関数に対する微分演算子sの作用

積分演算子: $l=\left\{ 1 \right\}$
微分演算子: $s=\frac{1}{l}=\frac{1}{\left\{ 1 \right\}}$
関数: $f(t), \, g(t)$
定数: $k$

以下が成立する．
$\left\{ f(t) \right\}+\left\{ g(t) \right\} = \left\{ f(t)+g(t) \right\}.$
$\left\{ f(t) \right\}\left\{ g(t) \right\} = \left\{ \int^t_0 f(x)g(t-x)dx \right\}.$
$k \left\{ 1 \right\}=lk=\left\{ k \right\}.$
$l\left\{ f(t) \right\}=\left\{ 1 \right\}\left\{ f(t) \right\}= \left\{ \int^t_0 1 \cdot f(x)dx \right\}.$
$ls=sl=\frac{\left\{ 1 \right\}}{\left\{ 1 \right\}}=1.$

微分 演算子の微分作用

$\left\{ f(t) \right\} =\left\{ f(t)-f(0)+f(0) \right\} = \left\{ \left[ f(x) \right]^{x=t}_{x=0}+ f(0) \right\} \\ = \left\{ \int^{t}_{0} f'(x) dx+ a(0) \right\} = \left\{ \int^{t}_{0} 1 \cdot f'(x) dx\right\}+\left\{ f(0) \right\} \\ = l \left\{ f'(t) \right\}+l\,f(0).$
辺々に微分演算子sを掛ける．
$s\left\{ f(t) \right\} = sl \left\{ f'(t) \right\}+sl\,f(0) = \left\{ f'(t) \right\}+f(0) .$