# Confusion about how a dual space element is also a mapping.

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).

• The spaces are not equal, they are isomorphic. It is the dual of $\ell^p$ up to isomorphism. Sep 17 '20 at 11:14

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.
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 $$l_p .$$