Limit of multiple integral?

Question

Let $~f:\mathbb{R}\to\mathbb{R}$ be continuous non-negative function and $\int_{-\infty}^{+\infty} f(x)dx = 1$

Denote : $$I_n(r) = {\idotsint}_{x_1^2+\dots x_n^2 \le r^2} \prod_{k = 1}^{n} f(x_k)dx_1\cdots dx_n$$

How to find the following limit : $\lim_{n\to\infty} I_n(r)$ for fixed r.

Answer 1

Let's do it two ways.

1. Probabilistic method.

This can be viewed as probability problem. If $X_1,X_2,\dots$ are IID with density $f$, then the integral in question is the probability that the random vector $(X_1,\dots,X_n)$ is in the ball of radius $r$ centered at the origin, that is, the integral is equal to

$$ P(X_1^2 + \dots + X_n^2 \le r^2)=(*)$$

But since $P(X_1=0)>0$ (simply because $X_1$ has a density), it follows that $P(X_1^2>\epsilon)>0$ for some $\epsilon>0$, and from this (Borel-Cantelli), we obtain $P(X_j^2 > \epsilon\mbox{ for infinitely many }j\mbox{'s})=1$. As a result $\lim_{n\to\infty} (X_1^2+\dots+X_n^2) =\infty$, a.s. which implies
$$\lim_{n\to\infty} (*)= 0.$$

2. Straightforward calculation.

$$(1)\quad I_{n+1} (r) = \int_{|z|\le r} I_n (\sqrt{r^2-z^2}) f(z)dz \le I_n (r) \int_{|z|\le r} f(z) dz.$$

where last inequality is because $I_n(\cdot)$ is nondecreasing. This implies that $I_{n+1}(r) \le I_n (r)$, hence $I_\infty(r)=\lim_{n\to\infty} I_n (r)$ exists for all $r$.

Fix $c\in (0,1)$. Then by continuity of the integral, there exists $r>0$ such that $\int_{|z|\le r} f(z) dz = c$. From $(1)$ we then obtain

$$ I_{n+1} (r) \le c I_n (r),$$

which in turn implies $I_\infty(r)=0$. Next, let

$$r_0 = \inf \{ r: I_\infty (r)>0\}.$$

If $r_0=\infty$, we're done (this is automatically the case if the support of $f$ is unbounded). Suppose $r_0<\infty$, and let $\epsilon \in (0,r_0)$.

From $(1)$ and the fact that $I_n(\cdot)$ is nondecreasing, we obtain

\begin{align*} I_n(r_0+\epsilon) &= \int_{|z|\le r_0+\epsilon} I_n (\sqrt{(r_0+\epsilon)^2 -z^2})f(z) dz\\ & \le I_n(r_0-\epsilon) \int_{\{z: z^2 > (r_0+\epsilon)^2 -(r_0-\epsilon)^2\}} f(z) dz + I_{n} (r_0+\epsilon) \int_{\{z: z^2\le (r_0+\epsilon)^2 - (r_0-\epsilon)^2\}} f(z) dz\\ &\to 0 + I_\infty(r_0+\epsilon) \int_{\{z:|z|^2 < 4r_0 \epsilon\}}f(z) dz. \end{align*} If in addition, we require $\epsilon>0$ to be small enough that the integral on the RHS above is equal $c\in (0,1)$, then the limit of the RHS is bounded above by $c I_\infty(r_0+\epsilon)$ for some $c\in (0,1)$, while the limit of the LHS side is equal to $I_\infty(r+\epsilon_0)$. That gives $$I_\infty (r_0+\epsilon) \le c I_\infty(r_0+\epsilon),$$ a contradiction, because $I_\infty(r_0+\epsilon)>0$. Hence $r_0=\infty$.

Answer 2

The answer by @Fnacool treats the more general case when $f\in L^1(\mathbb{R})$. If $f$ is bounded on every segment $J\subset \mathbb{R}$ then $\displaystyle\lim_{r\to\infty} I_n(r) = 0$ for all $r > 0$. Indeed, fix a radius $r$ and let $M = \sup_{|x| < r} |f(x)|$. We have $$ |I_n(r)| \le \iiint_{B_n(r)} M^n\,dx_1\ldots\,dx_n = M^n\cdot \text{Vol}_n(B_n(r)) = \frac{\pi^{n/2}r^nM^n}{\Gamma(n/2 + 1)}. $$ The latter tends to $0$ very fast as $n\to\infty$.