Convergence of Product of Convergent Scalar Sequence and Convergent Vector Sequence in Normed Vector Space

Theorem
Let $\Bbb F \in \set {\R, \C}$.

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

Let $\alpha \in \R$.

Let $x \in V$.

Let $\sequence {\alpha_n}_{n \mathop \in \N}$ be a real sequence in $\Bbb F$ such that:


 * $\alpha_n \to \alpha$

Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $V$ such that:


 * $x_n \to x$

in $\struct {V, \norm \cdot}$.

Then we have:


 * $\alpha_n x_n \to \alpha x$

in $\struct {V, \norm \cdot}$.

Proof
Let $\epsilon > 0$.

We have:

From Convergent Real Sequence is Bounded, there exists $M > 0$ such that we have:


 * $\size {\alpha_n} \le M$

for each $n \in \N$.

Then we have:


 * $\norm {\alpha_n x_n - \alpha x} \le M \norm {x_n - x} + \norm x \size {\alpha_n - \alpha}$

Suppose that:


 * $\norm x = 0$

then, from positive definiteness of $\norm \cdot$:


 * $x = 0$

In this case:


 * $x_n \to 0$

and:


 * $\norm {\alpha_n x_n} = \norm {\alpha_n x_n - \alpha x} \le M \norm {x_n - x} = M \norm {x_n}$

Then we can choose $N \in \N$ such that:


 * $\ds \norm {x_n} < \frac \epsilon M$

for $n > N$.

From which we have:


 * $\norm {\alpha_n x_n} < M$

for $n > N$.

So, from the definition of a convergent sequence, we have:


 * $\alpha_n x_n \to 0 = \alpha x$

in the case $\norm x = 0$.

Now suppose that:


 * $\norm x \ne 0$

Now suppose that:


 * $\norm x \ne 0$

Since:


 * $\alpha_n \to \alpha$

we can pick $N_1 \in \N$ such that:


 * $\ds \size {\alpha_n - \alpha} < \frac \epsilon {2 \norm x}$

for $n \ge N_1$, from the definition of convergence.

Since we also have:


 * $x_n \to x$

we can pick $N_2 \in \N$ such that:


 * $\ds \norm {x_n - x} < \frac \epsilon {2 M}$

from the definition of a convergent sequence.

Let:


 * $N = \max \set {N_1, N_2}$

We then have, for $n > N$:

So, from the definition of a convergent sequence, we have:


 * $\alpha_n x_n \to \alpha x$

in the case $x \ne 0$ also, hence the result.