# 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 ${n=1, 2, 3}$:

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.

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 ${n}$?