16.2 Convergence of sequences of vectors

$$g_n=\left(\begin{array}{c}{g}_n(1)\\ g_n(2)\end{array}\right)()=\left(\begin{array}{c}1+\frac{1}{n}\\ 2-\frac{1}{n}\end{array}\right)()$$ 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 $t\in \mathbb{R}$ .

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

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

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,

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.