A New Proof of King’s Rule Using Involution

The King's Rule is

$$\begin{array}{}\text{(1)}& {\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx.\end{array}$$

The traditional proof of King’s Rule is not understandable, purely counting and does not reval anything.

Now let us use involution to give a new proof.

Lemma Consider an $R$$R$-module $M$$M$ with $\mathrm{Char}(R)\ne 2$$\mathrm{Char}(R)\ne 2$, and $\ast \in {\mathrm{Aut}}_{R-\mathrm{Mod}}(M)$$*\in\mathrm{Aut}_{R-\mathrm{Mod}}(M)$ satisfying ${\ast }^2={\mathrm{Id}}_M$$*^2=\mathrm{Id}_M$. We have

$$\begin{array}{}\text{(2)}& M\cong \mathrm{Odd}\oplus \mathrm{Even},\end{array}$$

where $\mathrm{Odd}=\mathrm{Ker}(\ast +{\mathrm{Id}}_M)$$\mathrm{Odd}=\mathrm{Ker}(*+\mathrm{Id}_M)$ and $\mathrm{Even}=\mathrm{Ker}(\ast -{\mathrm{Id}}_M)$$\mathrm{Even}=\mathrm{Ker}(*-\mathrm{Id}_M)$.

Proof. Firstly, let us prove that $\mathrm{Odd}+\mathrm{Even}=M$$\mathrm{Odd}+\mathrm{Even}=M$. For any $x\in M$$x\in M$, we have:

$$\begin{array}{}\text{(3)}& x={\displaystyle \frac{x+x^{\ast }}{2}}+{\displaystyle \frac{x-x^{\ast }}{2}}.\end{array}$$

Secondly, let us prove that $\mathrm{Odd}\cap \mathrm{Even}=0$$\mathrm{Odd}\cap \mathrm{Even}=0$. For $x\in \mathrm{Odd}\cap \mathrm{Even}$$x\in \mathrm{Odd}\cap \mathrm{Even}$, we have $x^{\ast }=x$$x^*=x$ (from $\mathrm{Even}$$\mathrm{Even}$) and $x^{\ast }=-x$$x^*=-x$ (from $\mathrm{Odd}$$\mathrm{Odd}$). Thus:

$$\begin{array}{}\text{(4)}& x=x^{\ast }=-x⟹2x=0.\end{array}$$

Since $\mathrm{Char}(R)\ne 2$$\mathrm{Char}(R)\ne 2$, $2$$2$ is a unit, so $x=0$$x=0$. $◻$$\square$

Remark. The map $E(x)={\displaystyle \frac{x+x^{\ast }}{2}}:M\to \mathrm{Even}$$E(x)=\dfrac{x+x^*}{2}:M\to\mathrm{Even}$ is the projection onto $\mathrm{Even}$$\mathrm{Even}$ since for $x\in \mathrm{Even}$$x\in\mathrm{Even}$, $E(x)=x$$E(x)=x$. The kernel of $E$$E$ is obviously $\mathrm{Odd}:=\mathrm{Ker}(\ast +{\mathrm{Id}}_M)$$\mathrm{Odd}:=\mathrm{Ker}(*+\mathrm{Id}_M)$. Similarly, $O(x)={\displaystyle \frac{x-x^{\ast }}{2}}:M\to \mathrm{Odd}$$O(x)=\dfrac{x-x^*}{2}:M\to\mathrm{Odd}$. When $M$$M$ is an inner product space, $\mathrm{Odd}$$\mathrm{Odd}$ and $\mathrm{Even}$$\mathrm{Even}$ will be orthogonal to each other.

Lemma. Let $T:M\to N$$T:M\to N$ be a module homomorphism with $\mathrm{Odd}\subseteq \mathrm{Ker}(T)$$\mathrm{Odd}\subseteq\mathrm{Ker}(T)$. Then $Tx=T{x}^{\ast }$$Tx=Tx^*$ for all $x\in M$$x\in M$.

Proof.

$$\begin{array}{}\text{(5)}& Tx=T{x}^{\ast }⟺T(x-x^{\ast })=0.◻\end{array}$$

Let us consider an involution on $C[a,b]$$C[a,b]$, defined as:

$$\begin{array}{}\text{(6)}& f^{\ast }(x)=f(a+b-x).\end{array}$$

When $a=-b$$a=-b$, this becomes $f^{\ast }(x)=f(-x)$$f^*(x)=f(-x)$. Let ${\displaystyle T:={\int }_a^b(-)dx}$$\displaystyle T:=\int_a^b(-)dx$. Then it is easy to see that $\mathrm{Odd}\subseteq \mathrm{Ker}(T)$$\mathrm{Odd}\subseteq\mathrm{Ker}(T)$, since the odd part here consists of functions that are odd about the center ${\displaystyle \frac{a+b}{2}}.$$\dfrac{a+b}{2}.$

Hence we have:

$$\begin{array}{}\text{(7)}& {\int }_a^{b}f(x)dx={\int }_a^b{f}^{\ast }(x)dx.\end{array}$$