Absolute Value of Function is Composite with Absolute Value Function
Jump to navigation
Jump to search
This article needs proofreading. Please check it for mathematical errors. If you believe there are none, please remove {{Proofread}} from the code.To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Proofread}} from the code. |
Theorem
Let $S$ be a set.
Let $\R$ denote the real number line.
Let $f: S \to \R$ be real-valued function.
Let $\size f$ denote the absolute value of $f$, that is, $\size f$ is the mapping defined by:
- $\forall s \in S : \map {\size f} s = \size{\map f s}$
Then:
- $\size f = \size{\,\cdot\,} \circ f$
where:
- $\size{\,\cdot\,} : \R \to \R$ denotes the absoute value function
- $\size{\,\cdot\,} \circ f$ denotes the composite mapping of $\size{\,\cdot\,}$ and $f$
Proof
We have:
\(\ds \forall s \in S: \, \) | \(\ds \map {\size f } s\) | \(=\) | \(\ds \size{\map f s}\) | Definition of Absolute Value of Real-Valued Function | ||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren{\size{\,\cdot\,} \circ f} } s\) | Definition of Composite Mapping |
By definition of equality of mappings:
- $\size f = \size{\,\cdot\,} \circ f$
$\blacksquare$