# Can a subspace have a larger dual?

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?

• @avid19 how does one convince themself of this fact? Dec 10, 2015 at 7:27
• @avid19 What? No infinite dual is larger than the dual of $\Bbb R[X]$? Dec 10, 2015 at 7:27
• @HagenvonEitzen I should have been more precise.
– user223391
Dec 10, 2015 at 7:29
• 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. Dec 10, 2015 at 7:57
• @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. Dec 10, 2015 at 8:17

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.

• 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). Dec 10, 2015 at 8:02
• @Eric Wofsay whats does " dualizing a map" mean? Dec 10, 2015 at 11:50
• 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. Mar 23, 2019 at 14:50