Let $(K\subset \mathbb{R}^n,\mathcal{B}(K),\lambda)$ be a measure space, where $\lambda$ is the Lebesgue measure and $K$ is compact.

According to Wikipedia (with adapted notation),

The dual Banach space $L^\infty(K)^*$ is isomorphic to the space of finitely additive signed measures on $K$ that are absolutely continuous with respect to $\lambda$.

On the other hand, the Riesz-Markov theorem tells us that the dual of the (non-dense) subspace $C(K)\subset L^\infty(K)$ is the space of regular countably additive measures.

Is the former included in the latter, for $K$ compact? What is the difference of the former and $L^1$?

In any case, by the Hahn-Banach theorem, since we equip both $C(K)$ and $L^\infty$ with the same ($esssup$) norm, shouldn't we have an inclusion in the other way, i.e. shouldn't the dual of $L^\infty(K)$ be larger than that of $C(K)$? (Because every cts. functional on $C(K)$ can be extended to $L^\infty(K)$)

Our professor today said that by shrinking our attention from $L^\infty$ to $C$, the dual grows and we can find weak-* convergent subsequences in the duals (we found that $L^1\subset (L^\infty)^*$ doesn't always have weak-* convergent subsequences...). I know that the formal reason is Banach-Alaoglu and $C$ unlike $L^\infty$ is separable, but I'm curious for what is wrong about my professors statement about shrinking the space and growing the dual and what truth is in it.


If we have a Banach space $X$ with norm $\|\cdot\|$ and a (proper) closed subspace $Y\subset X$, then their duals are related via $X' \subset Y'$:

This is due to the fact that the dual of $X$ is the space of bounded linear functionals $f$ on $X$, i.e. the quantity $$\sup_{x \in X, \|x\|=1}|f(x)|$$ has to be finite in order for $f\in X'$ (assuming $f:X\rightarrow \mathbb R$ is linear).

On the other hand, in order to achieve $f\in Y'$, we just need to fulfill $$\sup_{x \in Y, \|x\|=1}|f(x)|.$$

Since $Y\subset X$, the latter condition is a weaker one and for all $f\in X'$, we trivially have $f\in Y'$, but not the other way around.

tl;dr: It's easier for $f$ to be continuous on a small set than on a larger set.


Fours year later, I hope I understand the situation a bit better and I am very confident to say that the current answer by @Roland is wrong:

Just look at $Y:=\mathbb{R}^2\subset X:=\mathbb{R}^3$. The duals are $Y'=(\mathbb{R}^2)'\cong\mathbb{R}^2$ and $X'=(\mathbb{R}^3)'\cong \mathbb{R}^3$ so we can certainly not find an embedding $X'\subset Y'$.

The flaw in the reasoning It's easier for $f$ to be continuous on a small set than on a larger set, hence the dual of the smaller space is larger is a bit tricky to understand and requires three observations:

  1. It is true that every continuous function on the larger space is also continuous on the smaller space. However, if we use this to identify an element of $X'$ with one of $Y'$ we get a map that is not injective. Thus, we do not have an imbedding $X'\subset Y'$

  2. What I wrote in the OP about Hahn-Banach was true: Every continuous functional on the smaller space can be extended to one on the larger space. This hints at the correctness of $Y'\subset X'$. Intuitively: There are more functions on a larger set than on a smaller set.

  3. However, one sometimes needs to be careful about what identifications one makes. For example, $(L^1)^*=L^{\infty}$ is false, the correct statement is: There is an isomorphism between $(L^1)^*$ and $L^{\infty}$. For another example, $C(\mathbb{R})\subset C(\mathbb{C})$ is false, the correct statement is: There is an injection of $C(\mathbb{R})$ into $C(\mathbb{C})$ Similarly, $Y'\subset X'$ is obviously wrong if taken literally. It is true that Hahn-Banach allows us to find a map $\phi: Y'\to X'$ such that $\phi(y')(y)=y'(y)$(i.e. $\phi$ is a right-inverse of the canonical map $X'\to Y'$ given by the dual of the inclusion map $Y\to X$); however, this map is not necessarily linear. With a bit more effort, one can show that $Y'$ is indeed linearly isomorphic to $X'/Y^{\bot}$, but again, this does not mean that $Y'\subset X'$

In summary, it is true that $\mathcal{M}(K)(\cong C(K)^*)$ can be identified with some quotient of the space of finitely additive $\lambda$-a.c. measure ($\cong (L^{\infty}(K))^*$) and is therefore in some sense smaller, but this abstract statement does not mean that the identity map on the space of measures embeds the countably additive measures, $\mathcal{M}(K)$, into the fin. add. $\lambda$-a.c. measures. (Indeed, the Dirac measure is in the former but not in the latter.)


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.