Let
$f:D\subseteq R\to R^N$
$g:D\subseteq R\to R$
where both functions are continuous at $p.$
I am asked to prove that $(fg)(x)$ is continuous at $p$.
Since $(fg)(x)=(f_i(x)g(x),...,f_n(x)g(x))$ where $f_i$ is the coordinate function.
I also know that $$\left\| X-Y\right\| ^2=\sum _{i=1}^n \left(x_i-y_i\right){}^2$$
Now this is what I thought:
To prove that $(fg)(x)$ is continuous it would mean that I could prove that for any $\epsilon$>0 there exists a K element of the natural numbers such that
$\left\| f (x_k)g(x_k) -f(p)g(p)\right\| <\epsilon$ whenever k>K.
Now that would mean that it would be sufficient to prove that:
$\left\| f (x_k)g(x_k) -f(p)g(p)\right\| {}^2=\sum _{i=1}^n \left(f _i(x_k)g(x_k)-f (p)g(p)\right){}^2<\epsilon{}$ whenever k>K
and as f and g are continuous I have that
$||f(x_k)-f(p)||<\epsilon{}$
$||g(x_k)-g(p)||<\epsilon{}$
whenever k>K
Next: choose any $\epsilon{}>0$
To each $i=1,...,n$ there corresponds a number $K_i$ element of the naturals such that $abs(f_i(x_k)-f_i(p))<\epsilon$
to each $i$ there exists a $K_i$ such that
$abs((f i (x k )-f i (p ))(g(x k )-g (p )))))<\epsilon$$
now if I choose $K=max(K_i)$ then I thought I could continue by constructing $\epsilon$ such that multipling by n and squaring would allow me to sum up however, I am stuck with the
$-f i (x k )g (p ) -f i (p )g(x k )$ and I cant move on. Could someone take me through the steps, preferable continue with where I left of, or a couple of steps earlier. However, if what I have done is completely fruitless ore wrong any solution would do.