# Analysis Proof - Convergence and scalar product

• Sep 4th 2008, 04:54 AM
Paperwings
Analysis Proof - Convergence and scalar product
Problem:
Let $\displaystyle { u_{k} }$ be a sequence in $\displaystyle \mathbb{R}^n$ and let $\displaystyle u \in \mathbb{R}^n$. Suppose that for every $\displaystyle v \in \mathbb{R}^n$,

$\displaystyle \lim_{k \to \infty} \left\langle u_{k}, v \right\rangle = \left\langle u, v \right\rangle$

Prove that $\displaystyle { u_{k}}$ converges to u.

========================
Let $\displaystyle p_{i}$ be the ith coordinate function $\displaystyle \mathbb{R}^n \rightarrow \mathbb{R}$ and let $\displaystyle u, v \in R^n$

For every index i with $\displaystyle 1 \leq i \leq n$ let $\displaystyle p_{i}(u) = \left\langle u, e_{i} \right\rangle$ where $\displaystyle e_{i} \in R^{n}$ and which the ith component is 1 and all other components are 0.

Since we are given that

$\displaystyle \lim_{k \to \infty} \left\langle u_{k}, v \right\rangle = \left\langle u, v \right\rangle$

$\displaystyle \implies \left\langle u , v \right\rangle = \sum_{i}^n {p_{i}(u)p_{i}(v)}$

From here, I do not know how to show that $\displaystyle \lim_{x \to \infty} u_{k} = u$

Thank you for reading. Any help is greatly appreciated.
• Sep 4th 2008, 06:55 AM
ThePerfectHacker
Quote:

Originally Posted by Paperwings
Problem:
Let $\displaystyle { u_{k} }$ be a sequence in $\displaystyle \mathbb{R}^n$ and let $\displaystyle u \in \mathbb{R}^n$. Suppose that for every $\displaystyle v \in \mathbb{R}^n$,

$\displaystyle \lim_{k \to \infty} \left\langle u_{k}, v \right\rangle = \left\langle u, v \right\rangle$

Prove that $\displaystyle { u_{k}}$ converges to u.

========================
Let $\displaystyle p_{i}$ be the ith coordinate function $\displaystyle \mathbb{R}^n \rightarrow \mathbb{R}$ and let $\displaystyle u, v \in R^n$

For every index i with $\displaystyle 1 \leq i \leq n$ let $\displaystyle p_{i}(u) = \left\langle u, e_{i} \right\rangle$ where $\displaystyle e_{i} \in R^{n}$ and which the ith component is 1 and all other components are 0.

Since we are given that

$\displaystyle \lim_{k \to \infty} \left\langle u_{k}, v \right\rangle = \left\langle u, v \right\rangle$

$\displaystyle \implies \left\langle u , v \right\rangle = \sum_{i}^n {p_{i}(u)p_{i}(v)}$

From here, I do not know how to show that $\displaystyle \lim_{x \to \infty} u_{k} = u$

Thank you for reading. Any help is greatly appreciated.

Pick $\displaystyle \bold{v} = \bold{e}_k$ where $\displaystyle \{ \bold{e}_1,...,\bold{e}_n\}$ is the standard basis.
Now write $\displaystyle \bold{u}=(a_1,...,a_n)$ then $\displaystyle \left< \bold{u},\bold{e}_k\right> = a_k$.
This means the sequence of $\displaystyle \{ \bold{u}_j\}$ at the $\displaystyle k$ coordinate converges to $\displaystyle a_k$.
Therefore the limit of $\displaystyle \{ \bold{u}_j\}$ is $\displaystyle (a_1,...,a_k)$.
• Sep 4th 2008, 07:01 AM
CaptainBlack
Quote:

Originally Posted by Paperwings
Problem:
Let $\displaystyle { u_{k} }$ be a sequence in $\displaystyle \mathbb{R}^n$ and let $\displaystyle u \in \mathbb{R}^n$. Suppose that for every $\displaystyle v \in \mathbb{R}^n$,

$\displaystyle \lim_{k \to \infty} \left\langle u_{k}, v \right\rangle = \left\langle u, v \right\rangle$

Prove that $\displaystyle { u_{k}}$ converges to u.

========================
Let $\displaystyle p_{i}$ be the ith coordinate function $\displaystyle \mathbb{R}^n \rightarrow \mathbb{R}$ and let $\displaystyle u, v \in R^n$

For every index i with $\displaystyle 1 \leq i \leq n$ let $\displaystyle p_{i}(u) = \left\langle u, e_{i} \right\rangle$ where $\displaystyle e_{i} \in R^{n}$ and which the ith component is 1 and all other components are 0.

Since we are given that

$\displaystyle \lim_{k \to \infty} \left\langle u_{k}, v \right\rangle = \left\langle u, v \right\rangle$

$\displaystyle \implies \left\langle u , v \right\rangle = \sum_{i}^n {p_{i}(u)p_{i}(v)}$

From here, I do not know how to show that $\displaystyle \lim_{x \to \infty} u_{k} = u$

Thank you for reading. Any help is greatly appreciated.

For fixed $\displaystyle i$ what is $\displaystyle \lim_{k \to \infty} p_i(u_k)$ ? So what is $\displaystyle \lim_{k \to \infty} p_i(u_k)p_i(v)$ ? And:

$\displaystyle \lim_{k \to \infty} \sum_{i=1}^n p_i(u_k)p_i(v)=\lim_{k \to \infty} \langle u_k,v \rangle$ ??

RonL
• Sep 4th 2008, 11:41 AM
Paperwings
To CaptainBlack, For $\displaystyle \lim_{k \to \infty} p_i(u_k)$, then $\displaystyle \lim_{k \to \infty} p_i(u_k) = p_{i}(u)$.

As a result, then $\displaystyle \implies \lim_{k \to \infty} p_i(u_k)p_i(v) = p_i(u) p_i(v) \neq \left\langle u, v \right\rangle$
• Sep 4th 2008, 01:13 PM
CaptainBlack
Quote:

Originally Posted by Paperwings
To CaptainBlack, For $\displaystyle \lim_{k \to \infty} p_i(u_k)$, then $\displaystyle \lim_{k \to \infty} p_i(u_k) = p_{i}(u)$.

As a result, then $\displaystyle \implies \lim_{k \to \infty} p_i(u_k)p_i(v) = p_i(u) p_i(v) \neq \left\langle u, v \right\rangle$

$\displaystyle \left\langle u, v \right\rangle=\sum_i p_i(u)p_i(v)=\sum_i \lim_{k \to \infty} p_i(u_k)p_i(v)=\lim_{k \to \infty} \sum_i p_i(u_k)p_i(v)=\lim_{k \to \infty} \left\langle u_k, v \right\rangle$

The sum and limit can be interchanged because the sum is a finite sum.

RonL