The positive and negative parts of a function

Let $f:X\to \bar{\mathbb{R}}$ with $\bar{\mathbb{R}}$ as in The extended real line $\bar{\mathbb{R}}=\mathbb{R}\cup \{-\mathrm{\infty },+\mathrm{\infty }\}$, its order, and the arithmetic that is left undefined. Its positive part and negative part are

$f^{+}:=\max \{f,0\},f^{-}:=\max \{-f,0\}\mathrm{.}$

All maxima, negations, and sums here are pointwise, using the total order, reflection and partial addition of The extended real line $\bar{\mathbb{R}}=\mathbb{R}\cup \{-\mathrm{\infty },+\mathrm{\infty }\}$, its order, and the arithmetic that is left undefined. Both parts take values in $[0,+\mathrm{\infty }]$. For an extended-real-valued function, define the magnitude function by $\mathrm{\mid }f\mathrm{\mid }(x):=\max \{f(x),-f(x)\}$. This is function notation; it does not introduce an absolute-value operation into the scalar arithmetic of that item. Likewise the notation $f^{+}-f^{-}$ below abbreviates the pointwise sum $f^{+}+(-f^{-})$, whose definedness is verified here.

At each point at least one of $f^{+}$ and $f^{-}$ is zero. If $f(x)\ge 0$, then $f^{+}(x)=f(x)$ and $f^{-}(x)=0$; if $f(x)\le 0$, then $f^{+}(x)=0$ and $f^{-}(x)=-f(x)$. These cases include both infinite values and show that the following sums are defined and satisfy

$f=f^{+}+(-f^{-}),\mathrm{\mid }f\mathrm{\mid }=f^{+}+f^{-}\mathrm{.}$

Thus the familiar notation $f=f^{+}-f^{-}$ retains exactly these pointwise values; no subtraction of two infinite nonzero values occurs.