Spring 2010 - Problem 1

Solution.

1. By definition, there exists $f \in L^p\p{\R}$ such that $\norm{f_n - f}_{L^p} \to 0$. Thus, there exists $n_1$ so that $\norm{f_{n_1} - f}_{L^p} \leq 2^{-1/p}$. By induction, there exists a subsequence $\set{n_k}_{k\geq1}$ such that $\norm{f_{n_k} - f}_{L^p} \leq 2^{-k/p}$. Now consider

   $$F_N = \sum_{k=1}^N \abs{f_{n_{k+1}} - f_{n_k}}^p \quad\text{and}\quad F = \sum_{k=1}^\infty \abs{f_{n_{k+1}} - f_{n_k}}^p.$$

   Since the terms are non-negative, $\set{F_N}_{N\geq1}$ is an increasing sequence of functions, so by the monotone convergence theorem,

   $$\norm{F}_{L^1} = \sum_{k=1}^\infty \norm{f_{n_k} - f}_{L^p}^p \leq \sum_{k=1}^\infty 2^{-k} = 1.$$

   Thus, $\abs{F} < \infty$ almost everywhere, so by the divergence test for series, it follows that $\abs{f_{n_k} - f} \to 0$ for almost every $x \in \R$, i.e., $\set{f_{n_k}}_{k\geq1}$ converges to $f$ almost everywhere.

2. Consider the typewriter sequence: for $n \geq 1$ and $1 \leq k \leq 2^{n-1}$, set

   $$I_{n,k} = \br{\frac{k - 1}{2^{n-1}}, \frac{k}{2^{n-1}}} \quad\text{and}\quad f_{n,k} = \chi_{I_{n,k}}.$$

   For each $n \geq 1$, $\set{I_{n,k}}_k$ partition the interval $\br{0, 1}$. Thus, for any $x$ and $n \geq 1$, there exists $k$ so that $f_{n,k}\p{x} = 1$, but $f_{n,k-1}\p{x} = 0$ if $k > 1$ or $f_{n,k+1}\p{x} = 0$ if $k = 1$, so $f_{n,k}$ converges nowhere pointwise. On the other hand,

   $$\norm{f_{n,k}}_{L^2}^2 = \frac{1}{2^{n-1}} \xrightarrow{n\to\infty} 0,$$

   so $f_{n,k} \to 0$ in $L^2\p{\R}$.