Analysis Problem of the Day 87

Today’s problem appeared as Problem 6 on the UCLA Fall 2023 Analysis Qual:

Problem 6. Define $f_n(x) = \frac{\sin(n^4 x)}{n^3 x}.$

a) Prove that $f_n$ does not converge to zero in $L^4(\mathbb{R}).$

b) Prove that $f_n$ converges to zero weakly in $L^4(\mathbb{R}),$ i.e. for all $g \in L^{\frac{4}{3}}(\mathbb{R}),$

$\lim_{n \to \infty} \int_{\mathbb{R}} f_n(x)g(x)dx =0.$

Solution: a) We consider the change of variables $y=n^4 x, dy = n^4 dx$ and write

$\|f_n\|_{L^4}^4 = \int_{\mathbb{R}} \frac{\sin^4(n^4 x)}{n^{12}x^4} dx = \int_\mathbb{R} \frac{\sin^4(y)}{y^4 n^{-4}} n^{-4} dy = \int_\mathbb{R} \frac{\sin^4(y)}{y^4}.$

Note that $g(y)=\frac{\sin y}{y}$ is a continuous function (if one replaces the removable singularity at $y=0$ with $g(y)=1$ ) that is not identically zero, so $g^4(y)$ is a continuous nonnegative nonzero function, so $\|f_n\|_{L^4}^4 = \|g\|_{L^4}^4 = C \not = 0.$ This shows that $f_n$ does not converge to zero in $L^4.$

b) We rely on a density argument. Since the span of characteristic functions of open intervals is dense in the set of simple functions in the $L^{\frac43}$ norm, and simple functions are dense in $L^{\frac43},$ by linearity it suffices to show that the above expression converges to zero for $g(x)=\chi_{(a,b)}(x).$ But indeed, by the same change of variables as above,

$\int_{\mathbb{R}} \frac{\sin(n^4 x)}{n^3 x} \chi_{(a,b)}(x)dx = \int_{n^4 a}^{n^4 b} \frac{\sin(y)}{n^{-1}y} n^{-4} dy = \frac{1}{n^3}\int_{n^4 a}^{n^4 b} \frac{\sin(y)}{y} dy.$

One may then recall that $\frac{\sin(y)}{y}$ is the classic example of an improper Riemann integrable function that is not Lebesgue integrable, with $\lim_{n \to \infty}\int_{-n}^n \frac{\sin y}{y} = \frac{\pi}{2}.$ Thus, the above integrals are certainly bounded for large enough $n,$ meaning $\int f_n(x) \chi_{(a,b)}(x)dx \lesssim n^{-3} \to 0$ as $n \to \infty.$ By the density argument, we therefore conclude that $f_n \rightharpoonup 0$ in $L^4,$ despite not converging to zero in norm.

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