Multiple Rule for Sequence in Normed Vector Space

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\Bbb F$ be a subfield of $\C$.

Let $\struct {X, \norm \cdot}$ be a normed vector space over $\Bbb F$.

Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$ converging to $x$.

Let $\lambda \in \Bbb F$.


Then:

$\lambda x_n \to \lambda x$


Proof

The case $\lambda = 0$ follows from Constant Sequence in Normed Vector Space Converges.

Now take $\lambda \ne 0$.

Since $\sequence {x_n}_{n \mathop \in \N}$ converges to $x$, for each $\epsilon > 0$ there exists $N \in \N$ such that:

$\ds \norm {x_n - x} < \frac \epsilon {\cmod \lambda}$ for all $n \ge N$.

Noting that:

$\ds \cmod \lambda \norm {x_n - x} = \norm {\lambda x_n - \lambda x}$

from the norm axioms, we have:

$\ds \norm {\lambda x_n - \lambda x} < \epsilon$ for all $n \ge N$.

Since $\epsilon$ was arbitrary, it follows that:

$\lambda x_n \to \lambda x$

$\blacksquare$