Pointwise Difference is Pointwise Addition with Negation

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Theorem

Let $S$ be a set.

Let $\R$ denote the real number line.

Let $f, g :S \to \R$ be real-valued functions.

Let $f - g$ denote the pointwise difference of $f$ and $g$, that is, $f - g$ is the mapping defined by:

$\forall s \in S : \map {\paren{f - g} } s = \map f s - \map g s$

Then:

$f - g = f + \paren{-g}$

where:

$-g$ denotes the pointwise negation of $g$

$f + \paren{-g}$ denotes the pointwise addition of $f$ and $-g$

Proof

We have:

\(\ds \forall s \in S: \, \)

\(\ds \map {\paren{f - g} } s\)

\(=\)

\(\ds \map f s - \map g s\)

Definition of Pointwise Difference of Real-Valued Functions

\(\ds \)

\(=\)

\(\ds \map f s + \paren{- \map g s}\)

\(\ds \)

\(=\)

\(\ds \map f s + \map {\paren{-g} } s\)

Definition of Pointwise Negation of Real-Valued Function

\(\ds \)

\(=\)

\(\ds \map {\paren{f + \paren{-g} } } s\)

Definition of Pointwise Addition of Real-Valued Functions

By definition of equality of mappings:

$f - g = f + \paren{-g}$

$\blacksquare$