Let $\Omega = \mathopen]0,1\mathclose[$ and let a function $A_n: \Omega \to \mathbb R$ defined as: $$A_n(x) = \begin{cases}\alpha &\text{if } k \epsilon \leq x < (k+\tfrac{1}{2}) \epsilon \\ \beta &\text{if } \big(k+\tfrac{1}{2}\big) \epsilon \leq x < (k+1) \epsilon \end{cases} $$ for $k=0$, $1,\ldots,n-1$ where $\alpha$, $\beta > 0$ and $\epsilon=1/n$.
How can I compute the weak star limit for $A_n$ in $L^{\infty}(\Omega)?$
Thanks for your help.
On
Since $L_\infty(\Omega)=L_1^*(\Omega)$ you need to show that for all $f\in L_1(\Omega)$ $$ \lim\limits_{n\to\infty}\langle A_n, f\rangle = \langle A, f\rangle\tag{1} $$ where $A$ is the desired limit. In fact it is enough to check $(1)$ only for some functions $f\in S$, where $L_1(\Omega)=\overline{\mathrm{span}S}$. Now consider $$ S=\left\{\chi_{[a,b)}:[a,b)\subset [0,1)\right\} $$ Then for all $f\in S$ you have $$ \lim\limits_{n\to\infty}\langle A_n,f\rangle=\frac{\alpha+\beta}{2}\int\limits_{(0,1)}f(t)d\mu(t) $$ Now you can suggest what is $A$.
Hint: It is enough to compute $$\lim_{n\to\infty}\int_0^1 A_n(x)f(x)\,dx$$ for continuous functions $f\colon[0,1]\to\mathbb{R}$, since these are dense in $L^1(\Omega)$. You will find that the limit is $$\frac{\alpha+\beta}{2}\int_0^1 f(x)\,dx\tag{1}$$ – just take the difference between the two integrals, consider the difference of integrals over each subinterval $[k\epsilon,(k+1)\epsilon]$, and use uniform continuity. (Perhaps you wish to insert a Riemann sum in the estimate.)
Edit: For more detail, note $$\int_0^1 A_n(x)f(x)\,dx=\sum_{k=0}^{n-1}\Bigl(\alpha\int_{k\epsilon}^{(k+1/2)\epsilon}f(x)\,dx+\beta\int_{(k+1/2)\epsilon}^{(k+1)\epsilon}f(x)\,dx\Bigr)$$ in which you replace $f(x)$ in the integrals on the right by $f(k\epsilon)$, carefully estimating the error you introduce by doing so. (For any $\eta>0$ you can pick $\epsilon>0$ so that $\lvert f(x)-f(k\epsilon)\rvert<\eta$ for $x\in[k\epsilon,(k+1)\epsilon]$ – this is uniform continuity.) Now do the same with the integral in (1) and subtract.