Can the weak-* topology in $L^{\infty}$ be characterized by convergence of with respect to simple test functions??

Question

Consider the weak-* topology in $L^{\infty}(X,\mu,\mathcal B)$ in which a net $\{f^{\alpha}\}\subseteq L^{\infty}$ converges to $f^*$ if and only if $$\int f^{\alpha} g d\mu \to \int f^{*} g d\mu,$$ for every $g\in L^1(X,\mu,\mathcal B)$.

Since the simple functions are dense in $L^{1}$ is this equivalent to $\{f^{\alpha}\}\subseteq L^{\infty}$ converges to $f^*$ if and only if $$\int f^{\alpha} g d\mu \to \int f^{*} g d\mu,$$ for every $g=\mathbb 1\{x\in B\}$ with $B\in \mathcal B$? I understand that this is true if we know that $f^{\alpha}$ is uniformly bounded. Is it true in general?

Answer 1

No to the question as stated. In fact

Suppose $X$ is a Banach space and $V$ is a proper subspace of $X$. There exists a net $(x_\alpha^*)\subset X^*$ such that $x_\alpha^*(x)\to0$ for every $x\in V$ but $(x_\alpha^*)$ is not weak* convergent.

Two proofs, one from the definitions and one using a well known not quite trivial theorem:

Proof 1: Say $z\in X\setminus V$.

If $V$ is finite-dimensional: Hahn-Banach shows there is a sequence $(x_n^*)\subset X^*$ with $x_n^*(x)=0$ for all $x\in V$ but $x_n^*(z)=n$.

If $V$ is infinite-dimensional: Say $(x_\beta)_{\beta\in B}$ is a (Hamel) basis for $V$. For every finite set $F\subset B$ let $V_F$ be the span of $\{z\}\cup(x_\beta)_{\beta\in F}$.

There exists a linear $\Lambda_F:V_F\to\Bbb C$ such that $$\Lambda_F(x_\beta)=0\quad(\beta\in F)$$and $$\Lambda_F(z)=|F|,$$ where $|F|$ is the cardinality of $F$. Hahn-Banach shows there exists $x_F^*\in X^*$ with $$x_F^*|_{V_F}=\Lambda_F$$.

Consider $(x_F)$ as a net in $X^*$, where the $F$ are ordered by inclusion. Then $x_F^*(x)\to0$ for every $x\in V$, but $x_F^*(z)\to\infty$.

Detail: Say $x\in V$. There exists a finite set $F_0$ with $x\in V_{F_0}$. Hence $x_F^*(x)=0$ for every $F$ with $F_0\subset F$; hence $x_F^*(x)\to0$.

Proof 2: If $W$ is a linear spaces of linear functionals on the vector soace $Y$ let $\tau_W$ be the weakest topology on $Y$ making every element of $W$ continuous. Recall:

Theorem $(Y,\tau_W)^*=W$.

Hence in the setting above we have $$(X^*,\tau_{X})^*\ne(X^*,\tau_V)^*,$$so $$\tau_X\ne\tau_V.$$

Detail: Hence $X$-convergence for nets in $X^*$ is not equivalent to $Y$-convergence; since the first implies the second, it must be that the second does not imply the first.

Answer 2

WARNING: This answer only consider sequences, not arbitrary nets. For nets, the result fails as David C.Ullrich proves in the other answer.

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This is a general fact of weak convergence. If $X$ is a Banach space, and $G\subset X$ is dense, a sequence $f_n\in X^\star$ converges weakly to $f\in X^\star$ if and only if it is uniformly bounded and $$ \langle f_n, g\rangle \to \langle f, g\rangle, \qquad \forall g\in G.$$