# $f_n \rightharpoonup f$, $g_n \to g$ in measure, $\|g_n\|_{L^\infty} \leq M$, then $f_n g_n \rightharpoonup fg$,

Let $f_n$, $f \in L^2(0, 1)$ be such that $f_n$ converges to $f$ weakly in $L^2(0, 1)$. Let $g_n, g : (0, 1) \to \mathbb{R}$ be measurable functions such that $g_n$ converges to $g$ in measure and $\|g_n\|_{L^\infty} \leq M$ for every $n$. Prove that

a) $g\in L^\infty(0,1)$

b) $f_ng_n$ converges to $fg$ weakly in $L^2(0, 1)$.

My attempt: I think that point a) it's ok but i don't know how to finish point b).

a) Since $g_n$ converges to $g$ in measure for all $\varepsilon>0$ we have $$\lim_{n\to\infty} \mu(\{|g-g_n|<\varepsilon\})=1$$ We define $A^\varepsilon_n:=\{|g-g_n|<\varepsilon\}$ and $A^\varepsilon:=\bigcup_n A^\varepsilon_n$. By definition $A^\varepsilon_n \subset A^\varepsilon$ for every $n$, then $\mu(A^\varepsilon_n) \leq \mu(A^\varepsilon)$ and, taking the limit for $n\to +\infty$, we get $\mu(A^\varepsilon)=1$. Defined $$A:=\{\,x\in (0,1)\,|\,|g_n(x)|\leq M\,\},$$ it follows $\mu(A)=1$ (and so $\mu(A^\varepsilon \cap A)=1$). I claim that there exists $C>0$ such that $|g(x)|\leq C$ for every $x\in A^\varepsilon \cap A$. Indeed for every $x\in A^\varepsilon \cap A$, there exists $\overline n=\overline n_x$ s.t. $x\in A^\varepsilon_{\overline n_x}$ and so $$|g(x)|\leq |g(x)-g_{\overline n_x}(x)|+|g_{\overline n_x}(x)|< \varepsilon + M.$$ In conclusion, we have proved $|g(x)|< \varepsilon + M$ a.e., that is $||g||_{L^\infty}\leq M+\varepsilon.$

b) Let $h$ be an arbitrary function in $L^2$, then $$\bigg|\int (f_n g_n - f g) h \, dx \bigg| \leq \bigg| \int (f_n g_n - f_n g)h \, dx\bigg| + \bigg|\int (f_ng - fg) h \, dx\bigg| =$$ $$= \bigg|\int f_n(g_n - g)h \, dx \bigg| + \bigg|\int (f_n - f)g h \, dx\bigg|$$ $$\leq (2M +\varepsilon)\bigg|\int f_n h \, dx \bigg| + \bigg|\int (f_n - f)g h \, dx\bigg|$$ And now ? The second integral goes to zero (since $gh\in L^2$ and $f_n \rightharpoonup f$), but what about the first one ? I know that convergence in measure implies other modes of convergence, but this only applies to subsequences.

1. Firstly, observe that $\sup_n\|f_{n}-f\|_2<\infty$ by uniform boundedness principle: Let $\mathcal{H}=L^2(0,1)$ be a Hilbert space with the inner product $\langle f,g\rangle=\int fg$. For each $n$, define a bounded linear functional $\theta_n:\mathcal{H}\rightarrow\mathbb{R}$ by $\theta_n(h)=\langle f_n-f,h\rangle$. By assumption, for each $h\in\mathcal{H}$, $\theta_n(h)\rightarrow0$. Hence $\{\theta_n(h)\mid n\in\mathbb{N}\}$ is bounded. By uniform boundedness principle, we have that $\sup_n\|\theta_n\|<\infty$. By Riesz Representation Theorem, $\mathcal{H}^{\ast}=\mathcal{H}$ and $\sup_n\|f_{n}-f\|_2=\sup_n\|\theta_n\|<\infty$.

