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Spring 2012 - Problem 6

Fourier analysis

Suppose $f \in L^2\p{\R}$ and that the Fourier transform obeys $\hat{f}\p{\xi} > 0$ for almost every $\xi$ . Show that the set of finite linear combinations of translates of $f$ is dense in the Hilbert space $L^2\p{\R}$ .

Solution.

For $a \in \R$ , let $f_a\p{x} = f\p{x - a}$ be the translate of $f$ by $a$ . First, notice that by a translation change of variables,

\hat{f_a}\p{\xi} = \int e^{-2\pi i\xi x} f\p{x - a} \,\diff{x} = \int e^{-2\pi i\xi \p{x+a}} f\p{x} \,\diff{x} = e^{-2\pi i\xi a} \hat{f}\p{\xi}.

Let $\mathcal{M}$ be the closure of the span of $\set{f_a}_{a\in\R}$ , and suppose $g \in \mathcal{M}^\perp$ . Then by Plancherel, we have for any $a \in \R$ that

0 = \inner{f_a, g} = \inner{\hat{f_a}, \hat{g}} = \int e^{-2\pi i\xi a} \hat{f}\p{\xi} \hat{g}\p{\xi} \,\diff\xi.

Let $h\p{\xi} = \hat{f}\p{\xi}\hat{g}\p{\xi}$ . Then the equality above implies that $\hat{h} = 0$ almost everywhere. By another application of Plancherel, we have $\norm{h}_{L^2} = \bigl\lVert \hat{h} \bigr\rVert_{L^2} = 0$ , so $h = 0$ almost everywhere as well. Since $\hat{f}\p{\xi} > 0$ almost everywhere, it follows that $\hat{g}\p{\xi} = 0$ almost everywhere, and so $\norm{g}_{L^2} = \norm{\hat{g}}_{L^2} = 0$ , i.e., $g = 0$ almost everywhere. We have shown that $\mathcal{M}^\perp = \set{0}$ , so $\mathcal{M}$ is all of $L^2\p{\R}$ . By definition, this means that the translates of $f$ are dense in $L^2\p{\R^3}$ , which completes the proof.