Finiteness and Lower Semi-Continuity of an Functional.

Question

Assume I have a probability density $\rho$ on $\mathbb{R}^n$ with finite second moment

$$ \int_{\mathbb{R}^n}\|x\|^2\rho(x) dx<C. $$

I'm now interested in the following functional

$$ F(\rho):=\int_{\mathbb{R}^n} f(x) \rho(x)dx, $$

for some $f\in C(\mathbb{R}^n)$ which is bounded from below : $f(x)\geq c$ for some $c\in\mathbb{R}$.

Two questions, under what conditions on $f$ do we need for

• 1. $F$ finite?
• 2. $F$ lower semi-continuous?

Is $f$ Lipschitz enough or Im guessing it WILL depend on $\rho$?

EDIT : maybe Theorem $2.38$ of Functions of bounded variation and free discontinuity problems by Luigi Ambrosio could help?

Answer 1

For $F$ to be finite you need that $\frac{f(x)}{\|x\|^2+1}$ is bounded.

This is clearly sufficient.

Lets check that it is necessary:

Suppose you have a sequence $x_n$ with $\frac{f(x_n)}{\|x_n\|^2+1}≥ 4^n$. By continuity of $f$ these $x_n$ will eventually escape every compactum, so you may assume $\|x_n- x_m\|>1$ for $n\neq m$ and $\|x_n\|>1$ for all $n$. Around each $x_n$ you have a ball on which $f(x)≥4^n\|x_n\|^2$ holds, wlog this ball has radius $<1$ so that all these balls around $x_n$ are disjoint.

On a given such ball look at some bump-function $\rho_n$ supported on the ball so that $\int \rho_n = \frac{2^{-n}}{\|x_n\|^2+1}$. It is clear that $\rho:=\sum_n \rho_n$ remains smooth, positive and $$\int \rho<\infty, \quad \int\|x\|^2\rho ≤\sum_n 2^{-n}=1$$ By rescaling $\rho$ you may make it into a probability measure, it is then of the class you are considering. Now $$F(\rho) =\sum_n \int f\, \rho_n ≥\sum_n 4^n\|x_n\|^2\int \rho_n = \sum_n 2^n \frac{\|x_n\|^2}{\|x_n\|^2+1}=\infty$$

Note that if $f$ is Lipschitz this condition is satisfied.

Now about lower-semi-continuity:

Fix some $\rho$ and consider a net $\rho_\alpha\to\rho$ weakly. Let $\psi_n$ be the characteristic function of the ball of radius $n$. By weak convergence you've got $\int \psi_n\rho_\alpha \to \int\psi_n \rho$ as $\alpha\to\infty$. But you've also got that $$\int \psi_n \rho\to 1$$ as $n\to\infty$. This means that for every $\epsilon$ you've got an $n(\epsilon)$ so that the mass of $\rho$ outside of the ball of radius $n$ is smaller than $\epsilon$. Additionally the two equations mean you've got an $A(\epsilon)$ so that for all $\alpha>A$ the mass of $\rho_\alpha$ outside of the ball of radius $n(\epsilon)$ is smaller than $2\epsilon$.

Now $f\cdot \psi_n$ is bounded and hence $$\int f \psi_n\, \rho_\alpha\to \int f\psi_n \,\rho$$ Make use of this, for $n=n(\epsilon)$ and $\alpha ≥ A(\epsilon)$ you've got: $$\int f \rho_\alpha = \int f \psi_n\,\rho_\alpha + \int f(1-\psi_n)\rho_\alpha ≥ \int f \psi_n \rho_\alpha + 2c \epsilon$$ the term on the right converges, as $\alpha\to\infty$, towards $$\int f\psi_n\, \rho+ 2c\epsilon$$ where $\epsilon$ is as small as you please although $n$ will get larger with smaller $\epsilon$. But as $n$ goes to infinity the integral $\int f\psi_n\,\rho$ converges to $\int f\rho$ (here you again use that $f$ is bounded below and something like Beppo-Levi).

All these words amount to: $$\limsup_\alpha \int f\,\rho_\alpha ≥ \int f\rho$$

So $F$ is lower semi-continuous if $f$ is bounded below (finiteness not necessary).