Prove that $\lim\limits_{n\to \infty} ⟨k_n, m_n⟩ = ⟨c, d⟩$.

Question

Let ${k_n}$ and ${m_n}$ be sequences in $\mathbb{R}^d$ such that $\lim\limits_{n\to \infty} k_n = c$ and $\lim\limits_{n\to \infty} m_n = d$. Prove that $\lim\limits_{n\to \infty} ⟨k_n, m_n⟩ = ⟨c, d⟩$.

My thinking: $$\lim_{n\to \infty} k_n m_n = cd$$ $$|k_nm_n-cd|<\epsilon$$ $$||k_nm_n||-||cd||\leq|k_nm_n-cd|<\epsilon$$

then ? edit. $⟨k_n, m_n⟩ = ⟨c, d⟩$ are inner product.

Answer 1

Suppose $\left( \mathbf{a_n} \right)_{n \in \mathbb{N}}$ and $\left( \mathbf{b_n} \right)_{n \in \mathbb{N}}$ are sequences in $\mathbb{R}^k$.

Then, for each $n \in \mathbb{N}$, we have $$ \mathbf{a}_n, \mathbf{b}_n \in \mathbf{R}^k, $$ and thus $$ \mathbf{a_n} = \left( \alpha_{n1}, \ldots, \alpha_{nk} \right) $$ and $$ \mathbf{b_n} = \left( \beta_{n1}, \ldots, \beta_{nk} \right), $$ where $$ \alpha_{ni}, \beta_{ni} \in \mathbb{R} $$ for $i = 1, \ldots, k$.

Then, for each $n \in \mathbb{N}$, we have $$ \left\langle \mathbf{a}_n, \mathbf{b}_n \right\rangle = \sum_{i=1}^k \alpha_{nk}\beta_{nk}. $$

Now let $$ \mathbf{a} = \left( \alpha_1, \ldots, \alpha_k \right) $$ and $$ \mathbf{b} = \left( \beta_1, \ldots, \beta_k \right) $$ be any elements of $\mathbb{R}^k$.

Then $$ \lim_{n \to \infty} \mathbf{a}_n = \mathbf{a} $$ if and only if $$ \lim_{n \to \infty} \alpha_{ni} = \alpha_i $$ for each $i = 1, \ldots, k$.

And, similarly, $$ \lim_{n \to \infty} \mathbf{b}_n = \mathbf{b} $$ if and only if $$ \lim_{n \to \infty} \beta_{ni} = \beta_i $$ for each $i = 1, \ldots, k$.

Therefore if $$ \lim_{n \to \infty} \mathbf{a}_n = \mathbf{a} \ \mbox{ and } \ \lim_{n \to \infty} \mathbf{b}_n = \mathbf{b}, $$ then we have $$ \lim_{n \to \infty} \alpha_{ni} = \alpha_i \ \mbox{ and } \ \lim_{n \to \infty} \beta_{ni} = \beta_i $$ for each $i = 1, \ldots, k$.

Hence we have $$ \begin{align} \lim_{n \to \infty} \left\langle \mathbf{a}_n, \mathbf{b}_n \right\rangle &= \lim_{n \to \infty} \left( \sum_{i=1}^k \alpha_{ni}\beta_{ni} \right) \\ &= \sum_{i=1}^k \left[ \lim_{n \to \infty} \left( \alpha_{ni} \beta_{ni} \right) \right] \\ &= \sum_{i=1}^k \left[ \left( \lim_{n \to \infty} \alpha_{ni} \right) \left( \lim_{n \to \infty} \beta_{ni} \right) \right] \\ &= \sum_{i=1}^k \alpha_i \beta_i \\ &= \langle \mathbf{a}, \mathbf{b} \rangle. \end{align} $$

Hope this helps.

Should have any confusion, or further questions, please feel free to inquire.