Exercise 1.1.12

Question

Prove Proposition 1.1.18.\par 

	extbf{Proposition 12.1.18:} Let $\mathbb{R}^{n}$ be a Euclidean space, and let $\left(x^{(k)}ight)_{k=m}^{\infty}$ be a sequence of points in $\mathbb{R}^{n}$. We write $x^{(k)}=\left(x_{1}^{(k)}, x_{2}^{(k)}, \ldots, x_{n}^{(k)}ight)$, i.e., for $j=1,2, \ldots, n, x_{j}^{(k)} \in \mathbb{R}$ is the $j^{	ext {th }}$ co-ordinate of $x^{(k)} \in \mathbb{R}^{n}$. Let $x=\left(x_{1}, \ldots, x_{n}ight)$ be a point in $\mathbb{R}^{n}$. Then the following four statements are equivalent:
ewline

(a) $\left(x^{(k)}ight)_{k=m}^{\infty}$ converges to $x$ with respect to the Euclidean metric $d_{l^{2}}$.
ewline

(b) $\left(x^{(k)}ight)_{k=m}^{\infty}$ converges to $x$ with respect to the taxi-cab metric $d_{l^{1}}$.
ewline

(c) $\left(x^{(k)}ight)_{k=m}^{\infty}$ converges to $x$ with respect to the sup norm metric $d_{l^{\infty}}$.
ewline

(d) For every $1 \leq j \leq n$, the sequence $\left(x_{j}^{(k)}ight)_{k=m}^{\infty}$ converges to $x_{j}$

Prakash Anand · Accepted Answer

We will prove this the following way: $a \Rightarrow b, b \Rightarrow d, d \Rightarrow c, c \Rightarrow a$ .

Proof. $(a \Rightarrow b)$

Given $𝜖 > 0$ take $𝜖^{'} = 𝜖 ∕ n$ . Then we know that there exists $N > m$ such that $d_{l^{2}} ({(x^{(k)})}_{k = m}^{\infty}, x) <$ $𝜖^{'} = 𝜖 ∕ n \forall k \geq N$ . But,

d_{l^{2}} ({(x^{(k)})}_{k = m}^{\infty}, x) = {(\sum_{j = 1}^{n} {| x_{j}^{(k)} - x |}^{2})}^{1 ∕ 2}

So,

\begin{aligned} {(\sum_{j = 1}^{n} {| x_{j}^{(k)} - x |}^{2})}^{1 ∕ 2} < 𝜖^{'} \\ \Rightarrow (\sum_{j = 1}^{n} {| x_{j}^{(k)} - x |}^{2}) < {(𝜖^{'})}^{2} \\ \Rightarrow {| x_{j}^{(k)} - x |}^{2} < {(𝜖^{'})}^{2} \forall j \\ \Rightarrow | x_{j}^{(k)} - x | < 𝜖^{'} \\ \Rightarrow \sum_{j = 1}^{n} | x_{j}^{(k)} - x | < n 𝜖^{'} = n \cdot 𝜖 ∕ n \\ \Rightarrow \sum_{j = 1}^{n} | x_{j}^{(k)} - x | < 𝜖 \\ \Rightarrow d_{l^{1}} (x^{(k)}, x) < 𝜖 \end{aligned}

Which was what we wanted.

$(b \Rightarrow d)$

We have that for any $𝜖 > 0$ there exists $N > m$ such that $d_{l^{l}} ({(x^{(k)})}_{k = m}^{\infty}, x) < 𝜖 \forall k \geq N$ . But,

d_{l^{'}} ({(x^{(k)})}_{k = m}^{\infty}, x) = \sum_{j = 1}^{n} | x_{j}^{(k)} - x |

So,

\sum_{j = 1}^{n} | x_{j}^{(k)} - x | < 𝜖

But,

| x_{j}^{(k)} - x | \geq 0 \forall j .

Thus,

| x_{j}^{(k)} - x | < 𝜖 \forall j .

Hence, ${(x_{j}^{(k)})}_{k = m}^{\infty}$ converges to $x_{j}$ , as desired.

$(d \Rightarrow c)$

We have that $\forall j$ , given $𝜖^{'} > 0 \exists N_{j} > m$ such that $| x_{j}^{(k)} - x_{j} | < 𝜖^{'} \forall k \geq N_{j} .$ So given $𝜖 > 0$ take $N = \max {N_{1}, N_{2}, \dots, N_{n}}$ then we have $| x_{j}^{(k)} - x_{j} | < 𝜖 \forall k \geq N$ and $\forall j = 1, \dots, n$ .

And since we have a finite number of $j$ ’s we have that:

$\sup {| x_{j}^{(k)} - x_{j} | : 1 \leq j \leq n} = \max {| x_{j}^{(k)} - x_{j} | : 1 \leq j \leq n} < 𝜖$ . Thus ${(x^{(k)})}_{k = m}^{\infty}$ converges to $x$ .

$(c \Rightarrow a)$

Given $𝜖 > 0$ take $𝜖^{'} = 𝜖 ∕ \sqrt{n} > 0$ Then there exists $N > m$ such that $d_{l∞} ({(x^{(k)})}_{k = m}^{\infty}, x) < 𝜖^{'} =$ $𝜖 ∕ \sqrt{n} \forall k \geq N$ . But,

d_{l} \infty (x^{(k)}, x) = \sup {| x_{j}^{(k)} - x | : 1 \leq j \leq n}

So,

\sup {| x_{j}^{(k)} - x | : 1 \leq j \leq n} < 𝜖^{'}

Also since we have a finite number $n$ the sup is just the max. Hence,

$\exists i$ such that $\forall j | x_{j}^{(k)} - x | \leq | x_{i}^{(k)} - x | = \max {| x_{j}^{(k)} - x | : 1 \leq j \leq n} = \sup {| x_{j}^{(k)} - x | : 1 \leq j \leq n}$ Thus,

\begin{aligned} | x_{i}^{(k)} - x | < 𝜖^{'} = 𝜖 ∕ \sqrt{n} \\ \Rightarrow \sqrt{n} | x_{i}^{(k)} - x | < 𝜖 \\ \Rightarrow n {| x_{i}^{(k)} - x |}^{2} < 𝜖^{2} \\ \Rightarrow \sum_{j = 1}^{n} {| x_{j}^{(k)} - x |}^{2} < n {| x_{i}^{(k)} - x |}^{2} < 𝜖^{2} \\ \Rightarrow {(\sum_{j = 1}^{n} {| x_{j}^{(k)} - x |}^{2})}^{1 ∕ 2} < 𝜖 \\ \Rightarrow d_{l^{2}} ({(x^{(k)})}_{k = m}^{\infty}, x) < 𝜖 \end{aligned}

Which is what we wanted.

So we have proved that $a \Rightarrow b \Rightarrow d \Rightarrow c \Rightarrow a$ , and this proves the proposition. □

Exercise 1.1.12

Answers

Comments

Exercise 1.1.12

Answers

Comments

Add answer