How do we prove the continuity of the inner product? Let $\lim x_{k}=a$, $\lim y_{k}=b$ in $\mathbb{R^n}$ then it must be proved that $\lim \langle x_{k}, y_{k}\rangle = \langle a, b\rangle$, where $\langle a, b \rangle$ is the inner product in $\mathbb{R^n}$
The idea I have is that $ \langle x_{k}, y_{k} \rangle$ $=$ $\sum_{i=1}^{n} x_{i}^k y_{i}^k$ then $\lim \langle x_{k},y_{k}\rangle = \lim \sum_{i=1}^{n} x_{i}^k y_{i}^k$ $=$ $\lim x_{1}^ky_{1}^k+...+\lim x_{n}^ky_{n}^k$
That would be fine and being like that how I continue?
 A: Proposition
Given a sequence $x_{k} = (x_{1}^{k},x_{2}^{k},\ldots,x_{n}^{k})\in\mathbb{R}^{n}$, it converges to $a = (a_{1},a_{2},\ldots,a_{n})$ iff $x_{j}^{k}$ converges to $a_{j}$.
Proof
Let us prove the implication $(\Rightarrow)$ first.
Suppose that $x_{k}\in\mathbb{R}^{n}$ converges to $a$. Let $\varepsilon > 0$. Then there exists a natural number $N_{\varepsilon}\in\mathbb{N}$ s.t.
\begin{align*}
k\geq N_{\varepsilon} \Rightarrow \|x_{k} - a\| \leq\varepsilon
\end{align*}
But we do also know that $|x_{j}^{k} - a_{j}| \leq \|x_{k} - a\|$. Hence we conclude that $x_{j}^{k}\to a_{j}$, and we are done.
Let us now prove the converse implication $(\Leftarrow)$.
We shall assume that $x_{j}^{k}$ converges to $a_{j}$. Let $\varepsilon > 0$. Then there exists $N_{\varepsilon_{j}}\in\mathbb{N}$ s.t.
\begin{align*}
k\geq N_{\varepsilon_{j}} \Rightarrow |x_{j}^{k} - a_{j}| \leq \varepsilon/n
\end{align*}
Consequently, if we take $N_{\varepsilon} = \max\{N_{\varepsilon_{1}},N_{\varepsilon_{2}},\ldots N_{\varepsilon_{n}}\}$, we conclude that
\begin{align*}
k\geq N_{\varepsilon} \Rightarrow \|x_{k} - a\| \leq |x_{1}^{k} - a_{1}| + |x^{k}_{2} - a_{2}| + \ldots + |x^{k}_{n} - a_{n}| \leq n\times\frac{\varepsilon}{n} = \varepsilon
\end{align*}
and we are done.
Proposition
Let $a_{k}\in\mathbb{R}$ and $b_{k}\in\mathbb{R}$ be sequences s.t $a_{k}\to a$ and $b_{k}\to b$. Then $a_{k} + b_{k}$ converges to $a + b$.
Proof
Let $\varepsilon > 0$. Then there are natural numbers $N_{\varepsilon_{1}}$ and $N_{\varepsilon_{2}}$ such that
\begin{align*}
\begin{cases}
k\geq N_{\varepsilon_{1}} \Rightarrow |a_{k} - a| \leq \varepsilon/2\\\\
k\geq N_{\varepsilon_{2}} \Rightarrow |b_{k} - b| \leq \varepsilon/2
\end{cases}
\end{align*}
Thus if we take $N_{\varepsilon} = \max\{N_{\varepsilon_{1}},N_{\varepsilon_{2}}\}$, one has that
\begin{align*}
k\geq N_{\varepsilon} \Rightarrow |a_{k} + b_{k} - a - b| \leq |a_{k} - a| + |b_{k} - b| \leq \varepsilon/2 + \varepsilon/2 = \varepsilon
\end{align*}
whence we conclude that $a_{k} + b_{k} \to a + b$.
Proposition
Let $a_{k}\in\mathbb{R}$ and $b_{k}\in\mathbb{R}$ be sequences such that $a_{k}\to a$ and $b_{k}\to b$. Then $a_{k}b_{k}$ converges to $ab$.
Proof
Let us consider the expression:
\begin{align*}
|a_{k}b_{k} - ab| = |a_{k}b_{k} - ab_{k} + ab_{k} - ab| & \leq |b_{k}||a_{k} - a| + |a||b_{k} - b|\\\\
& \leq |B||a_{k} - a| + |a||b_{k} - b|
\end{align*}
where $|b_{k}|\leq B$ (since $b_{k}$ converges). Since $a_{k}\to a$ and $b_{k}\to b$, let $\varepsilon > 0$. Then there are natural numbers $N_{\varepsilon_{1}}$ and $N_{\varepsilon_{2}}$ such that
\begin{align*}
\begin{cases}
k\geq N_{\varepsilon_{1}} \Rightarrow |a_{k} - a| \leq \displaystyle\frac{\varepsilon}{2(|B| + 1)}\\\\
k\geq N_{\varepsilon_{2}} \Rightarrow |b_{k} - b| \leq \displaystyle\frac{\varepsilon}{2(|a| + 1)}
\end{cases}
\end{align*}
If we take $N_{\varepsilon} = \max\{N_{\varepsilon_{1}},N_{\varepsilon_{2}}\}$, one has that
\begin{align*}
k\geq N_{\varepsilon} \Rightarrow |a_{k}b_{k} - ab| \leq \left(\frac{|B|}{2(|B| + 1)} + \frac{|a|}{2(|a| + 1)}\right)\varepsilon \leq \varepsilon
\end{align*}
whence we conclude that $a_{k}b_{k}\to ab$.
Solution
Based on the above-mentioned results, if $x_{k}\to a$ and $y_{k}\to b$, then $\langle x_{k},y_{k}\rangle$ converges to $\langle a,b\rangle$.
Hopefully this helps.
A: *

*Show that $f(x,y)=\|x+y\|^2$ and $g(x,y)=\|x-y\|^2$ are continuous from $(\Bbb R^n)^2$ to $\Bbb R.$


*The inner product $<x,y>=\frac {f(x,y)-g(x,y)}{4}.$
