3
$\begingroup$

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...

$\endgroup$
1
  • 1
    $\begingroup$ 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 '17 at 0:53
1
$\begingroup$

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 $x\in X$.

$\endgroup$
4
  • $\begingroup$ 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 '17 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 '17 at 0:49
  • $\begingroup$ Does $g_x$ mean a functional g evaluated at x? $\endgroup$
    – MathIsHard
    Oct 22 '17 at 0:50
  • 1
    $\begingroup$ 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 '17 at 0:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.