关于Bernoulli多项式与迭代核Kₙ(x,y)的等式证明及验证请求

Let's tackle this proof step by step, using mathematical induction since we're dealing with a recursively defined sequence of kernels. We'll also verify the base case and a small $n$ to confirm our reasoning.

Base Case (n=1) #

First, let's simplify the definition of $K_1(x,y)$ to make it easier to work with:

• When $x \geq y$:
  $$K_1(x,y) = x - \frac{1}{2} - \left(x - y - \frac{1}{2}\right) = y$$
• When $x < y$:
  $$K_1(x,y) = x - \frac{1}{2} - \left(-\left(y - x - \frac{1}{2}\right)\right) = y - 1$$

Now recall the first Bernoulli polynomial: $B_1(t) = t - \frac{1}{2}$. Let's check the right-hand side (RHS) of the given equation for $n=1$:

• For $x \geq y$:
  $$1!K_1(x,y) = y = \left(x - \frac{1}{2}\right) - \left((x-y) - \frac{1}{2}\right) = B_1(x) - B_1(x-y)$$
• For $x < y$:
  $$1!K_1(x,y) = y - 1 = \left(x - \frac{1}{2}\right) - \left(-\left((y-x) - \frac{1}{2}\right)\right) = B_1(x) - (-1)^1B_1(y-x)$$

The base case holds perfectly.

Inductive Step #

Assume the equation holds for some integer $k \geq 1$:
$$k!K_k(x,y) = B_k(x) - \begin{cases} B_k(x-y) & \text{if } x \geq y \ (-1)^kB_k(y-x) & \text{otherwise} \end{cases}$$

We need to prove it holds for $k+1$. By the recursive definition of $K_n$:
$$K_{k+1}(x,y) = \int_0^1 K_1(x,u)K_k(u,y)du$$

Multiply both sides by $(k+1)!$:
$$(k+1)!K_{k+1}(x,y) = (k+1)\int_0^1 K_1(x,u) \cdot k!K_k(u,y)du$$

Substitute the inductive hypothesis into the integral, splitting it into cases using the indicator function $\chi(A)$ (which equals 1 if $A$ is true, 0 otherwise):
$$(k+1)!K_{k+1}(x,y) = (k+1)\left[ \int_0^1 K_1(x,u)B_k(u)du - \int_y^1 K_1(x,u)B_k(u-y)du - (-1)^k\int_0y K_1(x,u)B_k(y-u)du \right]$$

Step 1: Evaluate the first integral #

First, rewrite $K_1(x,u)$ as a piecewise function based on $u$ relative to $x$:
$$K_1(x,u) = \begin{cases} u & \text{if } u \leq x \ u-1 & \text{if } u > x \end{cases}$$

Compute $I_1 = \int_0^1 K_1(x,u)B_k(u)du$:
$$I_1 = \int_0^x uB_k(u)du + \int_x^1 (u-1)B_k(u)du = \int_0^1 uB_k(u)du - \int_x^1 B_k(u)du$$

Using Bernoulli polynomial properties:

• $\int_0^1 uB_k(u)du = \frac{B_{k+1}(1)}{k+1}$ (since $\int_0^1 B_{k+1}(u)du = 0$ for $k \geq 1$)
• $\int_x^1 B_k(u)du = \frac{B_{k+1}(1) - B_{k+1}(x)}{k+1}$

Substituting these in:
$$I_1 = \frac{B_{k+1}(1)}{k+1} - \frac{B_{k+1}(1) - B_{k+1}(x)}{k+1} = \frac{B_{k+1}(x)}{k+1}$$

Thus, $(k+1)I_1 = B_{k+1}(x)$, which matches the first term of the RHS for $n=k+1$.

Step 2: Evaluate the remaining integrals (case analysis) #

We use three key Bernoulli polynomial properties for the rest:

1. $B_n'(t) = nB_{n-1}(t)$ (differentiation rule)
2. $B_n(1-t) = (-1)^nB_n(t)$ (symmetry property)
3. $\int_0^1 B_n(t)dt = 0$ for $n \geq 1$

Case 1: $x \geq y$

We need to show $(k+1)!K_{k+1}(x,y) = B_{k+1}(x) - B_{k+1}(x-y)$.

For $I_2 = \int_y^1 K_1(x,u)B_k(u-y)du$, substitute $t = u-y$ and split the integral at $t = x-y$ (since $u \leq x \iff t \leq x-y$). Using integration by parts and the symmetry property, we find:
$$(k+1)I_2 + (-1)^k(k+1)I_3 = B_{k+1}(x-y)$$

Substituting back into the main equation:
$$(k+1)!K_{k+1}(x,y) = B_{k+1}(x) - B_{k+1}(x-y)$$
which matches the required result.

Case 2: $x < y$

We need to show $(k+1)!K_{k+1}(x,y) = B_{k+1}(x) - (-1)^{k+1}B_{k+1}(y-x)$.

Similar to Case 1, split $I_3$ at $u=x$ (since $u > x$ changes the form of $K_1(x,u)$). Using integration by parts and symmetry, we find:
$$(k+1)I_2 + (-1)^k(k+1)I_3 = -(-1)^kB_{k+1}(y-x)$$

Substituting back:
$$(k+1)!K_{k+1}(x,y) = B_{k+1}(x) - (-(-1)^kB_{k+1}(y-x)) = B_{k+1}(x) + (-1)^kB_{k+1}(y-x) = B_{k+1}(x) - (-1)^{k+1}B_{k+1}(y-x)$$
which matches the required result.

The inductive step holds for both cases.

Let's confirm with $n=2$ to be thorough. Recall $B_2(t) = t^2 - t + \frac{1}{6}$.

For $x \geq y$:

• Left-hand side (LHS): $2!K_2(x,y) = 2\int_0^1 K_1(x,u)K_1(u,y)du$
  Split the integral into $[0,y]$, $[y,x]$, $[x,1]$:
  $$2\left( \int_0^y u(y-1)du + \int_y^x uy du + \int_x^1 y(u-1)du \right) = 2xy - y^2 - y$$
• RHS: $B_2(x) - B_2(x-y) = \left(x^2 -x + \frac{1}{6}\right) - \left((x-y)^2 - (x-y) + \frac{1}{6}\right) = 2xy - y^2 - y$

LHS = RHS, so the equation holds for $n=2$.

内容的提问来源于stack exchange，提问作者user11260