<Home>Qualifying Exams><Analysis>Fall 2017 - Problem 5

Fall 2017 - Problem 5

measure theory

Suppose $\func{f}{\R}{\R}$ is a bounded and measurable function satisfying $f\p{x + 1} = f\p{x}$ and $f\p{2x} = f\p{x}$ for almost every $x \in \R$ . Show that then there exists a constant $c \in \R$ such that $f\p{x} = c$ for almost every $x \in \R$ .

Solution.

First, observe that on any dyadic interval $\br{\frac{k}{2^n}, \frac{k+1}{2^n}} \subseteq \br{0, 1}$ , i.e., $0 \leq k \leq 2^n - 1$ and $n \geq 0$ , we have

\int_{k/2^n}^{\p{k+1}/2^n} f \,\diff{x} = \frac{1}{2^n} \int_k^{k+1} f \,\diff{x} = \frac{1}{2^n} \int_0^1 f \,\diff{x}.

Let $c = \int_0^1 f \,\diff{x}$ and consider $f_n = 2^{n} \int_{k/2^n}^{\p{k+1}/2^n} f \,\diff{x}$ on $\br{\frac{k}{2^n}, \frac{k+1}{2^n}}$ and $0$ otherwise. If we show that $f_n \to f$ in $L^1\p{\br{0, 1}}$ , then

\norm{f_n - f}_{L^1\p{\br{0,1}}} = \int_0^1 \abs{f_n - f} \,\diff{x} = \int_0^1 \abs{c - f} \,\diff{x} \xrightarrow{n\to\infty} 0,

so $\int_0^1 \abs{c - f} \,\diff{x} = 0$ and so $f = c$ for almost every $x \in \br{0, 1}$ . Thus, by the translation invariance of $f$ almost everywhere,

\int_k^{k+1} \abs{f - c} \,\diff{x} = \int_0^1 \abs{f - c} \,\diff{x} = 0,

so $f = c$ for almost every $x \in \R$ . Hence, it remains to prove the $L^1$ convergence.

Let $A_{k,n}\p{g} = \int_{k/2^n}^{\p{k+1}/2^n} \abs{g} \,\diff{x}$ . First, note that

A_{k,n}\p{1} = 2^n \int_{k/2^n}^{\p{k+1}/2^n} \,\diff{x} = 1.

Let $\epsilon > 0$ and let $g$ be a continuous function on $\br{0, 1}$ such that $\norm{f - g}_{L^1} < \epsilon$ . Then

\begin{aligned} \norm{f_n - f}_{L^1\p{\br{0,1}}} &= \int_0^1 \abs{f_n - f} \,\diff{x} \\ &= \sum_{k=0}^{2^n-1} \int_{k/2^n}^{\p{k+1}/2^n} \abs{2^n \int_{k/2^n}^{\p{k+1}/2^n} f\p{y} - f\p{x} \,\diff{y}} \,\diff{x} \\ &\leq \sum_{k=0}^{2^n-1} \int_{k/2^n}^{\p{k+1}/2^n} 2^n \int_{k/2^n}^{\p{k+1}/2^n} \abs{f\p{y} - f\p{x}} \,\diff{y} \,\diff{x} \\ &\leq \sum_{k=0}^{2^n-1} \int_{k/2^n}^{\p{k+1}/2^n} A_{k,n}\p{f - g} + A_{k,n}\p{f\p{x} - g\p{x}} + A_{k,n}\p{g - g\p{x}} \,\diff{x}. \end{aligned}

We estimate separately:

\begin{aligned} \int_{k/2^n}^{\p{k+1}/2^n} A_{k,n}\p{f - g} \,\diff{x} &= \int_{k/2^n}^{\p{k+1}/2^n} 2^n \int_{k/2^n}^{\p{k+1}/2^n} \abs{f\p{y} - g\p{y}} \,\diff{y} \,\diff{x} \\ &= \int_{k/2^n}^{\p{k+1}/2^n} \abs{f\p{y} - g\p{y}} \,\diff{y}. \end{aligned}

Similarly, by Fubini-Tonelli (everything is non-negative)

\begin{aligned} A_{k,n}\p{f\p{x} - g\p{x}} &= \int_{k/2^n}^{\p{k+1}/2^n} 2^n \int_{k/2^n}^{\p{k+1}/2^n} \abs{f\p{y} - g\p{y}} \,\diff{y} \,\diff{x} \\ &= \int_{k/2^n}^{\p{k+1}/2^n} 2^n \int_{k/2^n}^{\p{k+1}/2^n} \abs{f\p{y} - g\p{y}} \,\diff{x} \,\diff{y} \\ &= \int_{k/2^n}^{\p{k+1}/2^n} \abs{f\p{y} - g\p{y}} \,\diff{y}. \end{aligned}

Finally, by absolute continuity of $g$ on the compact dyadic square, there exists $\delta > 0$ so that $\abs{x - y} < \delta \implies \abs{g\p{x} - g\p{y}} < \epsilon$ . Choose $n$ large enough so that $\frac{1}{2^n} < \delta$ and we get

\begin{aligned} \int_{k/2^n}^{\p{k+1}/2^n} A_{k,n}\p{g - g\p{x}} \,\diff{x} &= \int_{k/2^n}^{\p{k+1}/2^n} 2^n \int_{k/2^n}^{\p{k+1}/2^n} \abs{g\p{x} - g\p{y}} \,\diff{y} \,\diff{x} \\ &\leq \int_{k/2^n}^{\p{k+1}/2^n} \epsilon \,\diff{x} \\ &= \frac{\epsilon}{2^n}. \end{aligned}

Putting it all together,

\begin{aligned} \norm{f_n - f}_{L^1\p{\br{0,1}}} &\leq \sum_{k=0}^{2^n-1} \int_{k/2^n}^{\p{k+1}/2^n} A_{k,n}\p{f - g} + A_{k,n}\p{f\p{x} - g\p{x}} + A_{k,n}\p{g - g\p{x}} \,\diff{x} \\ &\leq \sum_{k=0}^{2^n-1} \p{\int_{k/2^n}^{\p{k+1}/2^n} \abs{f\p{y} - g\p{y}} \,\diff{y} + \int_{k/2^n}^{\p{k+1}/2^n} \abs{f\p{y} - g\p{y}} \,\diff{y} + \frac{\epsilon}{2^n}} \\ &= 2\norm{f - g}_{L^1\p{\br{0,1}}} + \epsilon \\ &\leq 3\epsilon. \end{aligned}

Sending $\epsilon \to 0$ , we see $f_n \to f$ in $L^1$ , which completes the proof.