Spring 2009 - Problem 6

Solution.

1. Let $X = \set{f \in L^2\p{I} \mid \int_I f \,\diff{t} = 0}$. Because integration is linear, it's clear that this is a linear subspace of $L^2\p{I}$. We will show that $\set{h_{n,j}}$ is an orthonormal basis for $X$.

   First, notice that $h_{n,j} \in X$ by symmetry. Also,

   $$\begin{aligned} \norm{h_{n,j}} = \int_I \abs{h_{n,j}}^2 &= 2^n \int_I \p{\chi_{I_{n+1,2j}} + \chi_{I_{n+1,2j+1}}} \\ &= 2^n\p{\frac{1}{2^{n+1}} + \frac{1}{2^{n+1}}} \\ &= 1, \end{aligned}$$

   since $\chi_{I_{n+1,2j}}$ and $\chi_{I_{n+1,2j+1}}$ are disjoint.

   To calculate the inner products, consider $I_{n,j}$ and $I_{m,k}$, where without loss of generality $n \leq m$. Notice that $I_{n,j} \cap I_{m,k}$ has positive measure only when $k \in \br{2^{m-n}j, 2^{m-n}j + 2^{m-n} - 1}$. Indeed, if $k \leq 2^{m-n}j - 1$, then

   $$\p{k+1}2^{-m} \leq \p{2^{m-n}j}2^{-m} = j2^{-n},$$

   i.e., $I_{n,j} \cap I_{m,k}$ is either empty or contains a single point. Similarly, if $k \geq 2^{m-n}j + 2^{m-n}$, then

   $$k2^{-m} \geq \p{2^{m-n}j + 2^{m-n}}2^{-m} = \p{j+1}2^{-n},$$

   so the same thing happens. Thus, if $k \in \br{2^{m-n}j, 2^{m-n}j + 2^{m-n} - 1}$, we have

   $$\p{k+1}2^{-m} \leq \p{2^{m-n}j + 2^{m-n}}2^{-m} \leq \p{j+1}2^{-n},$$

   i.e., $I_{m,k} \subseteq I_{n,j}$. In particular, given $h_{n,j}$ and $h_{m,k}$ with $n \leq m$, the support of $h_{m,k}$ is contained entirely in either $I_{n+1,2j}$ or $I_{n+1,2j+1}$, so

   $$\inner{h_{n,j}, h_{m,k}} = \int_I h_{n,j}h_{m,k} = \int_{I_{n+1,2j+i}} 2^{n/2}h_{m,k} = 0,$$

   by symmetry. With this, we have shown that $\set{h_{n,j}}$ is an orthonormal set. Next, we will show that they are in fact a basis for $X$. It suffices to show that if $f \in X$ satisfies $\inner{f, h_{n,j}} = 0$ for all $n, j$, then $f = 0$ in $X$.

   We will show that $\int_{I_{n,j}} f = 0$ for all $n, j$. When $n = 1$, notice that $I = I_{1,0} \cup I_{1,1}$ is a disjoint union, so

   $$0 = \int_I f = \int_{I_{1,0}} f + \int_{I_{1,0}} f \implies \int_{I_{1,0}} f = -\int_{I_{1,1}} f.$$

   On the other hand, by assumption,

   $$0 = \int_I fh_{0,0} = 2^{1/2} \p{\int_{I_{1,0}} f - \int_{I_{1,1}} f} \implies \int_{I_{1,0}} f = \int_{I_{1,1}} f.$$

   Putting these together, we have $\int_{I_{1,0}} f = \int_{I_{1,1}}f = 0$, so we get the base case. We proceed by induction: suppose $\int_{I_{n,j}} f = 0$ for all $0 \leq j \leq 2^n - 1$. Then

   $$0 = \int_{I_{n,j}} f = \int_{I_{n+1,2j}} f + \int_{I_{n+1,2j+1}} f \implies \int_{I_{n+1,2j}} f = -\int_{I_{n+1,2j+1}} f,$$

   by the inductive hypothesis, and by assumption,

   $$0 = \int_I fh_{n,j} = \int_{I_{n+1,2j}} f - \int_{I_{n+1,2j+1}} f \implies \int_{I_{n+1,2j}} f = \int_{I_{n+1,2j+1}} f.$$

   This implies that $\int_{I_{n+1,2j}} f = \int_{I_{n+1,2j+1}} f = 0$ for all $0 \leq j \leq 2^n - 1$, which completes the induction. From Problem 5, we see that $E_nf = 0$ everywhere, and because $E_nf \to f$ almost everywhere, $f = 0$ almost everywhere.

