Limits of two sequences of vectors being parallel?

Question

If ${u_k}$ is a sequence in a vector space $X$ with $u_k \to 0$, and there exist $v, w \in X$ and two real sequences $t_k \to 0, s_k \to 0$, such that $\frac{u_k}{t_k} \to v$ and $\frac{u_k}{s_k} \to w$, must it be the case that $v$ and $w$ are colinear, i.e., $\exists \lambda \in \mathbb{R}$, s.t. $ v = \lambda w$?

I'm having trouble proving this, but also can't come up with counter-examples...

P.S. By $ x_k \to y$, of course I mean $\lim_{k\to\infty}x_k = y$. And I'm currently looking just at $X$ being the Euclidean space with the standard $\ell_2$ norm/metric.

Answer 1

This indeed holds. Let $(t_k)_{k\in \mathbb{N}}$ be a non-zero sequence, $t_k >0$ for all $k$ (the case of a negative sequence is similar), $(u_k)_{k\in \mathbb{N}} \in X, \lim_{k\to\infty} u_k = 0$, and $\lim_{k\to\infty} \frac{u_k}{t_k}=v\neq 0$ (the statement trivially holds if $v=0$).

We just need one simple lemma, a limit law for a real and a vector sequence (this was used implicitly in @Servaes's answer): Let $(a_k)_{k\in \mathbb{N}}$ be a sequence of reals, $(b_k)_{k\in \mathbb{N}}$ be a sequence in a vector space $X$, $a \in \mathbb{R}$, $b \in X$, such that $\lim_{k\to\infty} a_k = a, \lim_{k\to\infty} b_k = b$. Then $\lim_{k\to\infty} a_k b_k = ab$.

Now observe that $\lim_{k\to\infty} \frac{u_k}{t_k}=v$ implies $\lim_{k\to\infty} \frac{\|u_k\|}{t_k} = \lim_{k\to\infty} \frac{\|u_k\|}{|t_k|}=\|v\|$, so $\lim_{k\to\infty} \frac{t_k}{\|u_k\|} = \frac{1}{\|v\|}$ (since $v\neq 0$). By our lemma, we then have $$\lim_{k\to\infty} \frac{u_k}{\|u_k\|} = \lim_{k\to\infty} \frac{u_k}{t_k} \frac{t_k} {\|u_k\|} = (\lim_{k\to\infty} \frac{u_k}{t_k}) (\lim_{k\to\infty} \frac{t_k} {\|u_k\|}) = \frac{v}{\|v\|} $$

If we also have $\lim_{k\to\infty} \frac{u_k}{s_k}=w\neq 0$ (again assuming $s_k > 0$; you need a negative sign in the following limit if $s_k < 0$), then by the same argument as above, $$\lim_{k\to\infty} \frac{u_k}{\|u_k\|} = \frac{w}{\|w\|} $$

By the uniqueness of limits, we must have $$ \frac{v}{\|v\|} = \frac{w}{\|w\|}$$ (or, $\frac{v}{\|v\|} = - \frac{w}{\|w\|}$, when $t_k$ and $s_k$ have different signs), which shows that $u$ and $w$ are parallel (and in fact point in the same direction as $\lim_{k\to\infty} \frac{u_k}{\|u_k\|}$).

Answer 2

Let $(s_k)_{k\in\Bbb{N}}$ and $(t_k)_{k\in\Bbb{N}}$ be two nonzero real sequences and $(u_k)_{k\in\Bbb{N}}$ a sequence in $X$ such that $$\lim_{k\to\infty}s_k=\lim_{k\to\infty}t_k=0, \qquad \lim_{k\to\infty}u_k=0, \qquad \lim_{k\to\infty}\tfrac{u_k}{s_k}=v \qquad\text{ and }\qquad \lim_{k\to\infty}\tfrac{u_k}{t_k}=w.$$ If either $v=0$ or $w=0$ then $v$ and $w$ are collinear. So suppose that $v$ and $w$ are nonzero.

Because the inner product on $X$ is continuous we have \begin{eqnarray*} ||v||^2=\langle v,v\rangle &=&\lim_{k\to\infty}\langle\tfrac{u_k}{s_k},\tfrac{u_k}{s_k}\rangle =\lim_{k\to\infty}\tfrac{||u_k||^2}{s_k^2}\\ \langle v,w\rangle &=&\lim_{k\to\infty}\langle\tfrac{u_k}{s_k},\tfrac{u_k}{t_k}\rangle =\lim_{k\to\infty}\tfrac{||u_k||^2}{s_kt_k}, \end{eqnarray*} where $\langle v,v\rangle\neq0$. It follows that $\tfrac{\langle v,w\rangle}{||v||^2} =\lim_{k\to\infty}\tfrac{s_k}{t_k}$, and hence that $$\frac{\langle v,w\rangle}{||v||^2}v =\lim_{k\to\infty}\tfrac{s_k}{t_k}\lim_{k\to\infty}\tfrac{u_k}{s_k} =\lim_{k\to\infty}\tfrac{u_k}{t_k}=w,$$ which shows that $w$ is a scalar multiple of $v$.