16.2 Convergence of sequences of vectors

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This modules presents two common types of convergence, pointwise and norm, and discusses their properties, differences, and relationships with one another.

Convergence of vectors

We now discuss pointwise and norm convergence of vectors. Other types of convergence also exist, and one in particular, uniform convergence , can also be studied. For this discussion , we will assumethat the vectors belong to a normed vector space .

Pointwise convergence

A sequence $(n, \left\{1\right\})$ g n converges pointwise to the limit $g$ if each element of ${g}_{n}$ converges to the corresponding element in $g$ . Below are few examples to try and help illustrate this idea.

${g}_{n}=\left(\begin{array}{c}{g}_{n}(1)\\ {g}_{n}(2)\end{array}\right)()=\left(\begin{array}{c}1+\frac{1}{n}\\ 2-\frac{1}{n}\end{array}\right)()$ First we find the following limits for our two ${g}_{n}$ 's: $\lim_{n\to }n\to$ g n 1 1 $\lim_{n\to }n\to$ g n 2 2 Therefore we have the following, $\lim_{n\to }n\to$ g n g pointwise, where $g=\left(\begin{array}{c}1\\ 2\end{array}\right)$ .

$\forall t, t\in \mathbb{R}\colon {g}_{n}(t)=\frac{t}{n}$ As done above, we first want to examine the limit $\lim_{n\to }n\to$ g n t 0 n t 0 n 0 where ${t}_{0}\in \mathbb{R}$ . Thus $\lim_{n\to }n\to$ g n g pointwise where $g(t)=0$ for all $t\in \mathbb{R}$ .

Norm convergence

The sequence $(n, \left\{1\right\})$ g n converges to $g$ in norm if $\lim_{n\to }n\to$ g n g 0 . Here $(˙)$ is the norm of the corresponding vector space of ${g}_{n}$ 's. Intuitively this means the distance between vectors ${g}_{n}$ and $g$ decreases to $0$ .

${g}_{n}=\left(\begin{array}{c}1+\frac{1}{n}\\ 2-\frac{1}{n}\end{array}\right)$ Let $g=\left(\begin{array}{c}1\\ 2\end{array}\right)$

$({g}_{n}-g)=\sqrt{(1+\frac{1}{n}-1)^{2}+(2-\frac{1}{n})^{2}}=\sqrt{\frac{1}{n^{2}}+\frac{1}{n^{2}}}=\frac{\sqrt{2}}{n}$
Thus $\lim_{n\to }n\to$ g n g 0 Therefore, ${g}_{n}\to g$ in norm.

${g}_{n}(t)=\begin{cases}\frac{t}{n} & \text{if 0\le t\le 1}\\ 0 & \text{otherwise}\end{cases}$ Let $g(t)=0$ for all $t$ .

$({g}_{n}(t)-g(t))=\int_{0}^{1} \frac{t^{2}}{n^{2}}\,d t=(n, \left[0 , 1\right], \frac{t^{3}}{3n^{2}})=\frac{1}{3n^{2}}$
Thus $\lim_{n\to }n\to$ g n t g t 0 Therefore, ${g}_{n}(t)\to g(t)$ in norm.

Pointwise vs. norm convergence

For $\mathbb{R}^{m}$ , pointwise and norm convergence are equivalent.

Pointwise ⇒ norm

${g}_{n}(i)\to g(i)$ Assuming the above, then $({g}_{n}-g)^{2}=\sum_{i=1}^{m} ({g}_{n}(i)-g(i))^{2}$ Thus,

$\lim_{n\to }n\to$ g n g 2 n i m 1 g n i g i 2 i m 1 n g n i g i 2 0

Norm ⇒ pointwise

$({g}_{n}-g)\to 0$

$\lim_{n\to }n\to$ i m 1 g n i g i 2 i m 1 n g n i g i 2 0
Since each term is greater than or equal zero, all' $m$ ' terms must be zero. Thus, $\lim_{n\to }n\to$ g n i g i 2 0 forall $i$ . Therefore, ${g}_{n}\to g\text{pointwise}$

In infinite dimensional spaces the above theorem is no longer true. We prove this with counter examples shownbelow.

Pointwise ⇏ norm

We are given the following function: ${g}_{n}(t)=\begin{cases}n & \text{if 0< t< \frac{1}{n}}\\ 0 & \text{otherwise}\end{cases}$ Then $\lim_{n\to }n\to$ g n t 0 This means that, ${g}_{n}(t)\to g(t)$ where for all $t$ $g(t)=0$ .

Now,

$({g}_{n})^{2}=\int \,d t$ g n t 2 t 1 n 0 n 2 n
Since the function norms blow up, they cannot converge to any function with finite norm.

Norm ⇏ pointwise

We are given the following function: ${g}_{n}(t)=\begin{cases}1 & \text{if 0< t< \frac{1}{n}}\\ 0 & \text{otherwise}\end{cases}\text{if n is even}$ ${g}_{n}(t)=\begin{cases}-1 & \text{if 0< t< \frac{1}{n}}\\ 0 & \text{otherwise}\end{cases}\text{if n is odd}$ Then, $({g}_{n}-g)=\int_{0}^{\frac{1}{n}} 1()\,d t=\frac{1}{n}\to 0$ where $g(t)=0$ for all $t$ . Therefore, ${g}_{n}\to g\text{in norm}$ However, at $t=0$ , ${g}_{n}(t)$ oscillates between -1 and 1, and so it does not converge. Thus, ${g}_{n}(t)$ does not converge pointwise.

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