Compute limit of $a_n:=E[f(x_1)f(x_2)]$, for random $(x_1,\ldots,x_n)$ on unit-sphere in $R^n$ and any function $f$ with a jump discontinuity at $0$. Let $f:[-1,1] \to \mathbb R$ be continuous a.e (and assumed to be bounded, if that helps), with a jump-discontinuity at $0$ and set
$$
c:=\frac{f(0^-)+f(0^+)}{2}.
$$
For any integer $n \ge 2$, define $a_n$ by
$$
a_n := \mathbb E[f(x_1)f(x_2)],
$$
where $x=(x_1,\ldots,x_n)$ is uniform on the unit-sphere in $\mathbb R^n$.

Question. Is it true that $\lim_{n \to \infty} a_n = c^2$ ?

For example, if the function $f$ is given by
$$
f(x) := \begin{cases}\cos^{13}(x),&\mbox{ if }x > 0,\\x^2-5,&\mbox{ else,}\end{cases}
$$
then $f$ has a jump discontinuity at $0$ with $c=-2$, and numerical experiments show that $a_n \to 4=c^2$.

 A: Let $g$ be a function such that $g(x)=f(x)+f(-x)$  for all $x \ne 0$ and $g(0)=2c$. So $g$ is continuous at $0$.
Let $(Y_n)$ be a sequence of independent standard normal random variables.
For any $n$, let $(X_1^n,X_2^n,...,X_n^n)$ be a random vector uniformly distributed on $\mathbb{S}^{n-1}$.
Then
$$(X_1^n,X_2^n) \stackrel{\text{law}}{=}(\tilde{X}_1^n,\tilde{X}_2^n)$$
where $$\tilde{X}_i^n := \frac{Y_i}{\sqrt{Y_1^2+\dots +Y_n^2}} \quad \text{ for } i = 1,2$$
Besides, by law of large numbers, $\tilde{X}_1^n$ and $\tilde{X}_2^n$ converge almost surely to $0$. Thus,
$$4\mathbb{E}( f(X_1^n)f(X_2^n))= \mathbb{E}( g(X_1^n)g(X_2^n))=\mathbb{E}( g(\tilde{X}_1^n)g(\tilde{X}_2^n))\xrightarrow{n \rightarrow \infty} \mathbb{E}( g(0)^2) =c^2$$
A: Disclaimer. If I've not made obvious errors, the post below answers the question in the affirmative. Suggestions / comments welcome.

Define the function $F:[-1,1]^2 \to \mathbb R$ by $F(u, v) := f(u)f(v)$, and let $\mu_n$ be the marginal distribution of $(x_1,x_2)$ when $(x_1,x_2,\ldots,x_n)$ is uniform on the unit-sphere in $\mathbb R^n$. Thus, we see the limit of the sequence of numbers defined by $a_n = \int_{\mathbb R^2}Fd\mu_n$.
Now, define a sequence of functions $(f_n)_n:[-1,1] \to \mathbb R$ by
$$
f_n(t) := \begin{cases}f(t),&\mbox{ if }1/n < |t| \le 1,\\
g_n(t),&\mbox{ if }|t| \le 1/n,\end{cases}
$$
for any choice of functions $g_n:[-1,1] \to \mathbb R$ such that

*

*$g_n$ is continuous on $[-1/n,1/n]$,

*$g_n(-1/n) = f(-1/n)$,

*$g_n(0) = c$,

*$g_n(1/n) = f(1/n)$.

Define $F_n:[-1,1]^2 \to \mathbb R$ by $F_n(u,v):=f_n(u)f_n(v)$. We note the following facts:

*

*(1) $F_n \to F$ pointwise on $[-1,1]^2\setminus L$, where $L := \{(u,v) \mid u = 0\,\lor\,  v= 0\}$, a set with $\mu(L) = \mu_n(L) = 0$ for all $n$.

*(2) $F_n$ is continuous (and therefore bounded).

*(3) $\mu_n \to \mu := \delta_{(0,0)}$ weakly. For example, see the comments under this question https://mathoverflow.net/q/406716/78539, and combine with Scheffé's Lemma.

*(4) $\int_{\mathbb R^2} F_n d\mu_n \to c^2$.

One computes
$$
\left|a_n-\int_{\mathbb R^2}F_nd\mu_n\right| \le \int_{\mathbb R^2} |F-F_n| d\mu_n := A_n.
$$
Thanks to this post https://mathoverflow.net/a/406728/78539, we know that $A_n \to 0$.  We conclude from (4) that $a_n \to c^2$ as claimed. $\quad\quad\quad\Box$
