Fall 2010 - Problem 3

Solution.

1. Because $\br{0, 1}$ is $\sigma$-finite with respect to the Lebesgue measure, we know that $\p{L^1\p{\br{0,1}}}^* \simeq L^\infty\p{\br{0,1}}$. Thus, to prove weak convergence, it suffices to show that there exists $f \in L^1\p{\br{0,1}}$ such that

   $$\lim_{n\to\infty} \int f_ng \,\diff{x} = \int fg \,\diff{x}$$

   for any $g \in L^\infty\p{\br{0,1}}$. Observe that for such a $g$,

   $$\norm{g}_{L^1} \leq \norm{g}_{L^\infty}m\p{\br{0,1}} = \norm{g}_{L^\infty} < \infty,$$

   so $L^\infty\p{\br{0,1}} \subseteq L^1\p{\br{0,1}}$. We simplify the problem as follows:

   • Decompose $g = g^+ - g^-$, where $f^+ = \max\set{0, g}$ and $g^- = \max\set{0, -g}$. By approximating these separately, it suffices to approximate non-negative $g$.
   • Since there exists a sequence of simple functions which increase to such a $g$, we may reduce to approximating simple functions, in light of the monotone convergence theorem.
   • Simple functions are finite linear combinations of integrable characteristic functions, so it further suffices to prove the result for characteristic functions on a measurable set of finite measure.
   • Finally, by regularity of the Lebesgue measure, we may reduce to an integrable characteristic function on an interval. Hence, we may assume without loss of generality that $g = \chi_{\br{a,b}}$, with $-\infty < a < b < \infty$.

   To emphasize the fact that $f_n$ is $\frac{1}{n}$-periodic, we let

   $$\begin{aligned} a_n &= \sup\,\set{\frac{k}{n} \st a \leq \frac{k}{n},\ k = 0, 1, \ldots, n} \\ b_n &= \inf\,\set{\frac{k}{n} \st \frac{k}{n} \leq b,\ k = 0, 1, \ldots, n} \end{aligned}$$

   so that $a \leq a_n \leq b_n \leq b$ for all $n \geq 1$, $a_n \xrightarrow{n\to\infty} a$, and $b_n \xrightarrow{n\to\infty} b$. Notice also that $b_n - a_n$ is an integer multiple of $\frac{1}{n}$, so let $k_n = \p{b_n - a_n}n$ also. Thus,

   $$\begin{aligned} \int_0^1 f_ng \,\diff{x} &= \int_a^b \exp\p{\sin\p{2\pi nx}} \,\diff{x} \\ &= \int_a^{a_n} \exp\p{\sin\p{2\pi nx}} \,\diff{x} + \int_{a_n}^{b_n} \exp\p{\sin\p{2\pi nx}} \,\diff{x} + \int_{b_n}^b \exp\p{\sin\p{2\pi nx}} \,\diff{x} \\ &\leq \p{a_n - a}\exp{1} + \p{b_n - b}\exp{1} + \int_{a_n}^{a_n+\frac{k_n}{n}} \exp\p{\sin\p{2\pi nx}} \,\diff{x} \\ &= e\p{a_n - a} + e\p{b_n - b} + \frac{k_n}{n} \int_0^1 \exp\p{\sin\p{2\pi x}} \,\diff{x} \\ &= e\p{a_n - a} + e\p{b_n - b} + \p{b_n - a_n} \int_0^1 \exp\p{\sin\p{2\pi x}} \,\diff{x}. \end{aligned}$$

   We get the inequality since $f_n$ is uniformly bounded above by $\exp{1} = e$, and we transformed the third integral via the substitution $x \mapsto n\p{x - a_n}$ and using the fact that $f_n$ is $\frac{1}{n}$-periodic. Sending $n\to\infty$, we get

   $$\begin{aligned} \lim_{n\to\infty} \int_0^1 f_ng \,\diff{x} &= \p{b - a} \int_0^1 \exp\p{\sin\p{2\pi x}} \,\diff{x} \\ &= \p{\int_0^1 \exp\p{\sin\p{2\pi x}} \,\diff{x}}\p{\int_0^1 g \,\diff{x}}. \end{aligned}$$

   Hence, if we set $f = \int_0^1 \exp\p{\sin\p{2\pi x}} \,\diff{x}$, then $f \in L^1\p{\br{0,1}}$ and $f_n \to f$ weakly.

2. Notice that in (1), we proved a stronger statement: we showed that $\int f_ng \,\diff{x} \to \int fg \,\diff{x}$ for any $g \in L^1\p{\br{0,1}}$ instead of just $g \in L^\infty\p{\br{0,1}}$. This is precisely what it means for $f_n \to f$ weakly-* in $L^\infty\p{\br{0,1}}$, so we are done.