Closure of $C([0,1]^2)$ via weak*-topology [closed]

Question

Let $C([0,1]^2)$ denote the set of continuous functions on $[0,1]^2$. Let $L^1([0,1]^2)$ be the set of all Lebesgue integrable functions on $[0,1]^2$.

The dual space of $C([0,1]^2)$, denoted by $C^*([0,1]^2)$, consists of all bounded, linear functionals that map from $C([0,1]^2)$ to $\mathbb{R}$. Specifically, $$C^*([0,1]^2) = \left\{ \varphi: C([0,1]^2) \to \mathbb{R} \mid \varphi \text{ is linear and continuous} \right\}.$$ The norm of any $\varphi \in C^*([0,1]^2)$ is given by the weak* norm: $$\lVert \varphi \rVert_* = \sup_{\lVert f \rVert_C \leq 1} |\varphi(f)|.$$

We have $C\subseteq L_1\subseteq C^*$.

Question Does the closure of $C([0,1]^2)$ be $L_1([0,1]^2)$ (or a subset of $L_1([0,1]^2)$) when considering topology $\lVert \cdot \rVert_*$?

Answer 1

Let's look more closely at the inclusions you wrote. The first, $C\subset L^1$, is dense w.r.to the $L^1$-norm (for this, it is sufficient to note that the $\|\cdot\|_1$-closure of $C$ contains the characteristic functions of rectangles $[a,b]\times[c,d]$, which is clear, hence contains all characteristic functions of measurable sets, hence all integrable simple functions, hence the whole $L^1$).

The other inclusion, $L^1\subset C^*$, is isometric just because for every $f\in L^1$ $$\|f\|_{C^*}:= \sup_{g\in C, \|g\|_C\le 1}\int fg=\|f\|_1,$$ which follows from the preceding one, since you can approximate $\text{sgn} f$ in $L^1$ by a sequence $(g_k)_k$ in the unit ball of $C$. Being isometric, the inclusion $L^1\subset C^*$ is closed, because $L^1$ is complete.

But then the closure of any set $S\subset L^1$ into $C^*$ w.r.to the $\|\cdot\|_{C^*}$ norm is exactly the same as its closure in $L^1$ w.r.to the $\|\cdot\|_{L^1}$-norm; in particular the $\|\cdot\|_{C^*}$-closure of $C$ in $C^*$ is $L^1$.

As to the question in the title: The weak* closure of $C$ as a subset of $C^*$ is $C^*$: recall that a linear subspace $V$ of a dual $X^*$ is weak* dense iff its pre-annihilator $V_\perp:=\{x\in X: \forall f\in V\, \langle f,x\rangle=0\}$ is $(0)$, which is of course the case here.

Answer 2

By the Riesz–Markov theorem, your $C^*$ is the space of all finite signed Borel measures on $Q:=[0,1]^2$ endowed with the total variation norm $\|\cdot\|$, which is the same as your dual norm $\|\cdot\|_C$.

Let now $(f_n)$ be a sequence of functions in $L_1$ such that the corresponding measures converge to a signed measure $\mu\in C^*$ wrt the total variation norm $\|\cdot\|$, that is, $$d_n:=\int_Q|f_n(x)\,dx-\mu(dx)|\to0. \tag{1}\label{1}$$

Suppose for a second that the singular (wrt to the Lebesgue measure) component (say $\nu$) of $\mu$ is
nonzero, so that $\mu(dx)=f(x)\,dx+\nu(dx)$ for some $f\in L_1$, and for some Borel subset $B$ of $Q$ of Lebesgue measure $0$ we have $c:=|\nu|(B)>0$. Then $d_n\ge c>0$ for all $n$, which contradicts \eqref{1}. So, $\nu=0$ and $\mu(dx)=f(x)\,dx$.

Thus, indeed the closure of $C$ is a subset of $L_1$.

On the other hand, the restriction of the total variation norm $\|\cdot\|$ to $L_1$ is the $L_1$-norm. Also, $C$ is dense in $L_1$ wrt to the $L_1$-norm.

Thus, the closure of $C$ is $L_1$.