I'm having trouble solving the following problem.
Problem. Prove ${u_k}\to u$ given $\lim_{k\to\infty}\langle u_k,v\rangle=\langle u,v\rangle$ for $u\in \mathbb{R}^n$, $\forall v\in \mathbb{R}^n$
The textbook introduces the $i^{\text{th}}$ component function $p_{i} : \mathbb{R}^{n} \to \mathbb{R}$ for $1 \leq i \leq n$ by $p_{i}(u) = u_{i}$, where $u \in \mathbb{R}^{n}$.
Using this definition, I can express any vector $u \in \mathbb{R}^{n}$ by $u = (p_{1}(u), \ldots, p_{n}(u))$. I know that this function is linear. The book also tells us that a sequence $\{u_{k}\}$ converges to $u$ in $\mathbb{R}^{n}$ if and only if it converges componentwise (i.e. for each $1 \leq i \leq n$, $\lim_{k\to\infty} p_{i}(u_{k}) = p_{i}(u)).$
The book provides a hint to define the point $e_{i} \in \mathbb{R}^{n}$ whose $i^{\text{ith}}$ component is equal to $1$ and every other component equals $0$. This way, $p_{i}(u) = \langle u, e_{i}\rangle$ for each point $u \in \mathbb{R}^{n}$. I've been working with this component function and don't seem to be making any progress. Any help is appreciated
Note:$\langle u,v\rangle$ denotes $u\cdot v$, u and v are points in $\mathbb{R}^n$, $u_k$ is a sequence in $\mathbb{R}^n$.
My attempt: Letting $u_k=(u_1^k,u_2^k,...u_n^k)$ with the $k$ representing the $k$-th term in the sequence $u=(u_1,u_2,...u_n)$. Then $\lim_{k\to\infty}u_i^k=u_i$ $\forall i$.
$\lim_{k\to\infty}\langle u_k,v\rangle=\langle u,v\rangle$ for any given $v\in\mathbb{R}$ , let $v=e_i$ then $\lim_{k\to\infty}\langle u_k,e_i\rangle=\langle u,e_i\rangle$.
Before proceeding we establish $\langle e_i, e_i\rangle=1$ and $\langle e_i,e_j\rangle=0$ assuming $i\neq j$.
\begin{align*} \lim_{k\to\infty} \langle u_k,e_i\rangle &= \Big< \lim_{k\to\infty} u_k, e_i \Big> = \Big< \lim_{k\to\infty}(u_1^k, u_2^k, \cdots, u_n^k),e_i \Big> \\ &= \Big< \Big( \lim_{k\to\infty}u_1^k, \lim_{k\to\infty}u_2^k, \cdots, \lim_{k\to\infty}u_n^k \Big),e_i \Big> =\lim_{k\to\infty} u_i^k \end{align*}
Now we have
$$\lim_{k\to\infty}\langle u_k,e_i\rangle = \lim_{k\to\infty}u_i^k = \langle u,e_i\rangle=\langle (u_1,u_2,...u_n),e_i\rangle = u_i, \ \forall i \quad \Box$$
$$\lim_{k\to\infty} \langle u_k,v \rangle = \langle u,v \rangle$$ $$\lim_{k\to\infty} \langle u_k-u,v \rangle = 0$$ For all $\epsilon>0$ there exists an index K such that $|\langle u_k-u,v \rangle | < \epsilon $ for all $k \geq K$
Using the Cauchy-Schwarz Inequality $-\|u\|\|v\|\leq\langle u,v\rangle\leq\|u\|\|v\| $ $$-\|u_k-u\|.\|v\|\leq\langle u_k-u,v \rangle< \epsilon$$ $$-\|u_k-u\|<\epsilon/\|v\|$$ $$\lim_{k\to\infty} -\|u_k-u\|=0$$ $$-\lim_{k\to\infty} \|u_k-u\|=0$$ Correct me if I am wrong please.