Product Rule for Sequence in Normed Algebra
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Theorem
Let $\GF \in \set {\R, \C}$.
Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra over $\GF$.
Let $\sequence {a_n}_{n \in \N}$ and $\sequence {b_n}_{n \in \N}$ be sequences in $A$ converging to $a$ and $b$ respectively.
Then:
- $a_n b_n \to a b$
Proof
From Convergent Sequence in Normed Vector Space is Bounded, there exists $M > 0$ such that:
- $\norm {a_n} \le M$ for each $n \in \N$.
We have for $n \in \N$:
\(\ds \norm {a_n b_n - a b}\) | \(=\) | \(\ds \norm {a_n b_n - a_n b + a_n b - a b}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm {a_n \paren {b_n - b} + b \paren {a_n - a} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {a_n \paren {b_n - b} } + \norm {b \paren {a_n - a} }\) | Norm Axiom $\text N 3$: Triangle Inequality | |||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {a_n} \norm {b_n - b} + \norm b \norm {a_n - a}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds M \norm {b_n - b} + \norm b \norm {a_n - a}\) | ||||||||||||
\(\ds \) | \(\to\) | \(\ds 0\) |
So from Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence, we have:
- $a_n b_n \to a b$
$\blacksquare$