   Finally, given any $f \in X$, we have for all $n, j$ that

   $$\inner{f - \sum_{n,j} \inner{f, h_{n,j}}h_{n,j}, h_{m,k}} = \inner{f, h_{m,k}} - \inner{f, h_{m,k}} = 0,$$

   which implies

   $$\begin{aligned} 0 = \norm{f - \sum_{n,j} \inner{f, h_{n,j}}h_{n,j}}^2 &= \norm{f}^2 + \sum_{n,j} \abs{\inner{f, h_{n,j}}}^2 - 2\sum_{n,j} \abs{\inner{f, h_{n,j}}}^2 \\ &= \norm{f}^2 - \sum_{n,j} \abs{\inner{f, h_{n,j}}}^2 \\ \implies \norm{f}^2 &= \sum_{n,j} \abs{\inner{f, h_{n,j}}}^2, \end{aligned}$$

   which was what we wanted to show.

2. Let

   $$S_N\p{x} = \sum_{n=1}^N \sum_{j=0}^{2^n-1} \p{\int fh_{n,j} \,\diff{t}} h_{n,j}\p{x}.$$

   Notice that $\set{j2^{-n} \in I \mid n \geq 1,\ 0 \leq j \leq 2^n-1}$ is countable, hence with Lebesgue measure $0$. Pick an $x$ outside of this set so that for each $n \geq 1$, there exists a unique $j\p{n,x}$ such that $x \in I_{n,j\p{n,x}}$. Further let $\delta\p{j,x}$ be the unique number in $\set{0, 1}$ such that $x \in I_{n+1,2j\p{n,x}+\delta\p{n,x}}$. These satisfy

   $$j\p{n, x} = 2j\p{n-1, x} + \delta\p{n-1, x},$$

   since our indices are unique. Thus,

   $$\begin{aligned} S_N\p{x} &= \sum_{n=1}^N \sum_{j=0}^{2^n-1} \p{\int fh_{n,j} \,\diff{t}} h_{n,j}\p{x} \\ &= \sum_{n=1}^N \p{\int fh_{n,j\p{n,x}} \,\diff{t}} h_{n,j\p{n,x}}\p{x} \\ &= \sum_{n=1}^N 2^n\p{\int_{I_{n+1,2j\p{n,x}+\delta\p{n,x}}} f \,\diff{t} - \int_{I_{n+1,2j\p{n,x}+\p{1-\delta\p{n,x}}}} f \,\diff{t}} \\ &= \sum_{n=1}^N 2^n\p{\int_{I_{n+1,2j\p{n,x}+\delta\p{n,x}}} f \,\diff{t} - \p{\int_{I_{n,j\p{n,x}}} f \,\diff{t} - \int_{I_{n+1,2j\p{n,x}+\delta\p{n,x}}} f \,\diff{t}}} \\ &= \sum_{n=1}^N 2^n\p{2\int_{I_{n+1,2j\p{n,x}+\delta\p{n,x}}} f \,\diff{t} - \int_{I_{n,j\p{n,x}}} f \,\diff{t}} \\ &= \sum_{n=1}^N 2^{n+1}\int_{I_{n+1,2j\p{n,x}+\delta\p{n,x}}} f \,\diff{t} - 2^n\int_{I_{n,2j\p{n-1,x}+\delta\p{n-1,x}}} f \,\diff{t} \\ \end{aligned}$$

   where the third-to-last equality comes from the fact that $I_{n,j\p{n,x}}$ is the disjoint union of $I_{n+1,2j\p{n,x}}$ and $I_{n+1,2j\p{n,x}+1}$. This sum telescopes, and using the fact that $\int_I f = 0$, we're left with

   $$\begin{aligned} S_N\p{x} &= 2^{N+1} \int_{I_{N+1,2j\p{N,x}+\delta\p{N,x}}} f \,\diff{t} - 2\int_{I_{0,0}} f \,\diff{t} \\ &= 2^{N+1}\int_{I_{N+1,j\p{N+1,x}}} f \,\diff{t} \\ &= 2^{N+1}\p{\int_{I_{N+1,j\p{N+1,x}}} f \,\diff{t}} \chi_{I_{N+1,j\p{N+1,x}}}\p{x} \\ &= \sum_{j=0}^{2^N-1} 2^{N+1} \p{\int_{I_{N+1,j}} f \,\diff{t}} \chi_{I_{N+1,j}}\p{x} \\ &= E_{N+1}f\p{x}. \end{aligned}$$

   where the second-to-last equality comes from the fact that $\set{I_{N+1,j}}_j$ partition $I$ (ignoring endpoints). By Problem 5, it follows that

   $$\lim_{N\to\infty} \abs{S_N\p{x} - f\p{x}} = \lim_{N\to\infty} \abs{E_{N+1}\p{x} - f\p{x}} = 0,$$

   so we have the formula almost everywhere, as desired.