Suppose $X$ is a finite measure space, and $f_n$ is uniformly bounded and converges to $f$ in the weak-star topology of $L^\infty(X)$. This means $\int f_n\phi \to \int f\phi$ for all $\phi\in L^1(X)$. If also $g_n$ is uniformly bounded and converges to $g$ in the weak-star topology of $L^\infty(X)$, does the product $f_ng_n$ converge weak-star to $fg$?
Thank you.
On
This is my attempt to modify the answer given above by PhoemueX to provide a counterexample in the $L^\infty([0,1])$ case. For each positive integer $n$, define $$ f_n:[0,1]\to \{-1,1\}, x\mapsto \sum_{k=0}^{2^n-1} (-1)^k 1_{[k2^{-n},(k+1)2^{-n})}(x). $$ Then $f_nf_n=1_{[0,1)}$, but we will show that $f_n\to 0$ weak-star in $L^\infty([0,1])$. Since the linear span of characteristic functions of dyadic intervals is dense in $L^1([0,1])$, it is enough to show that $$ \int f_ng\to 0 $$ for $g=1_{[j2^{-m},(j+1)2^{-m})}=1_E$ ($E$ is a dyadic interval in $[0,1]$). We note that, for $n>m$, the measure of $E\cap [k2^{-n},(k+1)2^{-n})$ is either $0$ or is equal to $2^{-n}$. Therefore, when $n>m$, we have that $$ \int f_ng=\sum_{k=0}^{2^n-1} (-1)^k m\Big(E\cap{[k2^{-n},(k+1)2^{-n})}\Big)=0. $$ Hence $f_n\to 0$ weak-star, but $f_nf_n=1_{[0,1)}$.
What you are trying to show is false. In the following, I give a counterexample.
Define $$ f:\mathbb{R}\to\left\{ 1,-1\right\} ,x\mapsto\sum_{n=-\infty}^{\infty}\left(-1\right)^{n}1_{\left[n,n+1\right)}. $$ Note that $y^{2}=y$ for all $y\in\left\{ \pm1\right\} $, so that we get $\left(f\left(x\right)\right)^{2}=1$ for all $x\in\mathbb{R}$. Thus, $f_{n}\cdot f_{n}\equiv1$ for $$ f_{n}:\mathbb{R}\to\mathbb{R},x\mapsto f\left(nx\right). $$ Hence, if we can show $f_{n}\xrightarrow[n\to\infty]{\text{weak-}\ast\text{ in }L^{\infty}}0$, then we will have a counterexample to your claim (since your claim would imply $1=f_{n}\cdot f_{n}\xrightarrow[n\to\infty]{\text{weak-}\ast\text{ in }L^{\infty}}0\cdot0=0$, which is absurd).
Note that the sequence $\left(f_{n}\right)_{n\in\mathbb{N}}$ is uniformly bounded in $L^{\infty}\cong\left(L^{1}\right)^{\ast}$, so that it suffices to show $\int f_{n}\left(x\right)\cdot g\left(x\right)\,{\rm d}x\xrightarrow[n\to\infty]{}0$ for all $g\in C_{c}^{\infty}\left(\mathbb{R}\right)$, since those functions are dense in $L^{1}$. Given such a function, note that there are $M\in\mathbb{N}$ with ${\rm supp}\,g\subset\left(-M,M\right)$ and with $\left|g\left(x\right)-g\left(y\right)\right|\leq L\cdot\left|x-y\right|$ for all $x,y\in\mathbb{R}$ (note that $g$ has a bounded derivative and is thus Lipschitz continuous).
Hence, \begin{align*} \left|\int f_{n}\left(x\right)\cdot g\left(x\right)\,{\rm d}x\right| & =\left|\frac{1}{n}\cdot\int f\left(y\right)g\left(\frac{y}{n}\right)\,{\rm d}y\right|\\ \left[g\left(y/n\right)=0\text{ unless }y\in\left(-Mn,Mn\right)\right] & =\left|\frac{1}{n}\sum_{\ell=-Mn}^{Mn-1}\left(-1\right)^{\ell}\int_{\ell}^{\ell+1}g\left(\frac{y}{n}\right)\,{\rm d}y\right|\\ \left[k=\ell+Mn\right] & =\left|\frac{1}{n}\sum_{k=0}^{2Mn-1}\left(-1\right)^{k-Mn}\int_{k-Mn}^{k-Mn+1}g\left(y/n\right)\,{\rm d}y\right|. \end{align*}
Now, we will use that two consecutive terms (almost) cancel out. To this end, define $$ a_{k}:=\left(-1\right)^{k}\cdot\int_{k-Mn}^{k-Mn+1}g\left(y/n\right)\,{\rm d}y $$ and note \begin{align*} & a_{2k}+a_{2k+1}\\ & =\int_{k-Mn}^{k-Mn+1}g\left(y/n\right)\,{\rm d}y-\int_{k+1-Mn}^{k-Mn+2}g\left(z/n\right)\,{\rm d}z\\ \left[y=z-1\text{ in the 2nd integral}\right] & =\int_{k-Mn}^{k-Mn+1}g\left(\frac{y}{n}\right)-g\left(\frac{y+1}{n}\right)\,{\rm d}y. \end{align*} We conclude \begin{align*} \left|a_{2k}+a_{2k+1}\right| & \leq\int_{k-Mn}^{k-Mn+1}\left|g\left(\frac{y}{n}\right)-g\left(\frac{y+1}{n}\right)\right|\,{\rm d}y\\ & \leq\int_{k-Mn}^{k-Mn+1}\frac{L}{n}\,{\rm d}y=\frac{L}{n}. \end{align*}
All in all, we arrive at \begin{align*} \left|\int f_{n}\left(x\right)g\left(x\right)\,{\rm d}x\right| & =\left|\frac{1}{n}\sum_{k=0}^{2Mn-1}a_{k}\right|\\ & =\left|\frac{1}{n}\sum_{\ell=0}^{Mn-1}a_{2\ell}+a_{2\ell+1}\right|\\ & \leq\frac{1}{n}\sum_{\ell=0}^{Mn-1}\frac{L}{n}\\ & =\frac{LMn}{n^{2}}=\frac{LM}{n}\xrightarrow[n\to\infty]{}0, \end{align*} as desired.