Weak Star Convergence of $u_n=\sin(\frac{nx}{n+1})(1+\exp{(-n|y|)})$

Question

Consider the sequence

\begin{equation}u_n=\sin(\frac{nx}{n+1})(1+\exp{(-n|y|)}) \text{ where } (x,y)\in I=[-1,1]\times[-1,1], n\in\mathbb{N}.\end{equation}

(a) Study the equicontinuity of $(u_n)$ on $I$.

(b) Study convergence of $(u_n)$ in th weak* topology of $L^\infty$

My Attempt

(a) Since $u_n(x,y)\rightarrow$ $u= \begin{cases} 2\sin{x} & y=0 \\ \sin{x} & \text{else} \\ \end{cases}$ for all $(x,y)\in I$ where $u$ is not continous, $(u_n)$ is not Equicontinuous.

(b) We show \begin{equation} \int u_n(x,y)v(x,y)\rightarrow \int u(x,y)v(x,y), \quad\forall v\in L^1. \end{equation}

Please how to I show that? What if the function is modified to $u_n=(1+\sin(nx))(1+\exp(-n|y|)$?

Answer 1

Since $u_n(x,y)\rightarrow u(x,y)$ pointwise and the integrand \begin{equation}|u_n(x,y)||v(x,y)|\leq 2|v(x,y)|\end{equation} (is dominated) for all $v\in L^1$, we have by Dominated Convergence Theorem that

\begin{equation}\int u_n(x,y)v(x,y)\rightarrow \int \sin(x,y)v(x,y)\end{equation} on $I - [-1,1]\times \{0\}$ where $[-1,1]\times \{0\}$ has measure zero. Hence, $u_n$ converges in $L^\infty$ weak* to $\sin(x)$

Answer 2

Note that the family $C([-1,1]^2)$ of all continuous functions is dense in $(L^1([-1,1]^2),\|\cdot\|_1)$. Since $\{u_n\}$ is a uniformly bounded sequence in $L^\infty$, it suffices to show that $$ \begin{equation} \int u_n(x,y)v(x,y)\;dxdy\rightarrow \int u(x,y)v(x,y)\;dxdy \tag{*}\end{equation}$$ holds for all $v\in C([-1,1]^2)$.

Now observe that by Stone-Weierstrass theorem, the polynomial algebra generated by $\{x^ny^m\;|\;n,m\ge 0\}$ is dense in $C([-1,1]^2)$ with respect to the supremum norm, and hence with respect to the $L^1$-norm. By the same argument, it is sufficient to show $(*)$ holds for $v(x,y) = x^ny^m$, $n,m\ge 0$. Now, the conclusion follows from Fubini's theorem and Lebesgue's dominated convergence theorem. (note: for the modified one, we can apply Riemann-Lebesgue lemma about $x$.)