Analysis Problem of the Day 15

Today’s problem is Problem 4 on the UCLA Fall 2024 Analysis Qual.

Problem 4: Let $f_n \in L^1(\mathbb{R}^d)$ be a sequence of nonnegative functions with $\int f_n dx = 1$ for all $n$ and $\int_K f_n dx \to 0$ for all compact $K \subset \mathbb{R}^d \setminus \{0\}.$ Show that there exist $0 \leq \alpha \leq \beta \leq 1$ such that for all $g \in C_0(\mathbb{R}^d)$ , one has

$\liminf_{n \to \infty} \int f_n g dx = \alpha g(0), \quad \limsup_{n \to \infty} \int f_n g dx = \beta g(0).$

Solution: Intuitively, if the $L^1$ “mass” of $f_n$ eventually vanishes on every compact set avoiding zero, then the mass has to either escape to 0 or to infinity. With that in mind, we proceed directly. Define

$\alpha = \lim_{r \to 0} \liminf_{n \to \infty}\int_{|x|<r} f_ndx.$

Notice that for each $r,$ the limit infimum is well-defined, and is strictly decreasing as $r \to 0,$ so this limit is well-defined. Likewise, define

$\beta = \lim_{r \to 0} \limsup_{n \to \infty} \int_{|x|<r} f_n dx.$

Clearly, $0 \leq \alpha \leq \beta \leq 1.$ I claim that $\alpha$ and $\beta$ satisfy the identities given in the problem. Indeed, for $0 \leq g \in C_0(\mathbb{R}^d),$ we first consider the splitting

$\int f_n g dx = \int_{|x|<r}f_n g dx+\int_{|x|\geq r} f_n g dx.$

I claim that the limit as $n \to \infty$ of the second term exists and equals 0. Indeed, pick $K \subset B(0,r)^c$ compact and large enough so that $|g|<\epsilon$ on $K^c.$ Then,

$\int_{|x| \geq r} f_n g dx\right \leq \int_K f_n g dx + \int_{K^c} f_n g dx < \epsilon \|g\|_\infty + \epsilon$

as $n \to \infty$ since $g$ is bounded, $\int_K f_n \to 0,$ and $\int f_n dx = 1$ for all $n.$ Then, taking the limit infimum as $n \to \infty$ gets rid of the second term in the sum.

Finally, for small enough $r,$ by continuity of $g$ there exists small enough $\epsilon >0$ such that

$(g(0)-\epsilon) \liminf_{n \to \infty} \int_{|x| < r} f_n dx \leq \liminf_{n \to \infty} \int_{|x|<r} f_n g dx \leq (g(0)+\epsilon) \liminf_{n \to \infty} \int f_n dx,$

so taking the limit as $r \to 0$ yields

$\liminf_{n \to \infty} \int f_n g dx = \lim_{r \to 0} \liminf_{n \to \infty} \int_{|x|<r} f_n g dx = \alpha g(0).$

An analogous statement holds for the limit supremum.

Remark: The originally presented proof of this fact was incorrect, as it made the incorrect claim that the map $h \to \liminf_{n \to \infty} \int f_n h dx$ is a linear functional, which it is not (as it is only subadditive).

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