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Spring 2020 - Problem 4

weak convergence

Fix f∈L1(R)f \in L^1\p{\R}. Show that

lim⁡n→∞∫02f(x)sin⁡(xn) dx=0.\lim_{n\to\infty} \int_0^2 f\p{x} \sin\p{x^n} \,\diff{x} = 0.
Solution.

Notice that on [0,1)\pco{0, 1}, we have sin⁡(xn)→0\sin\p{x^n} \to 0, and we also have ∣f(x)sin⁡(xn)∣≤∣f(x)∣∈L1(R)\abs{f\p{x}\sin\p{x^n}} \leq \abs{f\p{x}} \in L^1\p{\R}, so by dominated convergence,

lim⁡n→∞∫01f(x)sin⁡(xn) dx=0.\lim_{n\to\infty} \int_0^1 f\p{x} \sin\p{x^n} \,\diff{x} = 0.

Hence, it suffices to show that the integral on [1,2]\br{1, 2} vanishes. By linearity of the problem and density of step functions in L1(R)L^1\p{\R}, it suffices to prove this for f=χ[a,b]f = \chi_{\br{a,b}} with 1≤a≤b≤21 \leq a \leq b \leq 2. Integrating by parts, we have

∫absin⁡(xn) dx=∫ab1nxn−1ddx(−cos⁡(xn)) dx=−cos⁡(xn)nxn−1∣ab−n−1n∫abcos⁡(xn)xn−1 dx.\begin{aligned} \int_a^b \sin\p{x^n} \,\diff{x} &= \int_a^b \frac{1}{nx^{n-1}} \deriv{}{x} \p{-\cos\p{x^n}} \,\diff{x} \\ &= \left. -\frac{\cos\p{x^n}}{nx^{n-1}} \right\rvert_a^b - \frac{n - 1}{n} \int_a^b \frac{\cos\p{x^n}}{x^{n-1}} \,\diff{x}. \end{aligned}

Notice that the first term is bounded by

1nan−1+1nbn−1→n→∞0,\frac{1}{na^{n-1}} + \frac{1}{nb^{n-1}} \xrightarrow{n\to\infty} 0,

and for the second term, the integrand is bounded by 1an−1∈L1([0,1])\frac{1}{a^{n-1}} \in L^1\p{\br{0,1}} and tends to 00 for almost every x∈[a,b]x \in \br{a, b}. Hence, by dominated convergence, the integral tends to 00 as well, so overall,

lim⁡n→∞∫absin⁡(xn) dx=0,\lim_{n\to\infty} \int_a^b \sin\p{x^n} \,\diff{x} = 0,

as desired.