{k-1}k(k+1)/2}, we arrive at

{\displaystyle S_n'(-1) = (-1)^{n-1} \frac{n(n^2+3n+2)}{4}}

which is negative if {n} is even, ruling out this point as a minimum.

What about odd {n}, then: is it true that {K_n \ge -(n+1)/2} on the square {Q}?

{n=1}: yes, {K_1(x, y) = (3xy+1)/2 \ge (-3+1)/2 = -1} is clear enough.

{n=3}: the inequality {K_3\ge -2} is also true… at least numerically. It can be simplified to {35(xy)^3 + 9(xy)^2 + 15xy \ge (21x+21y+3)(x^2+y^2)} but I do not see a way forward from there. At least on the boundary of {Q} it can be shown without much work:

{\displaystyle K_3(x, 1) + 2 = \frac{5}{4}(x+1)(7x^2-4x+1)}

The quadratic term has no real roots, which is easy to check.

{n=5}: similar story, the inequality {K_5\ge -3} appears to be true but I can only prove it on the boundary, using

{\displaystyle K_5(x, 1)+3 = \frac{21}{16}(x + 1)(33 x^4 - 18x^3 - 12x^2 + 2x + 3)}

The quartic term has no real roots, which is not so easy to check.

{n=7}: surprisingly, {K_7(4/5, 1) = -2229959/500000} which is about {-4.46}, disproving the conjectural bound {K_7\ge -4}. This fact is not at all obvious from the plot.

K7
Degree 7 kernel

Questions:

  • Is {K_n \ge -Cn} on the square {Q = [-1, 1]\times [-1, 1]} with some universal constant {C}?
  • Is the minimum of {K_n} on {Q} always attained on the boundary of {Q}?
  • Can {M_n(x) = \sup\{|K(x, y)| \colon y\in [-1, 1]\}} be expressed in closed form, at least for small degrees {n}?