Analysis Problem of the Day 52

Today’s problem appeared as Problem 6 on the UCLA Fall 2021 Analysis Qual:

Problem 11. Let $K$ be a continuous function on $\mathbb{R}^2$ periodic in both coordinates, i.e. $K(x+1,y)=K(x,y)$ and $K(x,y+1)=K(x,y).$ For any $F \in L^1([0,1] \times [0,1]),$ show that

$\int_{[0,1]^2} K(x,y+nx) F(x,y)d\mu(x,y) \to \int_0^1 \left(\int_0^1 K(x,s)ds\right)\left(\int_0^1 F(x,y)dy\right)dx$

as $n \to \infty,$ where $\mu$ is the two-dimensional Lebesgue measure.

Solution: As with all such kinds of complicated and scary looking expressions, we try to approximate $F$ by the simplest functions we know – in our case, products of characteristic functions of the form $\chi_{(a,b)}(x)\chi_{(c,d)}(y).$ Note that the presence of the $n$ will make $K$ oscillate faster and faster in the second coordinate. This reminds us of weak convergence – namely, the result that

$\int_a^b g(ny) f(y) dy \to \left(\frac{1}{b-a}\int_a^b g(s) ds\right) \int_a^b f(y)dy$

whenever the integrals make sense. Motivated by this, we perform the change of variables $ny'=y+nx$ and use Fubini and periodicity of $K$ to rewrite the integral so that the inner integral with respect to $y$ becomes

$n \int_{x}^{x+\frac{1}{n}} K(x,ny') \chi_{(\frac{c}{n}+x,\frac{d}{n}+x)}(y') dy'= \int_{x}^{x+1} K(x,ny') \sum_{i=0}^{n-1}\chi_{(\frac{c+i}{n}+x,\frac{d+i}{n}+x)}(y') dy'.$

For fixed $x,$ the integral with respect to $y$ converges pointwise (by the above result) to

$\int_x^{x+1} K(x,y') dy' \int_{x}^{x+1} \sum_{i=0}^{n-1} \chi_{(\frac{c+i}{n}+x,\frac{d+i}{n}+x)}(y') dy' = \int_0^1 K(x,s)ds (d-c)$

by the periodicity of $K$ in the $y$ variable. Moreover, all these functions are uniformly bounded in $x,$ so by dominated/bounded convergence theorem one has that the integral with respect to $x$ converges to

$\int_0^1 \int_0^1 K(x,s)ds (d-c) \chi_{(a,b)}(x)dx = \int_0^1 \int_0^1 K(x,s)ds \int_0^1 \chi_{(a,b)}(x) \chi_{(c,d)}(y)dy dx.$

Since the span of such functions is dense in the span of characteristic functions of rectangles, and the span of the latter is dense in $L^1,$ it follows that this convergence extends to arbitrary $F \in L^1([0,1] \times [0,1]).$

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