I am an engineering student and I am reading the book "Introductory Functional Analysis " by Kreyszig and am lost in the proof of finding the dual space of the $l^{1}$ space. Here is how author proves it:

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My Questions : Now my understanding of the dual space is that dual space,denoted by $X^{*}$, of a normed space $X$ is the set of all bounded linear functionals on $X$. Now assuming the dual space of $l^{1}$ is $l^{\infty}$ , this means that every element of the set of all bounded linear functionals on $l^{1}$ is an infinite sequence in the normed space $l^{\infty}$. So in equation (6), I can see that $f(x)$ is a real valued function, but how would $f$ look like given it has to be both linear and a bounded infinite sequence?

Next, my understanding of author's complete derivation of the dual of $l^{1}$ norm is as follows: He assumes a bounded linear functional $f(x)$ on $l^{1}$ and he eventually wants to prove that $||f||$ is given by the supremum of a sequence and hence $f$ must belong to the $l^{\infty}$ because the supremum norm is the norm of $l^{\infty}$. Is my understanding here correct?

Further, To prove this point he uses the boundedness property of $f$ to first show the inequality (As in equation 7) And then making use of $7(a)$ and $7(b)$ he proves the equality as in Equation 8 which would mean the $\|f\|$ is given as if $f \in l^{\infty}$ .

Finally if this is what the author's main proof , then how is the text between Equation 7 and Equation 7a relevant to proof ?

If my assumption(s) are wrong . Please correct me .

  • 1
    $\begingroup$ The objects in $\ell^{\infty}$ are sequences, not functionals on $\ell^1$. There is correspondence between the two, which will turn out to be equation (6). To prove the correspondence you need to prove three things: 1) That starting with a bounded functional $f$ the sequence of coefficients $\gamma_k$ form an element of $\ell^\infty$. 2) That starting with an element of $\ell^\infty$ and using them as the coefficients in (6) produces a bounded linear functional $f$. This is what is being argued between (7) and (7.a). $\endgroup$
    – user647486
    Apr 28 '19 at 3:02
  • $\begingroup$ 3) That the norm of $f$ and the norm of its corresponding element in $\ell^\infty$ are the same value. This is done from the sentence before (7.a) onward. $\endgroup$
    – user647486
    Apr 28 '19 at 3:03

Let us define a mapping $\varphi \colon {\ell^1}^\prime \to \ell^\infty$ as follows: Let $f \in {\ell^1}^\prime$; that is, let $f$ be a bounded linear functional defined on $\ell^1$. Then let us define $$ \varphi(f) \colon= \left( \ f\left(e_1\right), f \left( e_2 \right), \ldots \right).$$ Then Kreyszig's proof is all about saying that this map is an isomorphism between the spaces ${\ell^1}^\prime$ and $\ell^\infty$, which means that, corresponding to each element of the space ${\ell^1}^\prime$, there is one and only one element of the space $\ell^\infty$, and also that this correspondence preserves the vector space operations of addition and scalar multiplication and the operation of taking the norm.

Therefore for the purposes of the theory of normed spaces and bounded linear operators (and functionals) on normed spaces the spaces ${\ell^1}^\prime$ and $\ell^\infty$ can be regarded as identical.

Please also read through Sec. 2.8 in Kreyszig, specifically from the paragraph on page 108 that begins with "In our work we are concerned with various spaces. ... "

Hope this helps.


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