Combination Theorem for Sequences/Normed Division Ring/Product Rule/Proof 2

From ProofWiki
Jump to: navigation, search

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:

$\displaystyle \lim_{n \mathop \to \infty} x_n = l$
$\displaystyle \lim_{n \mathop \to \infty} y_n = m$

Then:

$\sequence {x_n y_n}$ is convergent to the limit $\displaystyle \lim_{n \mathop \to \infty} \paren {x_n y_n} = l m$


Proof

Because $\sequence {x_n}$ is convergent, it is a bounded by Convergent Sequence in Normed Division Ring is Bounded.

Suppose $\norm {x_n} \le K$ for $n = 1, 2, 3, \ldots$.

Then for $n = 1, 2, 3, \ldots$:

\(\displaystyle \norm {x_n y_n - l m}\) \(=\) \(\displaystyle \norm {x_n y_n - x_n m + x_n m - l m}\) $\quad$ $\quad$
\(\displaystyle \) \(\le\) \(\displaystyle \norm {x_n y_n - x_n m} + \norm {x_n m - l m}\) $\quad$ Axiom (N3) of norm (Triangle Inequality) $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \norm {x_n} \norm {y_n - m} + \norm {x_n - l} \norm m\) $\quad$ Axiom (N2) of norm (Multiplicativity) $\quad$
\(\displaystyle \) \(\le\) \(\displaystyle K \norm {y_n - m} + \norm m \norm {x_n - l}\) $\quad$ since $\sequence {x_n}$ is bounded by $K$ $\quad$
\(\displaystyle \) \(=:\) \(\displaystyle z_n\) $\quad$ $\quad$

We note that $\sequence {z_n}$ is a real sequence.

But $x_n \to l$ as $n \to \infty$.

So $\norm {x_n - l} \to 0$ as $n \to \infty$ from Definition:Convergent Sequence in Normed Division Ring.

Similarly $\norm {y_n - m} \to 0$ as $n \to \infty$.

From the Combined Sum Rule for Real Sequences:

$\displaystyle \lim_{n \mathop \to \infty} \paren {\lambda x'_n + \mu y'_n} = \lambda l' + \mu m'$, $z_n \to 0$ as $n \to \infty$

By applying the Squeeze Theorem for Sequences of Complex Numbers (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:

$\displaystyle \lim_{n \mathop \to \infty} \paren {x_n y_n} = l m$

$\blacksquare$


Sources