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


*

*$Y$ is a closed subspace of $X$ (and is consequently equipped with the same norm).

*$Y$ is a dense, proper subspace of $X$, but is a Banach space with respect to stronger norm.


What happens?


*

*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 $Y^*$.

*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, $Y^*$ is larger than $Y^*$.
Examples


*

*Let me give some examples for the first case.


*

*$\mathbb{R}^n$ 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.

*$C([0,1]) \subset L^\infty(0,1)$. 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 $C([0,1])$.


*Examples for the second case:


*

*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 $\mu_f(A) = \int_A f \, \mathrm{d}x$.

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


