Group Homomorphism/Examples/Pointwise Addition on Continuous Real Functions on Closed Unit Interval/Example 5
Example of Group Homomorphism
Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr C} J$ denote the set of all continuous real functions from $J$ to $\R$.
Let $G = \struct {\map {\mathscr C} J, +}$ be the group formed on $\map {\mathscr C} J$ by pointwise addition.
Let $\struct {\R, +}$ denote the additive group of real numbers.
From Pointwise Addition on Continuous Real Functions on Closed Unit Interval forms Group we have that $G$ is indeed a group.
Let $\phi: \struct {\map {\mathscr C} J, +} \to \struct {\R, +}$ be the mapping defined as:
- $\forall f \in \map {\mathscr C} J: \map \phi f = \ds \int_0^1 \map \cos {\dfrac {\pi \map f x} 6} \rd x$
Then $\phi$ is not a homomorphism.
Proof
Let $f, g \in \map {\mathscr C} J$ be arbitrary.
As both $f$ and $g$ are continuous real functions on $J$, both $\map \cos {\dfrac {\pi \map f x} 6}$ and $\map \cos {\dfrac {\pi \map g x} 6}$ are integrable on $\closedint 0 1$.
We have:
\(\ds \map \phi {f + g}\) | \(=\) | \(\ds \int_0^1 \map \cos {\dfrac {\pi \map {\paren {f + g} } x} 6} \rd x\) | Definition of $\phi$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^1 \map \cos {\dfrac {\pi \paren {\map f x + \map g x} } 6} \rd x\) | Definition of Pointwise Addition of Real-Valued Functions | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^1 \paren {\map \cos {\dfrac {\pi \map f x} 6} \map \cos {\dfrac {\pi \map g x} 6} - \map \sin {\dfrac {\pi \map f x} 6} \map \sin {\dfrac {\pi \map g x} 6} } \rd x\) | Cosine of Sum |
However:
\(\ds \map \phi f + \map \phi g\) | \(=\) | \(\ds \int_0^1 \map \cos {\dfrac {\pi \map f x} 6} \rd x + \int_0^1 \map \cos {\dfrac {\pi \map g x} 6} \rd x\) | Definition of $\phi$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^1 \paren {\map \cos {\dfrac {\pi \map f x} 6} + \map \cos {\dfrac {\pi \map g x} 6} } \rd x\) | Linear Combination of Definite Integrals |
Thus:
- $\map \phi {f + g} \ne \map \phi f + \map \phi g$
and so $\phi$ is not a homomorphism by definition.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.18 \ \text {(e)}$