Prove that the trilinear map $\varphi(c,d,e) = \sum_{n \in \mathbb{N}} c_n d_n e_n$ is not continuous. Let $(E, \| \cdot \|_\infty)$ be the Normed Vector Space consisting of all sequences $\{ c= \{c_n\}_{n \in \mathbb{N}} \mid c_n \in \mathbb{R}, \text{only a finite number of them are} \neq 0\}$ with the sup-norm $\|c\|_\infty=\sup|c_n|$.
I'm not sure where to start with this. Maybe chose a $C^k$, $D^k$, $E^k$ that are the basis of their respective sequences?
 A: $\newcommand{\norm}[1]{\lVert#1\rVert}$Let $\norm{\cdot}$ be the supermum norm on $E^3$. That is,
$$ \norm{(c,d,e)} = \max\{\norm{c}_\infty,\norm{d}_\infty,\norm{e}_\infty\}
$$
Then $\norm{\cdot}$ generates the product topology on $E^3$. (You should verify that $\norm{\cdot}$ is indeed a norm when $E^3$ is equipped with component-wise linearity) Given $n \in \mathbb{N}$, let
$$c_k^n = d_k^n = e_k^n = [k \leq n]
$$
That is, $c^n$ is the sequence for which the first $n$ terms are $1$, and the rest are $0$. Then
$$ |\varphi(c^n,d^n,e^n)| = n
$$
while $\norm{(c^n,d^n,e^n)} = \max\{1,1,1\} = 1$. Hence $\varphi$ is not a bounded linear map, and discontinuous.
A: A trilinear form $\phi$ is continuous if there exists an $M >0$ so that we have
$$|\phi(c,d,e)|\le M \cdot \|c\|\cdot\|d\|\cdot \|e\|$$
Take $n>0$ natural and $c=d=e$ a sequence with first $n$ entries $1$ and the others $0$. We have $\phi(c,d,e)=n$, while $\|c\|=\|d\|=\|e\|=1$. Since $n$ can be arbitrary it is easy to see that such $M$ does not exist. 
A: Let $c$ be this sequence:
\begin{align}
& (1,0,0,0,0,0,0,\ldots) \\[5pt]
& ( 0, \underbrace{1/\sqrt[\large3]2, 1/\sqrt[\large3]2,}_\text{2 terms}\, 0,0,0,0,0,0,\ldots) \\
& (0,0,0,\underbrace{1/\sqrt[\large3]3, 1/\sqrt[\large 3] 3, 1/\sqrt[\large 3] 3,}_\text{3 terms} \, 0,0,0,0,\ldots) \\
& (0,0,0,0,0,0,\underbrace{1/\sqrt[\large3]4, 1/\sqrt[\large3]4, 1/\sqrt[\large 3]4, 1/\sqrt[\large 3] 4,}_\text{4 terms}\, 0,0,0,0,\ldots) \\
& \quad \vdots
\end{align}
And let $d$ and $e$ also be that same sequence.
Then $\displaystyle\sum_{n=1}^\infty c_n d_n e_n$ will always remain $1$ as you move along the sequence displayed above.
But the sequence displayed above approaches the zero sequence.  Thus $(c,d,e)$ approaches $0$ but $\varphi(c,d,e)$ does not.
