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Following Kreyszig's Introductory Functional Analysis, I am introduced to the dual space as the set of all bounded linear functionals mapping from the origin space X to $R$, which is endowed with a vector structure, making it a vector space.

Right after that, to find as an example the dual space to $l_p$ space, we show that $l_p$ is isomorphic with $l_q$, where $1/p+1/q=1$, and it is said that $l_q$ is hence the dual space.

But I don't understand how an infinite sequence (an element of space $l_q$) can be a mapping from $l_p$ to $R$ (an element of the dual space).

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    $\begingroup$ The spaces are not equal, they are isomorphic. It is the dual of $\ell^p$ up to isomorphism. $\endgroup$
    – Qi Zhu
    Sep 17 '20 at 11:14
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There is a simple way an element of $\ell_q$ can be thought of as a functional. This is at the heart of the isomorphism and I expect is set out in your text book. If $x \in \ell_p$, and $y \in \ell_q$ then $y$ corresponds to the functional $f$ defined by $$ f(x) = \sum_{n=1}^\infty x_n y_n $$ The right hand is always bounded by $\lVert x \rVert_p \cdot \lVert y \rVert_q $ for $x \in \ell_p$, $y \in \ell_q$ by virtue of Holder's inequality, so that every $y$ creates a bounded linear functional. That every bounded functional can be so represented is more difficult: given a bounded functional $f$ we can create a likely candidate for $y \in \ell_q$ by setting $$y= \Big( f(1,0,0,\cdots), f(0,1,0\cdots), \cdots \Big).$$ The challenge then is show that this $y$ always lies in $\ell_q$, which I trust is handled in in your text book.

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Any element $x\in l^q $ defines a functional $x^{*} : l_p \to \mathbb{R} (\mathbb{C} )$ in a such way $$x^{*} (y) =\sum_{j=1}^{\infty} x_j \overline{y_j }.$$

Moreover such correspondence $(x\to x^{*} ) $ is linear and $||x^{*} || =||x||_q $ so we can idntify elements of $l_q$ with linear functionals on $l_p .$

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