Product Rule for Sequence in Normed Algebra

From ProofWiki
Jump to navigation Jump to search


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.


$a_n b_n \to a b$


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:

$a_n b_n \to a b$