(1)

\begin{align*} H_{a}\left(x\right)-H_{b}\left(x\right) & =\begin{cases} a-b & x=0\\ 0 & x\ne0 \end{cases}\\ & =\left(a-b\right)\delta_{0,x} \end{align*} より、
\[ H_{a}\left(x\right)=H_{b}\left(x\right)+\left(a-b\right)\delta_{0,x} \]

(2)

(1)より\(\delta\)について解くと導出できる。

(3)

\begin{align*} H_{a}\left(x\right) & =H_{b}\left(x\right)+\left(a-b\right)\delta_{0,x}\\ & =H_{b}\left(x\right)+\frac{a-b}{c-d}\left(H_{c}\left(x\right)-H_{d}\left(x\right)\right) \end{align*}

(4)

\begin{align*} H_{a}\left(x\right) & =H_{0}\left(x\right)+a\delta_{0,x}\\ & =H_{0}\left(x\right)+a\left(H_{1}\left(x\right)-H_{0}\left(x\right)\right)\\ & =aH_{1}\left(x\right)+\left(1-a\right)H_{0}\left(x\right) \end{align*}