The Positive and Negative Parts of a Function

# The Positive and Negative Parts of a Function

 Definition: Let $f$ be a function on the interval $I$. The Positive Part of $f$ denoted $f^+$ is the function defined for all $x \in I$ by $f^+(x) = \max \{ f(x), 0 \}$. Similarly, the Negative Part of $f$ denoted $f^-$ is the function defined for all $x \in I$ by $f^-(x) = \max \{ -f(x), 0 \}$.

For example, if the graph of $f$ is:

The the graphs of the positive part, $f^+$, and negative part, $f^-$ are:

It is important to note that $f^+(x), f^-(x) \geq 0$ for all $x \in I$. We will look more thoroughly into some of the properties of these functions in the following theorem.

 Theorem 1: Let $f$ be a function defined on the interval $I$. Then $f = f^+ - f^-$.
• Proof: Let $f$ be defined on $I$. Define $P = \{ x \in I : f(x) \geq 0 \}$ and $N = \{ x \in I : f(x) < 0 \}$. Then $I = P \cup N$.
• On $P$ we have that $f(x) = f^+(x)$ and $f^+(x) = 0$ on $N$. Furthermore, on $N$ we have that $f(x) = -f^-(x)$ and $-f^-(x) = 0$ on $P$. So on all of $I$ we have that $f(x) = f^+(x) - f^-(x)$, i.e., $f = f^+ - f^-$. $\blacksquare$
 Theorem 2: Let $f$ be a function defined on the interval $I$. Then $\mid f \mid = f^+ + f^-$.
• Proof: Let $f$ be defined on $I$ and let $P$ and $N$ be as above. Then $\mid f(x) \mid = f^+(x)$ on $P$ and $f^+(x) = 0$ on $N$. Furthermore, $\mid f(x) \mid = f^{-}(x)$ on $N$ and $f^{-}(x) = 0$ on $P$.
• So on all of $I$ we have that $\mid f(x) \mid = f^+(x) + f^-(x)$, i.e., $\mid f \mid = f^+ + f^-$. $\blacksquare$