# Proving ${u_k}\to u$ given $\lim_{k\to\infty}\langle u_k,v\rangle=\langle u,v\rangle$ for $u\in \mathbb{R}^n,\forall v\in \mathbb{R}^n$

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

$$\lim_{k\to\infty}\langle u_k,e_i\rangle = \lim_{k\to\infty}u_i^k = \langle u,e_i\rangle=\langle (u_1,u_2,...u_n),e_i\rangle = u_i, \ \forall i \quad \Box$$

• You should use \langle $\langle$ and \rangle $\rangle$ instead of < and >. The former are delimiters, so the spacing for them is similar to that given for parentheses. The spacing for the latter is that of operators. That’s why it looks so bad when you use them. Compare $<e_1,e_2>$ with $\langle e_1,e_2\rangle$. Feb 3 '19 at 6:04
• Well, if you know that $u_i^k \to u_i$ as $k\to\infty$ for each $i=1,\cdots,n$, then the proof is done. And this follows from the assumption by plugging $v = e_i$ for each $i$. Your intermediate computation is puzzling me because $\lim_k \langle u_k, e_i \rangle = \lim_k u_i^k$ holds simply because $\langle u_k, e_i \rangle = u_i^k$ for all $k$ and $i$, so I am not sure why you made a significant detour for it. Feb 3 '19 at 6:14
• @Sangchul Lee How can you just assume $v = e_{i}$; wouldn't you lose generality? We are trying to prove it for all $v$
– user614735
Feb 3 '19 at 16:21

$$\lim_{k\to\infty} \langle u_k,v \rangle = \langle u,v \rangle$$ $$\lim_{k\to\infty} \langle u_k-u,v \rangle = 0$$ 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.