2. We assume that (a) has been proved. Choose another $M$ (for simplicity, we still denote it by $M$) such that $\|g_n\|_{L^\infty}\leq M$ and $\|g\|_{L^\infty}\leq M$. Let $h\in\mathcal{H}$ be arbitrary. Then \begin{eqnarray*} \int(f_n g_n-fg)h & = & \int(f_n-f)gh+\int f_n(g_n-g)h\\ & = & \langle f_n-f,gh\rangle+\int f_n h(g_n-g). \end{eqnarray*} Now $\langle f_n-f,gh\rangle\rightarrow0$ because $gh\in\mathcal{H}$. By 1, we have $\sup_n\|f_n\|_2<\infty$. By Cauchy-Schwarz inequality, we have $\int|f_n h|\leq\|f_n\|_2\|h\|_2$ and hence $\sup_n\int|f_n h|<\infty$. Let $\varepsilon>0$. Let $B_n=\{x\mid|g_n(x)-g(x)|\leq\varepsilon\}$. Consider \begin{eqnarray*} & & \int f_n h(g_n-g)\\ & = & \int_{B_n}f_n h(g_n-g)+\int_{B_n^c}f_n h(g_n-g). \end{eqnarray*} Observe that \begin{eqnarray*} & & \left|\int_{B_n}f_{n}h(g_n-g)\right|\\ & \leq & \varepsilon\int\left|f_n h\right|\\ & \leq & \varepsilon\cdot\sup_n \int|f_n h| \end{eqnarray*} For the second integral, by Cauchy-Schwarz inequality, we have \begin{eqnarray*} & & \left|\int_{B_n^c}f_{n}h(g_{n}-g)\right|\\ & \leq & 2M\int|f_n\cdot1_{B_n^c}h|\\ & \leq & 2M\|f_n\|_2\|1_{B_n^c}h\|_2\\ & \leq & 2M\cdot\sup_n\|f_n\|_2\cdot\|1_{B_n^c}h\|_2 \end{eqnarray*} Lastly, we show that $\int_{B_n^c}h^2\rightarrow0$. For each $m\in\mathbb{N}$, let $\xi_m=\min(h^2,m)$. Note that $\xi_m\rightarrow h^2$ and $|\xi_m|\leq h^2$. By Dominated Convergence Theorem, we have $\int|\xi_{m}-h^2|\rightarrow0$ as $m\rightarrow\infty$. For any $m,n\in\mathbb{N}$, we have \begin{eqnarray*} & & \left|\int_{B_n^c}h^2\right|\\ & \leq & \int_{B_n^c}|h^2-\xi_m|+\int_{B_n^c}|\xi_m|\\ & \leq & \int|h^2-\xi_m|+\int_{B_n^c}|\xi_m|\\ & \leq & \int|h^2-\xi_m|+m\mu(B_n^c). \end{eqnarray*} Fix $m$ and take $\limsup_{n\rightarrow\infty}$, we have $\limsup_n\left|\int_{B_n^c}h^2\right|\leq\int|h^2-\xi_m|$. Let $m\rightarrow\infty$, we further have $\limsup_n\left|\int_{B_n^c}h^2\right|=0$. This show that $\int_{B_n^c}h^2\rightarrow0$.

• +1 very nice answer. Jun 30, 2018 at 6:47
• Nice proof. Thanks. Do you agree with my solution of point a) ? Jun 30, 2018 at 10:05

One may use some existing properties of weakly convergence to prove such a result.

By definition, it suffices to prove that for any $h\in L^2$, $$\int f_ng_n h\rightarrow \int fgh$$ Now, we first fix $h$.

Let $\Omega=(0,1)$. Since $g_n$ converges to $g$ in measure, there exists a subsequence $g_{k(n)}$ of $g_n$ such that $g_{k(n)}$ converges to $g$ almost everywhere. By Egorov' theorem (since $\Vert g_n\Vert_{\infty}\leq M$), for each $\varepsilon>0$, there exists a set $A$ with $\mu(A)\leq \varepsilon$ such that $g_{k(n)}$ converges to $g$ uniformly on $\Omega\setminus A$.

On the other hand, we have the following theorem: Let $B$ be a bilinear continuous functional from $X\times Y$ to $\mathbb{R}$ where $X$ and $Y$ are normed space. If $x_n \rightharpoonup x$ in $X$ and $y_n\rightarrow y$ in $Y$, then we have $$B(x_n,y_n)\rightarrow B(x,y)$$ Observe that $B_h(x,y):=\int_{\Omega\setminus A} xy h$ is a bilinear continuous functional from $L^2(\Omega\setminus A)\times L^\infty(\Omega\setminus A)$ to $\mathbb{R}$, $g_{k(n)}\rightarrow g$ in $L^\infty(\Omega\setminus A)$ and $f_n \rightharpoonup f$ in $L^2(\Omega\setminus A)$ by assumption. Then above theorem implies that $$\int_{\Omega\setminus A}f_n g_{k(n)}h\rightarrow\int_{\Omega\setminus A}fgh$$ Thus, \begin{aligned} \left\vert \int_\Omega f_ng_{k(n)}h-\int_\Omega fgh\right\vert&\leq \left\vert\int_{\Omega\setminus A}(f_ng_{k(n)}h-fgh)\right\vert+\left\vert\int_A(f_ng_{k(n)}h-fgh)\right\vert\\ (\text{take n large enough})\quad &\leq \varepsilon+\mu(A)(\Vert f_n\Vert_{L^2}+\Vert f\Vert_{L^2})2M\Vert h\Vert_{L^2}\\ (\text{weak convergence implies boundedness})\quad &\leq \varepsilon+C\cdot \varepsilon \end{aligned} Thus, we have $$\int_{\Omega}f_n g_{k(n)}h\rightarrow\int_{\Omega}fgh$$ Since the limit is independent of the choice of subsequence, we have $$\int_{\Omega}f_n g_{n}h\rightarrow\int_{\Omega}fgh$$ which completes the proof.

• Nice proof. Thanks. When you say "limit is independent of the choice of subsequence" you mean that "since for each subsequence $g_{n_k}$ we can extract a new subsequence $g_{n_{k_{j}}}$ converging to the same limit then all the sequence $g_n$ must converge" ? Jun 30, 2018 at 10:09
• Do you agree with my solution of point a) ? Thanks. Jun 30, 2018 at 10:11
• @Ef_Ci Yes, that's what I mean. And your proof of a) seems ok for me.
– Q-Y
Jun 30, 2018 at 10:32