Sequence of points in $\mathbb{R}^{n}$

Question

Suppose that $(u_{k})$ is a sequence of points in $\mathbb{R}^{n}$ that converges to a point $u$. Assume $u_{k} \neq \textbf{0}$ for all $k$ and $u \neq \textbf{0}$. Prove that the sequence $w_{k} = \frac{1}{||u_{k}||} u_{k}$ converges to $w = \frac{1}{||u||} u$

Attempt: I tried to show $(w_{k})$ converges to $w$ component-wise. However I am having trouble working on it. Here is what I have so far:

Since $\lim_{k \rightarrow \infty}(u_{k}) = u$, by component wise convergence $\lim_{k \rightarrow \infty}p_{i}(u_{k}) = p_{i}(u)$.

So,

$||p_{i}(w_{k}) - p_{i}(w)|| =\Big| \Big| \frac{1}{||u_{k}||} p_{i}(u_{k}) - \frac{1}{||u||} p_{i}(u)\Big| \Big| = \Big| \Big| \frac{||u||*p_{i}(u_{k}) - ||u_{k}|| p_{i}(u)}{||u_{k}||*||u||} \Big| \Big| $.

I am not sure what to do here. Do I use Cauchy-Schwartz? Am I on the right track at least?

Thank you for your help!!

Answer 1

$$\begin{align} \lVert w_k-w\rVert &= \frac{1}{\lVert u\rVert\cdot\lVert u_k\rVert}\big\lVert\lVert u\rVert u_k-\lVert u_k\rVert u\big\rVert\\ &=\frac{1}{\lVert u\rVert\cdot\lVert u_k\rVert}\big\lVert\lVert u\rVert u_k-\lVert u\rVert u+\lVert u\rVert u-\lVert u_k\rVert u\big\rVert\\ &\le\frac{1}{\lVert u\rVert\cdot\lVert u_k\rVert}\left(\lVert u\rVert\cdot\lVert u_k-u\rVert+\big\lvert\lVert u\rVert-\lVert u_k\rVert\big\rvert\cdot \lVert u\rVert\right)\\ &\le\frac{1}{\lVert u\rVert\cdot\lVert u_k\rVert}\left(\lVert u\rVert\cdot\lVert u_k-u\rVert+\lVert u-u_k\rVert\cdot \lVert u\rVert\right)\\ &=\frac{2\lVert u_k-u\rVert}{\lVert u_k\rVert}\to\frac{2\cdot 0}{\lVert u\rVert}=0 \end{align}$$

Answer 2

You don't need to work component wise. If $(X, \| \cdot \|)$ is a real normed space, then we know

1. $u_k \to u \implies \| u_k \| \to \| u \|$
2. $\alpha_k \to \alpha$ and $u_k \to u$ $\implies$ $\alpha_k u_k \to \alpha u$

Here $(X, \| \cdot \|) = (\mathbb{R}^n, \| \cdot \|_2)$ and $\alpha_k = \frac{1}{\| u_k \|}$. The hypotheses $u_k, u \neq 0$ are to ensure that everything is defined.

Proof of 1.: We have \begin{align} \| x \| &= \| (x - y) + y \| \\ &\leq \| x - y \| + \| y \| && \text{triangle inequality} \end{align} hence $$ \| x \| - \| y \| \leq \| x - y \| \tag{1} $$

Similarly \begin{align} \| y \| - \| x \| &\leq \| y - x \| \\ &= \| (-1)(x - y) \| \\ &= \| x - y \| && \text{since }\|\alpha x\| = |\alpha| \| x \| \tag{2} \end{align}

From $(1)$ and $(2)$, it follows that $$ \Bigg| \| x \| - \| y \| \Bigg| \leq \| x - y \| \tag{3} $$

Now, by $(3)$, it is clear that \begin{align} u_k \to u &\iff \| u_k - u \| \to 0 \\ &\,\implies \Bigg| \| u_k \| - \| u \| \Bigg| \to 0 \\ &\iff \| u_k \| \to \| u \| \end{align}

Proof of 2.: Note that \begin{align} \alpha_k u_k - \alpha u &= \alpha_k (u_k - u + u) - \alpha u \\ &=\alpha_k (u_k - u) + (\alpha_k - \alpha) u \end{align}

hence \begin{align} \| \alpha_k u_k - \alpha u \| &\leq |\alpha_k| \| u_k - u \| + |\alpha_k - \alpha| \| u \| \\ &\leq M \| u_k - u \| + |\alpha_k - \alpha| \| u \| \tag{4}\\ &\to 0 \tag{5} \end{align}

where $(4)$ follows from the fact that $(\alpha_k)$ is a convergent sequence hence a bounded one and $(5)$ follows from the fact that $\| u \|$ is a constant.

Answer 3

Lemma 1: If $u_n$ converges to $u,$ then $\|u_n\|$ converges to $\|u\|.$

Proof: for by triangle inequality $|\|u\| - \|u_n\|| \leq \|u - u_n\| \to 0.$

Lemma 2: If $\lambda_n \in \Bbb R$ converges to $\lambda$ and $u_n$ to $u$ then $\lambda_n u_n$ converges to $\lambda u.$

Proof: as $\|u_n\|$ converges to $\|u\|,$ by Lemma 1, there is some $n_0$ for which the relation "no matter what $n \geq n_0$ should be, $\|u_n\| \leq \|u\| + 1 = \mathrm{A}$" is true. Now, by adding a convenient zero $\| \lambda u - \lambda_n u_n \| \leq |\lambda| \|u - u_n\| + |\lambda - \lambda_n| \| u_n \| \leq |\lambda| \|u - u_n\| + \mathrm{A} |\lambda - \lambda_n| \to 0.$

To your exercise. Define $\lambda_n = \dfrac{1}{\|u_n\|}$ which converges towards $\lambda = \dfrac{1}{\|u\|}$ by Lemma 1 (& basic laws os sequences) and proceed in the obvious way ($\lambda_n u_n$ converges to $\lambda u$).