Understanding the duals of $L^{\infty}(K)$ and $C(K)$ and weak-* compactness

Question

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.

Answer 1

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.

Answer 2

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.)