Group Homomorphism/Examples/Pointwise Addition on Continuous Real Functions on Closed Unit Interval/Example 8/Kernel

From ProofWiki
Jump to navigation Jump to search

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 $G = \struct {\map {\mathscr C} J, +}$ denote the group formed on $\map {\mathscr C} J$ by pointwise addition.

Let $\struct {\R, +}$ denote the additive group of real numbers.

Let $I_J$ denote the identity mapping on $J$:

$\forall x \in J: \map {I_J} x = x$


Let $\phi: \struct {\map {\mathscr C} J, +} \to \struct {\R, +}$ be the homomorphism defined as:

$\forall f \in \map {\mathscr C} J: \map \phi f = \ds \int_0^1 \int_0^1 \map f {x y} \rd y \rd x$


The kernel of $\phi$ is given by:

$\map \ker \phi = I_J - f_m$

where:

$f_m: \R \to \R$ denotes the constant mapping on $\R$
$m = \dfrac 1 4$
$I_J$ denotes the identity mapping on $J$.


Proof

From Group Homomorphism: Example 8, we have that $\phi$ is indeed a homomorphism.

For all $c \in \R$, let $f_c: \R \to \R$ be the constant mapping:

$\forall x \in \R: \map {f_c} x = c$


First we show that:

$\forall c \in \R: \map \phi {f_c} = c$


Let $c \in \R$ be arbitrary.

We have:

\(\ds \map \phi {f_c}\) \(=\) \(\ds \int_0^1 \int_0^1 \map {f_c} {x y} \rd y \rd x\) Definition of $\phi$
\(\ds \) \(=\) \(\ds \int_0^1 \int_0^1 c \rd y \rd x\) Definition of Constant Mapping
\(\ds \) \(=\) \(\ds \int_0^1 \bigintlimits {c y} 0 1 \rd x\) Primitive of Constant
\(\ds \) \(=\) \(\ds \int_0^1 c \rd x\)
\(\ds \) \(=\) \(\ds \bigintlimits {c x} 0 1\) Primitive of Constant
\(\ds \) \(=\) \(\ds c\)

$\Box$


Then we show that there exists a unique $m \in \R$ such that:

$\map \phi {I_J - f_m} = 0$

where in this case:

$m = \dfrac 1 4$


We have:

\(\ds \map \phi {I_J - f_m}\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \int_0^1 \int_0^1 \map {\paren {I_J - f_m} } {x y} \rd y \rd x\) \(=\) \(\ds 0\) Definition of $\phi$
\(\ds \leadsto \ \ \) \(\ds \int_0^1 \int_0^1 \map {I_J} {x y} \rd y \rd x - \int_0^1 \int_0^1 \map {f_m} {x y} \rd y \rd x\) \(=\) \(\ds 0\) Definition of Pointwise Addition of Real-Valued Functions
\(\ds \leadsto \ \ \) \(\ds \int_0^1 \int_0^1 x y \rd y \rd x - m\) \(=\) \(\ds 0\) Definition of Identity Mapping, a priori
\(\ds \leadsto \ \ \) \(\ds \int_0^1 \intlimits {\dfrac {x y^2} 2} {y \mathop = 0} {y \mathop = 1} \rd x\) \(=\) \(\ds m\) Primitive of Power
\(\ds \leadsto \ \ \) \(\ds \int_0^1 \dfrac x 2 \rd x\) \(=\) \(\ds m\)
\(\ds \leadsto \ \ \) \(\ds \intlimits {\dfrac {x^2} 4} {x \mathop = 0} {x \mathop = 1}\) \(=\) \(\ds m\) Primitive of Power
\(\ds \leadsto \ \ \) \(\ds m\) \(=\) \(\ds \dfrac 1 4\)

Hence the result by definition of kernel.

$\blacksquare$


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 $13.18 \ \text {(h)}$