Bilinear map on the set of finite sequences Let $X = \{ x = (x_n)_{n=1} ^ {\infty}\subset \mathbb{R} \ \ | \ \ \exists N \in \mathbb{N} : \forall n>N : x_n=0 \}$
Let the norm on $X$ be $||x|| = \sum _{n=1} ^{\infty} |x_n|$ (which is fine, because the sequences have only finitely many nonzero terms).
Given $a = (a_n) _{n=1} ^{\infty}$ we define a bilinear map:
$B : X \times X \ni (x,y) \rightarrow \sum_{n=1} ^{\infty} a_nx_ny_n \in \mathbb{R}$
Prove that $B(x, \cdot)$ is continuous for any $x \in X$.
What is the necessary and sufficient condition for $B$ to be continuous?
For $u,w \in X :||u - w|| < \delta$ we have: $\sum _{n=1} ^{\infty} |u_n-w_n|< \delta$
$|B(x, u) - B(x, w)| = |\sum _1 ^{\infty} a_nx_n(u_n-w_n)| \le \sum _1 ^{\infty} |a_nx_n(u_n-w_n)| \le \max \{a_nx_n \ | n \in \mathbb{N} \} \cdot \delta < \epsilon$ 
if for any given $\epsilon$ we take $\delta = \frac{\epsilon}{\max...}$
Isn't this approach a bit naive. It seems to me that since there are only finitely many nonzero terms in every sequence in $X$ we can choose such a maximum product $a_nx_n$.
Could you tell me if I'm right and help me with the rest?
Thank you!
 A: For the first part, the continuity of $B(x,\,\cdot\,)$ for any fixed $x$, you have exactly the right approach, only you forgot to take the absolute value in the maximum. You can't really do anything else, there are sequences $u \in X$ with
$$\lvert B(x,u)\rvert = \lVert u\rVert\cdot \max \{ \lvert a_n x_n\rvert : n \in \mathbb{N}\},$$
and thus by linearity, you have for all $u\in X$
$$\sup \left\{\lvert B(x,w) - B(x,u)\rvert : \lVert w-u\rVert < \delta \right\} = \delta\cdot \max \{ \lvert a_n x_n\rvert : n \in \mathbb{N}\}.$$
Choosing $\delta = \varepsilon/\max \{ \lvert a_n x_n\rvert : n \in \mathbb{N}\}$ hence works for all $u$, and for no $u$ there is a larger choice that works.

Isn't this approach a bit naive.

No. It's the straightforward and correct (after taking absolute values) approach. The (bi)linearity ensures that there are no hidden traps that would make anything more complicated necessary.

It seems to me that since there are only finitely many nonzero terms in every sequence in $X$ we can choose such a maximum product $a_nx_n$.

Right. Since there are only finitely many nonzero terms, the maximum exists. If there were infinitely many nonzero terms, you would need to replace the maximum with a supremum, and a necessary condition on the sequence $(a_n)$ for $B(x,\,\cdot\,)$ to be continuous would then be that $\sup \{ \lvert a_nx_n\rvert : n\in\mathbb{N}\}$ is finite. That would not be satisfied for every sequence $(a_n)$ and all $(x_n)$ with $\sum \lvert x_n\rvert < \infty$.
For the joint continuity in both variables, let us first look at what is needed for $B$ to be continuous in $(0,0)$. Given any $\varepsilon > 0$, there must be a neighbourhood $U$ of $(0,0)$ such that $\lvert B(x,y)\rvert \leqslant \varepsilon$ for all $(x,y) \in U$. A neighbourhood of $(0,0)$ is by definition a set that contains the product of two open balls, each centered at $0$, so there must be $\delta_1,\delta_2 > 0$ such that $\lVert x\rVert < \delta_1,\, \lVert y\rVert < \delta_2 \Rightarrow \lvert B(x,y)\rvert < \varepsilon$.
By bilinearity, we then have for $\lVert\xi\rVert < 1,\,\lVert\eta\rVert < 1$
$$\lvert B(\xi,\eta)\rvert = \left\lvert B\left(\frac{\delta_1\xi}{\delta_1},\frac{\delta_2\eta}{\delta_2}\right)\right\rvert = \frac{1}{\delta_1\delta_2}\lvert B(\delta_1\xi,\delta_2\eta)\rvert < \frac{\varepsilon}{\delta_1\delta_2},$$
since $\lVert\delta_1\xi\rVert < \delta_1$ and $\lVert \delta_2\eta\rVert < \delta_2$.
Thus if $B \colon X\times X\to \mathbb{R}$
is continuous in $(0,0)$, we have
$$\lVert B\rVert := \sup \left\{\lvert B(x,y)\rvert : \lVert x\rVert < 1,\, \lVert y\rVert < 1\right\} < \infty.$$
If that condition is fulfilled, then bilinearity gives the estimate
$$\lvert B(x,y)\rvert \leqslant \lVert B\rVert\cdot\lVert x\rVert\,\lVert y\rVert$$
for all $x,y\in X$. Now you can use that estimate and
$$B(x,y) - B(x_0,y_0) = B(x_0,y-y_0) + B(x-x_0,y_0) + B(x-x_0,y-y_0)$$
to deduce the continuity of $B$ in an arbitrary $(x_0,y_0) \in X\times X$.
So $B$ is continuous if and only if
$$\sup \left\{\lvert B(x,y)\rvert : \lVert x\rVert < 1,\, \lVert y\rVert < 1\right\} < \infty.\tag{1}$$
Now find the condition on the sequence $(a_n)$ that ensures $(1)$.
