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Math Help - Analysis Proof - Convergence and scalar product

  1. #1
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    Analysis Proof - Convergence and scalar product

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

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

    Prove that  { u_{k}} converges to u.

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

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

    Since we are given that

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

     \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  \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  { u_{k} } be a sequence in \mathbb{R}^n and let  u \in \mathbb{R}^n . Suppose that for every  v \in \mathbb{R}^n ,

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

    Prove that  { u_{k}} converges to u.

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

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

    Since we are given that

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

     \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  \lim_{x \to \infty} u_{k} = u

    Thank you for reading. Any help is greatly appreciated.
    Pick \bold{v} = \bold{e}_k where \{ \bold{e}_1,...,\bold{e}_n\} is the standard basis.
    Now write \bold{u}=(a_1,...,a_n) then \left< \bold{u},\bold{e}_k\right> = a_k.
    This means the sequence of \{ \bold{u}_j\} at the k coordinate converges to a_k.
    Therefore the limit of \{ \bold{u}_j\} is (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  { u_{k} } be a sequence in \mathbb{R}^n and let  u \in \mathbb{R}^n . Suppose that for every  v \in \mathbb{R}^n ,

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

    Prove that  { u_{k}} converges to u.

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

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

    Since we are given that

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

     \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  \lim_{x \to \infty} u_{k} = u

    Thank you for reading. Any help is greatly appreciated.
    For fixed i what is \lim_{k \to \infty} p_i(u_k) ? So what is \lim_{k \to \infty} p_i(u_k)p_i(v) ? And:

    \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  \lim_{k \to \infty} p_i(u_k), then  \lim_{k \to \infty} p_i(u_k) = p_{i}(u).

    As a result, then \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  \lim_{k \to \infty} p_i(u_k), then  \lim_{k \to \infty} p_i(u_k) = p_{i}(u).

    As a result, then \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
    \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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