Combination Theorem for Sequences/Normed Division Ring/Product Rule
Theorem
Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring.
Let $\sequence {x_n}$, $\sequence {y_n} $ be sequences in $R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent in the norm $\norm {\,\cdot\,}$ to the following limits:
- $\ds \lim_{n \mathop \to \infty} x_n = l$
- $\ds \lim_{n \mathop \to \infty} y_n = m$
Then:
- $\sequence {x_n y_n}$ is convergent to the limit $\ds \lim_{n \mathop \to \infty} \paren {x_n y_n} = l m$
Proof 1
By Convergent Sequence in Normed Division Ring is Bounded, $\sequence {x_n}$ is bounded.
Suppose $\norm {x_n} \le K$ for $n = 1, 2, 3, \ldots$.
Let $M = \max \set {K, \norm m}$.
Then:
- $\norm m \le M$
and:
- $\forall n: \norm{x_n} \le M$
Let $\epsilon > 0$ be given.
Then $\dfrac \epsilon {2 M} > 0$.
As $\sequence {x_n}$ converges to $l$, we can find $N_1$ such that:
- $\forall n > N_1: \norm {x_n - l} < \dfrac \epsilon {2 M}$
Similarly, for $\sequence {y_n}$ we can find $N_2$ such that:
- $\forall n > N_2: \norm {y_n - m} < \dfrac \epsilon {2 M}$
Now let $N = \max \set {N_1, N_2}$.
Then if $n > N$, both the above inequalities will be true.
Thus $\forall n > N$:
| \(\ds \norm {x_n y_n - l m}\) | \(=\) | \(\ds \norm {x_n y_n - x_n m + x_n m - l m}\) | ||||||||||||
| \(\ds \) | \(\le\) | \(\ds \norm {x_n y_n - x_n m} + \norm {x_n m - l m}\) | Axiom (N3) of norm (Triangle Inequality). | |||||||||||
| \(\ds \) | \(=\) | \(\ds \norm {x_n \paren {y_n - m } } + \norm {\paren {x_n - l } m}\) | ||||||||||||
| \(\ds \) | \(\le\) | \(\ds \norm {x_n} \cdot \norm {y_n - m} + \norm {x_n - l} \cdot \norm m\) | Axiom (N2) of norm (Multiplicativity). | |||||||||||
| \(\ds \) | \(\le\) | \(\ds M \cdot \norm {y_n - m} + \norm {x_n - l} \cdot M\) | since $\sequence {x_n}$ is bounded by $M$ and $\norm m \le M$ | |||||||||||
| \(\ds \) | \(\le\) | \(\ds M \cdot \dfrac \epsilon {2 M} + \dfrac \epsilon {2 M} \cdot M\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac \epsilon 2 + \dfrac \epsilon 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \epsilon\) |
Hence:
- $\sequence {x_n y_n}$ is convergent.
It follows that:
- $\ds \lim_{n \mathop \to \infty} \paren {x_n y_n} = l m$
$\blacksquare$
Proof 2
By Convergent Sequence in Normed Division Ring is Bounded, $\sequence {x_n}$ is bounded.
Suppose $\norm {x_n} \le K$ for $n = 1, 2, 3, \ldots$.
Then for $n = 1, 2, 3, \ldots$:
| \(\ds \norm {x_n y_n - l m}\) | \(=\) | \(\ds \norm {x_n y_n - x_n m + x_n m - l m}\) | ||||||||||||
| \(\ds \) | \(\le\) | \(\ds \norm {x_n y_n - x_n m} + \norm {x_n m - l m}\) | Norm Axiom $\text N 3$: Triangle Inequality | |||||||||||
| \(\ds \) | \(=\) | \(\ds \norm {x_n} \norm {y_n - m} + \norm {x_n - l} \norm m\) | Norm Axiom $\text N 2$: Multiplicativity | |||||||||||
| \(\ds \) | \(\le\) | \(\ds K \norm {y_n - m} + \norm m \norm {x_n - l}\) | as $\sequence {x_n}$ is bounded by $K$ | |||||||||||
| \(\ds \) | \(=:\) | \(\ds z_n\) |
We note that $\sequence {z_n}$ is a real sequence.
But $x_n \to l$ as $n \to \infty$.
So from Definition:Convergent Sequence in Normed Division Ring:
- $\norm {x_n - l} \to 0$ as $n \to \infty$
Similarly $\norm {y_n - m} \to 0$ as $n \to \infty$.
From the Combined Sum Rule for Real Sequences:
- $\ds \lim_{n \mathop \to \infty} \paren {\lambda x'_n + \mu y'_n} = \lambda l' + \mu m'$, $z_n \to 0$ as $n \to \infty$
By applying the Squeeze Theorem for Complex Sequences (which applies as well to real as to complex sequences):
- $\sequence {\norm {x_n y_n - l m}}$ converges to $0$ in $\R$.
By definition of a convergent sequence in a normed division ring:
- $\sequence{x_n y_n}$ is convergent in $R$
It follows that:
- $\ds \lim_{n \mathop \to \infty} \paren {x_n y_n} = l m$
$\blacksquare$