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Fall 2019 - Problem 2

Fourier analysis

Let μ\mu be a finite Borel measure on R\R with μ({x})=0\mu\p{\set{x}} = 0 for all x∈Rx \in \R and let φ(t)=∫Reitx dμ(x)\phi\p{t} = \int_\R e^{itx} \,\diff\mu\p{x}. Prove that

lim⁡T→∞12T∫−TT∣φ(t)∣2 dt=0.\lim_{T\to\infty} \frac{1}{2T} \int_{-T}^T \abs{\phi\p{t}}^2 \,\diff{t} = 0.
Solution.

Since eitxe^{itx} is bounded and μ\mu is a finite measure, we may apply Fubini's theorem in the following:

12T∫−TT∣φ(t)∣2 dt=12T∫−TT(∫Reitx dμ(x))(∫Reity dμ(y))‾ dt=12T∫−TT∫R∫Reit(x−y) dμ(x) dμ(y) dt=12T∫R∫R∫−TTeit(x−y) dt dμ(x) dμ(y)=∫R∫Rsin⁡(T(x−y))T(x−y) dμ(x) dμ(y).\begin{aligned} \frac{1}{2T} \int_{-T}^T \abs{\phi\p{t}}^2 \,\diff{t} &= \frac{1}{2T} \int_{-T}^T \p{\int_\R e^{itx} \,\diff\mu\p{x}} \conj{\p{\int_\R e^{ity} \,\diff\mu\p{y}}} \,\diff{t} \\ &= \frac{1}{2T} \int_{-T}^T \int_\R \int_\R e^{it \p{x-y}} \,\diff\mu\p{x} \,\diff\mu\p{y} \,\diff{t} \\ &= \frac{1}{2T} \int_\R \int_\R \int_{-T}^T e^{it\p{x-y}} \,\diff{t} \,\diff\mu\p{x} \,\diff\mu\p{y} \\ &= \int_\R \int_\R \frac{\sin\p{T\p{x-y}}}{T\p{x - y}} \,\diff\mu\p{x} \,\diff\mu\p{y}. \end{aligned}

Again, because the integrand is continuous and μ\mu is finite, we may apply dominated convergence:

lim⁡T→∞12T∫−TT∣φ(t)∣2 dt=∫R∫Rlim⁡T→∞sin⁡(T(x−y))T(x−y) dμ(x) dμ(y).\lim_{T\to\infty} \frac{1}{2T} \int_{-T}^T \abs{\phi\p{t}}^2 \,\diff{t} = \int_\R \int_\R \lim_{T\to\infty} \frac{\sin\p{T\p{x-y}}}{T\p{x - y}} \,\diff\mu\p{x} \,\diff\mu\p{y}.

If x≠yx \neq y, then clearly the integrand tends to 00. If x=yx = y, then we get 11, so this becomes

lim⁡T→∞12T∫−TT∣φ(t)∣2 dt=∫R∫Rχ{x=y}(x,y) dμ(x) dμ(y)=∫Rμ({y}) dμ(y)=0,\begin{aligned} \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^T \abs{\phi\p{t}}^2 \,\diff{t} &= \int_\R \int_\R \chi_{\set{x = y}}\p{x, y} \,\diff\mu\p{x} \,\diff\mu\p{y} \\ &= \int_\R \mu\p{\set{y}} \,\diff\mu\p{y} \\ &= 0, \end{aligned}

since singletons have μ\mu-measurable zero, and this is what we wanted to show.