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Spring 2017 - Problem 5

density argument, Lebesgue-Radon-Nikodym derivative

Let $\diff\mu$ be a finite complex Borel measure on $\br{0, 1}$ such that

\hat{\mu}\p{n} = \int_0^1 e^{2\pi inx} \,\diff\mu\p{x} \to 0 \quad\text{as}\quad n \to \infty.

Let $\diff\nu$ be a finite complex Borel measure on $\br{0, 1}$ that is absolutely continuous with respect to $\diff\mu$ . Show that

\hat{\nu}\p{n} \to 0 \quad\text{as}\quad n \to \infty.

Solution.

Let $\epsilon > 0$ .

Since we are working with finite measures on a compact metric space, continuous functions are dense in $L^1$ .

By the Lebesgue-Radon-Nikodym theorem, there exists $f \in L^1\p{\br{0,1}}$ such that $\diff\nu = f \,\diff\mu$ . By Stone-Weierstrass, trigonometric polynomials are dense in $L^1\p{\br{0,1}}$ , so pick $g\p{x} = \sum_{k=1}^N a_k e^{2\pi ikx}$ such that $\norm{f - g}_{L^1} < \epsilon$ .

\begin{aligned} \abs{\hat{\nu}\p{n}} &= \abs{\int_0^1 e^{2\pi inx} \,\diff\nu\p{x}} \\ &= \abs{\int_0^1 e^{2\pi inx}f\p{x} \,\diff\mu\p{x}} \\ &= \abs{\int_0^1 e^{2\pi inx}\p{f\p{x} - g\p{x}} \,\diff\mu\p{x} + \int_0^1 e^{2\pi inx} g\p{x} \,\diff\mu\p{x}} \\ &\leq \int_0^1 \abs{f\p{x} - g\p{x}} \,\diff\mu\p{x} + \sum_{k=1}^N \abs{a_k \int_0^1 e^{2\pi inx} e^{2\pi ikx} \,\diff\mu\p{x}} \\ &= \norm{f - g}_{L^1} + \sum_{k=1}^N \abs{a_k} \abs{\hat{\mu}\p{k+n}}. \end{aligned}

Sending $n \to \infty$ and applying the assumption, we get

\limsup_{n\to\infty}\,\abs{\hat{\nu}\p{n}} \leq \epsilon.

Sending $\epsilon \to 0$ , we get the claim.