In $l^p$ the map $x\longrightarrow \sum_{n=1}^\infty x_ny_n$ is well-defined.

Question

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,

Answer 1

To prove that the map is well defined you have to prove convergence of the series. For this recall that $\sum |x_ny_n|\leq (\sum |x_n|^{p})^{1/p}(\sum |y_n|^{q})^{1/q}$. The same inequality shows that your map is a bounded operator whose norm is at the most ($\sum |y_n|^{q})^{1/q}$. This implies continuity of the map.

Answer 2

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 $f(x)\neq \infty$.