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I am trying to understand the second dual space and the notation in my book. The book Kreyszig defines the dual space as follows:

We define a functional $g_x$ on $X'$ by choosing a fixed $x \in X$ and setting $$g_x(f)=f(x) $$ $$(f \in X' variable)$$

What I don't understand is how $g_x (f)$ works. Is it the composition of functionals? If so, how does the composition equal $f(x)$ because wouldn't $f(x)$ give a scalar value? Thank you for any help...

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    $\begingroup$ Perhaps an example would help. Let $X=c_0$, so $X'=\ell^1$. Let $x=(x_k)_{k\in\mathbb N}\in c_0$ be given. Then we can define a functional $g_x$ on $\ell^1$ such that for any $y=(y_k)_{k\in\mathbb N}$, we have $$g_x(y)=y(x)=\sum_{k=1}^\infty x_ky_k.$$ That is, $g_x$ is just evaluation of the functional at $x$. $\endgroup$
    – Aweygan
    Oct 22, 2017 at 0:53

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You want to define a functional on $X'$. That is, a mapping $X' \to \mathbb F$. For each $f\in X'$ you have to output a scalar. So $g_x$ does the job: you input a $f\in X'$, $g_x(f)$ returns a scalar, which is $f(x)$ (note that $f : X\to \mathbb F$, so $f(x)$ is a scalar).

Note that in the above discussion I am fixing a $x\in X$.

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  • $\begingroup$ Thank you. So can I think of $g_x(f)$ as $g_x(f)(x)$? I guess I just don't see how $g_x(f)$ gives a scalar since they are both functionals... $\endgroup$
    – MathIsHard
    Oct 22, 2017 at 0:47
  • $\begingroup$ No, not at all. Is it not a composition. Like I said, $x$ is fixed. so $f(x)$ is not a function, but a scalar. $\endgroup$
    – user99914
    Oct 22, 2017 at 0:49
  • $\begingroup$ Does $g_x$ mean a functional g evaluated at x? $\endgroup$
    – MathIsHard
    Oct 22, 2017 at 0:50
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    $\begingroup$ No. The $x$ is fixed. For each fixed $x$, you define a mappint $X'\to \mathbb F$. @Math4Life (However, one can think of $x\mapsto g_x$ as a mapping $X \to X''$. $\endgroup$
    – user99914
    Oct 22, 2017 at 0:51

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