# 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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