(1)

\begin{align*} xH_{c}\left(x\right) & =\begin{cases} x & 0<x\\ 0 & x\leq0 \end{cases}\\ & =\max\left(x,0\right) \end{align*}

(1)

\begin{align*} H\left(x\right) & =\frac{\sgn\left(x\right)+1}{2}\\ & =\frac{\sgn\left(x\right)x+x}{2x}\\ & =\frac{\left|x\right|+x}{2x} \end{align*}

(2)

(1)より、
\begin{align*} xH_{c}\left(x\right) & =\max\left(x,0\right)\\ & =\frac{x+0+\left|x-0\right|}{2}\cmt{\because\max\left(a,b\right)=\frac{a+b+\left|a-b\right|}{2}}\\ & =\frac{x+\left|x\right|}{2} \end{align*} となるので与式は成り立つ。

(3)

\begin{align*} H_{\frac{1}{2}}\left(\pm x\right) & =\frac{\sgn\left(\pm x\right)+1}{2}\\ & =\frac{1\pm\sgn\left(x\right)}{2} \end{align*}

(3)-2

\(x\ne0\)とする。
\begin{align*} H\left(\pm x\right) & =\frac{\left|\pm x\right|\pm x}{\pm2x}\\ & =\frac{\left|x\right|\pm x}{\pm2x}\\ & =\frac{\pm\left|x\right|+x}{2x}\\ & =\frac{1\pm\frac{\left|x\right|}{x}}{2}\\ & =\frac{1\pm\sgn x}{2} \end{align*} \(x=0\)のとき右辺は\(\frac{1}{2}\)になるので、
\[ H_{\frac{1}{2}}\left(\pm x\right)=\frac{1\pm\sgn x}{2} \]

(4)

\begin{align*} 1\pm1 & =1+\sgn\left(\pm1\right)\\ & =1+2H\left(\pm1\right)-1\\ & =2H\left(\pm1\right) \end{align*}

(4)-2

\begin{align*} 1\pm1 & =1+\left\{ H\left(\pm1\right)-H\left(\mp1\right)\right\} \\ & =H\left(\pm1\right)+1-H\left(\mp1\right)\\ & =H\left(\pm1\right)+H\left(\pm1\right)\\ & =2H\left(\pm1\right) \end{align*}

(5)

\begin{align*} xH_{c}\left(x\right)-xH_{c}\left(-x\right) & =\begin{cases} x & 0\leq x\\ -x & x<0 \end{cases}\\ & =\left|x\right| \end{align*}