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
$g=\left(\begin{array}{c}1\\ 2\end{array}\right)$ .
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$$\forall t, t\in \mathbb{R}\colon {g}_{n}(t)=\frac{t}{n}$$ 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
$t\in \mathbb{R}$ .
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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
$(\dot{})$ 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.
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$${g}_{n}(t)=\begin{cases}\frac{t}{n} & \text{if $0\le t\le 1$}\\ 0 & \text{otherwise}\end{cases}$$ 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.
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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.
Counter examples
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.
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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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