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.