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Thread: Analysis Proof - Convergence and scalar product

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by Paperwings View Post
    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 $\displaystyle (a_1,...,a_k)$.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Paperwings View Post
    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
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  4. #4
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    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 $\displaystyle \implies \lim_{k \to \infty} p_i(u_k)p_i(v) = p_i(u) p_i(v) \neq \left\langle u, v \right\rangle$
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Paperwings View Post
    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
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