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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,

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  • $\begingroup$ I will just mention that you can use things such as $\|x\|\|y\|$ $\|x\|\|y\|$ or $\|x\|\cdot\|y\|$ $\|x\|\cdot\|y\|$. That might make some part of your post a bit more readable. For example, in the original version you have $\sum_{n=1}^\infty x^a_ny_n\leq|||x^a_n| ||y_n||$ $\sum_{n=1}^\infty x^a_ny_n\leq|||x^a_n| ||y_n||$; it's a bit difficult to say just from this equation what you mean there. (To me it seems that the number of absolute value signs which are "left" and "right" aren't the same in that expression.) $\endgroup$ Nov 3 '21 at 6:41
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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.

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  • $\begingroup$ I remember I encountered this statement and it is not clear why if it is bounded is continuous. $\endgroup$
    – Mihai.Mehe
    Oct 27 '21 at 6:14
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    $\begingroup$ @Mihai.Mehe To show this you will need to use the linearity of the map. $\endgroup$
    – jakobdt
    Oct 27 '21 at 6:14
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    $\begingroup$ Linear maps between normed linear spaces are bounded if and only if they are continuous. @Mihai.Mehe $\endgroup$ Oct 27 '21 at 6:19
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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 $f(x)\neq \infty$.

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  • $\begingroup$ My map is bounded above by $\sum_{n=1}^\infty x^a_ny_n\leq ||x_a|||y||$. If this is enough, is the well defined part ok? $\endgroup$
    – Mihai.Mehe
    Oct 27 '21 at 6:09
  • $\begingroup$ @Mihai.Mehe That bound is not correct. $\endgroup$
    – 5xum
    Oct 27 '21 at 6:13
  • $\begingroup$ $||xy||_{l^1}\leq ||x||_{l^q}|y||_{l^p}$ is this not so? I thought I can get the bound from this statement or maybe I made an error somewhere. $\endgroup$
    – Mihai.Mehe
    Oct 27 '21 at 6:21

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