シャープネスの計算


\newcommand{\then}{\mathop{?} }%
\newcommand{\else}{\mathop{:} }%
(\Delta_B \circ f)(\beta, \beta' | a) \\
= \sum_{b\in B} \Delta_B(\beta, \beta' | b) f(b | a) \\
= \sum_{b\in B} eq_B(\beta, \beta', b) f(b | a) \\
= (\, (\beta = \beta')\then \sum_{b\in B} eq(\beta, b)f(b | a) \else 0 \,) \\
= (\, (\beta = \beta')\then f(\beta | a) \else 0 \,) \\


( (f\otimes f)\circ \Delta_A)(\beta, \beta' | a) \\
=  \sum_{(\alpha, \alpha')\in A\times A}(f\otimes f)(\beta, \beta' | \alpha, \alpha')\Delta_A(\alpha, \alpha' | a) \\
=  \sum_{(\alpha, \alpha')\in A\times A}f(\beta | \alpha)f(\beta' | \alpha')\Delta_A(\alpha, \alpha' | a) \\
=  \sum_{(\alpha, \alpha')\in A\times A}f(\beta | \alpha)f(\beta' | \alpha')eq_A(\alpha, \alpha', a) \\
=  \sum_{\alpha \in A}f(\beta| \alpha )f(\beta'| \alpha )eq(\alpha, a) \\
= f(\beta| a )f(\beta'| a )

以上より:


(\, (\beta = \beta')\then f(\beta | a) \else 0 \,) = f(\beta| a )f(\beta'| a ) \\
\mbox{i.e} \\
\forall a\in A.\\
\forall \beta, \beta' \in B.\\
\:\:\beta = \beta' \Rightarrow  f(\beta | a) = f(\beta | a)^2
\:\:\beta \ne \beta' \Rightarrow f(\beta | a)f(\beta' | a) = 0\\
\mbox{i.e} \\
\forall a\in A, \beta \in B.\\
\:\:f(\beta | a) = f(\beta | a)^2\\
\forall a\in A. \forall \beta, \beta' \in B.\\
\:\:\beta \ne \beta' \Rightarrow  f(\beta | a) = 0 \lor f(\beta' | a) = 0 \\

もう少し書き換えると:


\forall a\in A, \beta \in B.\\
\:\:f(\beta | a) = f(\beta | a)^2\\
\Leftrightarrow \\
\forall a\in A, \beta \in B.\\
\:\:f(\beta | a) = 0 \lor f(\beta | a) = 1\\

成分は0か1に限る。


\forall a\in A. \forall \beta, \beta' \in B.\\
\:\:\beta \ne \beta' \Rightarrow  f(\beta | a) = 0 \lor f(\beta' | a) = 0 \\
\Leftrightarrow \\
\forall a\in A. \forall \beta, \beta' \in B.\\
\:\: \beta = \beta' \lor ( f(\beta | a) = 0 \lor f(\beta' | a) = 0 )\\
\Leftrightarrow \\
\forall a\in A. \forall \beta, \beta' \in B.\\
\:\: \beta \ne \beta' \land  f(\beta | a) \ne 0 \Rightarrow f(\beta' | a) = 0 )

固定したaに対して、1になれるβは高々ひとつだけ。

まとめると:

  1. 成分は1か0だけ。
  2. 固定したaに対して、1になれるβは高々ひとつだけ。

シャープ射は部分関数。