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Fall 2020 - Problem 2

Lp spaces, weak convergence

Show that there is a constant c∈Rc \in \R so that

lim⁡n→∞∫01f(x)cos⁡(sin⁡(nπx)) dx=c∫01f(x) dx\lim_{n\to\infty} \int_0^1 f\p{x} \cos\p{\sin\p{n\pi x}} \,\diff{x} = c \int_0^1 f\p{x} \,\diff{x}

for every f∈L1([0,1])f \in L^1\p{\br{0,1}}. The limit is taken over those n∈Nn \in \N.
Solution.

Notice that if g∈L1([0,1])g \in L^1\p{\br{0,1}}, then

∫01f(x)cos⁡(sin⁡(nπx)) dx=∫01g(x)cos⁡(sin⁡(nπx)) dx+∫01(f(x)−g(x))cos⁡(sin⁡(nπx)) dx,\int_0^1 f\p{x} \cos\p{\sin\p{n\pi x}} \,\diff{x} = \int_0^1 g\p{x} \cos\p{\sin\p{n\pi x}} \,\diff{x} + \int_0^1 \p{f\p{x} - g\p{x}} \cos\p{\sin\p{n\pi x}} \,\diff{x},

and the second term is bounded by ∥f−g∥L1\norm{f - g}_{L^1}. Hence, it suffices to prove the claim on a dense subset of L1([0,1])L^1\p{\br{0,1}}, so we may assume that ff is a step function. By the linearity of the problem, it further suffices to assume that f=χ[a,b]f = \chi_{\br{a,b}} for some 0≤a≤b≤10 \leq a \leq b \leq 1. For each nn, let kn,ℓn∈Nk_n, \ell_n \in \N be such that

kn−1n≤a≤knn≤ℓnn≤b≤ℓn+1n.\frac{k_n - 1}{n} \leq a \leq \frac{k_n}{n} \leq \frac{\ell_n}{n} \leq b \leq \frac{\ell_n + 1}{n}.

Then

∫01χ[a,b]cos⁡(sin⁡(nπx)) dx=∫abcos⁡(sin⁡(nπx)) dx=∫kn/nℓn/ncos⁡(sin⁡(nπx)) dx+∫akn/ncos⁡(sin⁡(nπx)) dx+∫ℓn/nbcos⁡(sin⁡(nπx)) dx=∑j=0⌊(ℓn−kn)/2⌋−1∫(kn+2j)/n(kn+2(j+1))/ncos⁡(sin⁡(nπx)) dx+∫kn+2⌊(ℓn−kn)/2⌋/nℓncos⁡(sin⁡(nπx)) dx+∫akn/ncos⁡(sin⁡(nπx)) dx+∫ℓn/nbcos⁡(sin⁡(nπx)) dx.\begin{aligned} \int_0^1 \chi_{\br{a,b}} \cos\p{\sin\p{n\pi x}} \,\diff{x} &= \int_a^b \cos\p{\sin\p{n\pi x}} \,\diff{x} \\ &= \int_{k_n/n}^{\ell_n/n} \cos\p{\sin\p{n\pi x}} \,\diff{x} + \int_a^{k_n/n} \cos\p{\sin\p{n\pi x}} \,\diff{x} + \int_{\ell_n/n}^b \cos\p{\sin\p{n\pi x}} \,\diff{x} \\ &= \sum_{j=0}^{\floor{\p{\ell_n-k_n}/2}-1} \int_{\p{k_n+2j}/n}^{\p{k_n+2\p{j+1}}/n} \cos\p{\sin\p{n\pi x}} \,\diff{x} + \int_{k_n+2\floor{\p{\ell_n-k_n}/2}/n}^{\ell_n} \cos\p{\sin\p{n\pi x}} \,\diff{x} + \int_a^{k_n/n} \cos\p{\sin\p{n\pi x}} \,\diff{x} + \int_{\ell_n/n}^b \cos\p{\sin\p{n\pi x}} \,\diff{x}. \end{aligned}

By dominated convergence, every term after the sum vanishes as n→∞n \to \infty, so we only need to concern ourselves with the sum:

∑j=0⌊(ℓn−kn)/2⌋−1∫(kn+2j)/n(kn+2(j+1))/ncos⁡(sin⁡(nπx)) dx=2n∑j=0⌊(ℓn−kn)/2⌋−1∫01cos⁡(sin⁡(2πx−πkn)) dx(x↦nx−kn2−j)=2n∑j=0⌊(ℓn−kn)/2⌋−1∫01cos⁡(sin⁡(2πx)) dx(cos⁡x is even)=2n⌊ℓn−kn2⌋∫01cos⁡(sin⁡(2πx)) dx.\begin{aligned} \sum_{j=0}^{\floor{\p{\ell_n-k_n}/2}-1} \int_{\p{k_n+2j}/n}^{\p{k_n+2\p{j+1}}/n} \cos\p{\sin\p{n\pi x}} \,\diff{x} &= \frac{2}{n} \sum_{j=0}^{\floor{\p{\ell_n-k_n}/2}-1} \int_0^1 \cos\p{\sin\p{2\pi x - \pi k_n}} \,\diff{x} && \p{x \mapsto \frac{nx - k_n}{2} - j} \\ &= \frac{2}{n} \sum_{j=0}^{\floor{\p{\ell_n-k_n}/2}-1} \int_0^1 \cos\p{\sin\p{2\pi x}} \,\diff{x} && \p{\cos{x} \text{ is even}} \\ &= \frac{2}{n} \floor{\frac{\ell_n - k_n}{2}} \int_0^1 \cos\p{\sin\p{2\pi x}} \,\diff{x}. \end{aligned}

To complete the proof, notice that if ℓn−kn\ell_n - k_n is even, then

2n⌊ℓn−kn2⌋=ℓn−knn→n→∞b−a,\frac{2}{n} \floor{\frac{\ell_n - k_n}{2}} = \frac{\ell_n - k_n}{n} \xrightarrow{n\to\infty} b - a,

and if it's odd,

2n⌊ℓn−kn2⌋=ℓn−kn−1n→n→∞b−a.\frac{2}{n} \floor{\frac{\ell_n - k_n}{2}} = \frac{\ell_n - k_n - 1}{n} \xrightarrow{n\to\infty} b - a.

Hence,

lim⁡n→∞∫01χ[a,b]cos⁡(sin⁡(nπx)) dx=(b−a)∫01cos⁡(sin⁡(2πx)) dx,\lim_{n\to\infty} \int_0^1 \chi_{\br{a,b}} \cos\p{\sin\p{n\pi x}} \,\diff{x} = \p{b - a} \int_0^1 \cos\p{\sin\p{2\pi x}} \,\diff{x},

so c=∫01cos⁡(sin⁡(2πx)) dxc = \int_0^1 \cos\p{\sin\p{2\pi x}} \,\diff{x} works.