In $l^p$ the map $x\longrightarrow \sum_{n=1}^\infty x_ny_n$ is well-defined. For the following I have proof ideas but they are uncertain:
Prove that for every $y\in l^q$ the map $x\longrightarrow \sum_{n=1}^\infty x_ny_n$ is well-defined, linear and continuous on $l^p$.
Proof ideas:
Linearity follows from definition: $f(\alpha x+\beta y)=\alpha f(x)+\beta f(y)$.
WELL-defined:
Given $l_p=\{(x_1,x_2,...,x_n,...): \sum_{k}|x_k|^p<\infty\}$ we know that $‖x‖_{l^p}=(\sum_{n=1}^\infty|x_n|^p)^{1/p}$.
To prove our map is well defined means that there cannot be $f(x^a)\neq f(x^b)$ where $x^a=x^b\in l^p$.
So $|f(x_a)-f(x_b)|=|\sum_{n=1}^\infty x^a_ny_n-\sum_{n=1}^\infty x^b_ny_n|$ for some $y\in l^q$.
Then $|f(x_a)-f(x_b)|=|\sum_{n=1}^\infty (x^a_n- x^b_n)y_n|=\sum_{n=1}^\infty |(x^a_n- x^b_n)||y_n|>0$ WLOG assuming  $f(x_a)>f(x_b)$.
This means that there is at least one $n$ such that $x_n^a\neq x_n^b$ and so $x^a\neq x^b$. This means it is well defined.
CONTINUITY:
There is an $\epsilon$ $\delta$ criterion somewhere.... since $\sum_{n=1}^\infty x^a_ny_n\leq|||x^a_n| ||y_n||$
$|f(x_a)-f(x_b)|\leq|||x^a_n| ||y_n||-||x^b_n| ||y_n|||=...$ for some $y\in l^q$.
Thanks and regards,


To prove our map is well defined means that there cannot be $f(x^a)\neq f(x^b)$ where $x^a=x^b\in l^p$.

That is only part of what well defined means.
There is more, and that is that you have to prove that for $x\in \ell^p,$ the value $f(x)$ is in $\mathbb R$, i.e. that $f(x)\neq \infty$.
