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I cant manage to figure this out, for instance $L^{1}[0,1]$ has $L^{\infty}[0,1]$ as dual and $C[0,1]$ (a sub space of $L^{1}[0,1]$ have the signed measures of bdd. varitaion as dual. I cant mange to prove anything regarding cardinality realtions between the measures and $L^{\infty}[0,1]$ tho. Intuitively it feels like we have more measures.

Hints?

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  • $\begingroup$ @avid19 how does one convince themself of this fact? $\endgroup$
    – user123124
    Commented Dec 10, 2015 at 7:27
  • $\begingroup$ @avid19 What? No infinite dual is larger than the dual of $\Bbb R[X]$? $\endgroup$
    – Hagen von Eitzen
    Commented Dec 10, 2015 at 7:27
  • $\begingroup$ @HagenvonEitzen I should have been more precise. $\endgroup$
    – user223391
    Commented Dec 10, 2015 at 7:29
  • $\begingroup$ Can you state the question more precisely? The particular case of $L^1[0,1]$ and $C[0,1]$ is simple enough (they are both separable so their duals have cardinality $2^{\aleph_0}$, and in fact $L^\infty[0,1]$ canonically embeds in the space of measures via $f\mapsto fd\mu$ where $\mu$ is Lebesgue measure), but it is unclear what general question you're asking. $\endgroup$
    – Eric Wofsey
    Commented Dec 10, 2015 at 7:57
  • $\begingroup$ @EricWofsey The general question arose from an exersice, show that for $\ell^{1} \supset Y = \{ x ; lim n^{2}x_{n} $is bdd$ \} $ any linear functional has atmost one extension to $\ell^{1}$. Then I took the dual $(\ell^{\infty} )$ picked an element and show we get different values if we tamper with it. But for that argument to work I need that $\ell^{\infty} $ is a whole dual on $Y$ aswell or a subset of the whole dual atleast. $\endgroup$
    – user123124
    Commented Dec 10, 2015 at 8:17

1 Answer 1

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Let $X$ be a Banach space and let $Y \subset X$ be a subspace. When you compare the duals, two important situations appear:

  1. $Y$ is a closed subspace of $X$ (and is consequently equipped with the same norm).
  2. $Y$ is a dense, proper subspace of $X$, but is a Banach space with respect to stronger norm.

What happens?

  1. In this case, we get $Y^* \subset X^*$ in the following sense (Here, $Y^*$ are the linear functionals on $Y$ which are continuous w.r.t. the norm in $X$):

Let $y^* \in Y^*$ be given. Then, $y^*$ is a continuous map on a subspace of $X$, and we can extend it by Hahn-Banach to a functional in $X^*$ (with the same norm). If $Y$ is a proper subspace, then this extension is not unique. Thus, $X^*$ is larger than $Y^*$.

  1. In this case, we have $X^* \subset Y^*$ in the following sense (Here, $Y^*$ are the linear functionals, which are continuous w.r.t. the stronger norm of $Y$).

Since $\|y\|_X \le C \, \|y\|_Y$, we get for $x^* \in X^*$: $$|x^*(y)| \le \|x^*\|_{X^*} \, \| y \|_X \le C \, \|x^*\|_{X^*} \, \|y\|_Y.$$ Hence, $x^* \in Y^*$. Moreover, if we have to different functionals in $X^*$, there values on $Y$ differ (since $Y$ is dense in $X$). Thus, $X^*$ is larger than $Y^*$.

Examples

  1. Let me give some examples for the first case.

  2. $\mathbb{R}^n$ can be treated as a subspace of $\mathbb{R}^m$ for $n \le m$ (identify $x \in \mathbb{R}^n$ with $(x_1, \ldots, x_n, 0,\ldots,0) \in \mathbb{R}^m$). Then, $\mathbb{R}^n \subset \mathbb{R}^m$ and we get the same inclusion for the dual spaces.

  3. $C([0,1]) \subset L^\infty(0,1)$. The dual of $C([0,1])$ are regular, signed Borel measures. The dual of $L^\infty(0,1)$ consists of less regular (thus more) measures (namely finitely additive measures). This situation is also a little bit delicate: The dirac $\delta_{1/2}$ lives in the dual of $C([0,1])$, but cannot be applied to arbitrary functions in $L^\infty(0,1)$. However, we can extend it by Hahn-Banach to a finitely additive measure in the dual of $L^\infty(0,1)$ which coincides with $\delta_{1/2}$ on the subspace $C([0,1])$.

  4. Examples for the second case:

  5. You already had a good example: $C([0,1])$ is a dense subspace of $L^1(0,1)$. Note that each function $f$ in $L^\infty(0,1)$ induces a measure via $\mu_f(A) = \int_A f \, \mathrm{d}x$.

  6. $H_0^1(0,1) \subset L^2(0,1)$ and the converse embedding holds for the duals.

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    $\begingroup$ In case (1) you really shouldn't say $Y^*\subset X^*$, especially since $Y$ might not be complemented in $X$. What you should say is there is a surjection $X^*\to Y^*$. It is also worth mentioning that in both cases, you are just dualizing the (continuous linear) inclusion map $Y\to X$ to get a map $X^*\to Y^*$, which happens to be surjective in case (1) and injective in case (2). $\endgroup$
    – Eric Wofsey
    Commented Dec 10, 2015 at 8:02
  • $\begingroup$ @Eric Wofsay whats does " dualizing a map" mean? $\endgroup$
    – user123124
    Commented Dec 10, 2015 at 11:50
  • $\begingroup$ @User1: Taking the adjoint. $\endgroup$
    – gerw
    Commented Dec 10, 2015 at 14:06
  • $\begingroup$ @EricWofsey Whats the difference between surjection of inclusion map and the subset relation? $\endgroup$
    – user123124
    Commented Dec 11, 2015 at 5:25
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    $\begingroup$ I am suspicous about the case 1. Hanh Banach does not provide a unique extension of function $y^*\in Y^*$ to $X^*$ . But it is easy to check that, $X^*\subset Y^*.$ in all cases. $\endgroup$
    – Guy Fsone
    Commented Mar 23, 2019 at 14:50

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