Let $h:\mathbb R \to \mathbb R$ be a sufficiently integrable function. For fixed unit vectors $u,v \in \mathbb R$, and $X=(X_1,\ldots,X_d)$ uniformly random  on the unit-sphere in $\mathbb R^d$, let $F_h(u,v) := E[h(X^\top u)h(X^\top v)]$. Because of the rotation-invariance of $X$, one has the identity $F_h(u,v) = E[h(X_1)h(qX_1 + \sqrt{1-q^2} X_2)] =: f_h(q)$, where $q:=u^\top v$.

Now, suppose $f_{d,h}$ is smooth around $0$ and define the constants
$$
\alpha_d(h) := f_{h,d}(0),\,\beta_d(h) := f_{h,d}'(0),\,\gamma_d(h) := f_{h,d}(1).
$$

Further suppose $h$ is smooth and for $G \sim N(0,1)$, define more constants
$$
a(h):= (E[h(G)])^2, \, b(h) = E[h'(G)]^2, \, c(h) := E[h(G)^2].
$$

>**Question.** In the limit $d \to \infty$, is it true that $\alpha(h) \to a(h)$, $\beta(h) \to b(h)$, and $\gamma(h) \to c(h)$ for any "sufficiently integrable" function $h$ ? 

Solution with added restriction that $h$ is Lipschitz continuous
---
>*Below, I do some computations which seem to suggest the result is true under some additional smoothness constraints on $h$. I'm not 100% sure of my calculations. I'd be grateful if someone else would double-check. Thanks in advance!*

So, suppose $h$ satisfies
- (1) $h$ is smooth with $\|h'\|_\infty < \infty$.
- (2) $h$ is square-integrable w.r.t $N(0,1)$ and the marginal distribution of $X_1$.

>**Claim 1.** Under the above assumptions, the following hold in the limit $d \to \infty$,
$$
\begin{split}
|\alpha(h)-a(h)| &= \mathcal O(1/d),\\
|\beta(h)-b(h)| &= \mathcal O(1/d).
\end{split}
$$

Under the additional assumption that

- $h'$ is Lipschitz continuous,

we have

>**Claim 2.** $|\gamma(h)-c(h)| = \mathcal O(1/d)$.


*Proof of Claim 1.* WLOG, suppose $\|h\|_{Lip} = 1$. For a square-integrable random variable $R$, let $\|R\|_2 := (E[R^2])^{1/2}$. Note that if $R_1$ and $R_2$ have bounded moments of order $2k$, then $\|R_1-R_2\|_k \lesssim \|R_1-R_2\|_1$.

- Now, let $(G_1,G_2) \sim N(0,I_2)$, independent of $X$. Let $W_p(X_1,G_1)$ be the Wasserstein distance of order $p$, between $X_1$ and $G_1$. For $W_2$-optimal couplings $(X_1,G_1)$ and $(X_2,G_2)$, one computes
$$
\begin{split}
&|\alpha(h)-a(h)| = |E[h(X_1)h(X_2) - h(G_1)h(G_2)]|\\
&\quad= |E[h(X_1)(h(X_2)-h(G_2))] + E[h(G_2)(h(X_1)-h(G_1))]|\\
&\quad\le |E[h(X_1)(h(X_2)-h(G_2))]| + |E[h(G_2)(h(X_1)-h(G_1))]|\\
&\quad\le \|h(X_1)\|_2\|h(X_2)-h(G_2)\|_2 + \|h(G_2)\|_2\|h(X_1)-h(G_1)\|_2\\
&\quad\lesssim \|X_2-G_2\|_2 + \|X_1-G_1\|_2\\
&\quad\lesssim W_2(X_2,G_2) + W_2(X_1,G_1)\\
&\quad\lesssim W_2(X_1,G_1).
\end{split}
\tag{1}
$$
The first inequality is the triangle inequality, the second is Cauchy-Schwarz, and the third is thanks to the Lipschitz continuity of $h$.
Recall the following fact (established here https://mathoverflow.net/a/401444/78539)

>**Fact 1.** $W_p(X_1,G_1) = \mathcal O(1/d)$ for all $ p \ge 1$.

We deduce from (1) that $|\alpha(h)-a(h)| = \mathcal O(1/d)$ as claimed.

- Next, we prove that $|\gamma(h)-c(h)| = \mathcal O(1/d)$. Indeed,
$$
\begin{split}
|\gamma(h)-c(h)| &= |E[h(X_1)^2 - h(G_1)^2| = |E[(h(X_1)+h(G_1))(h(X_1)-h(G_1)]|\\
&\le \|h(X_1)+h(G_1)\|_2\|h(X_1)-h(G_1)\|_2\\
&\le (\|h(X_1)\|_2+\|h(G_1)\|_2)\|X_1-G_1\|_2\\
& \lesssim \|X_1-G_1\|_2 = W_2(X_1,G_1) = \mathcal O(1/d).
\end{split}
$$
This completes the proof of **Claim 1.** $\quad\quad\Box$

*Proof of Claim 2.*
Consider the function $r:\mathbb R \to \mathbb R$ defined by $r(x) := xh(x)$. One computes
$$
\begin{split}
|\beta(h)-b(h)| &= |E[r(X_1)h'(X_2) - r(G_1)h'(G_2)]|\\
&= |E[r(X_1)(h'(X_2)-h'(G_2)) + h'(G_2)(r(X_1)-r(G_1))]|\\
&\le \|r(X_1)\|_2\|h'(X_2)-h'(G_2)\|_2 + \|h'(G_2)\|_2\|r(X_1)-r(G_1)\|_2\\
&\le \|X_2-G_2\|_2 + \|r(X_1)-r(G_1)\|_2\\
&\le W_2(X_2,G_2) + \|r(X_1)-r(G_1)\|_2\\
&= \|r(X_1)-r(G_1)\|_2+\mathcal O(1/d),
\end{split}
$$
where the second inequality is thanks to the Lipschitzness of $h'$. It only remains to bound the $\|r(X_1)-r(G_1)\|_2$. To this end, not that 
$$
\begin{split}
\|r(X_1)-r(G_1)\|_2 &= |X_1h(X_1)-G_1h(G_1)|\\
&= \|X_1(h(X_1)-h(G_1))+h(G_1)(X_1-G_1)\|_2\\
&\le \|X_1\|_4\|h(X_1)-h(G_1)\|_4+\|h(G_1)\|_4\|X_1-G_1\|_4\\
&\lesssim W_4(X_1,G_1) = \mathcal O(1/d).
\end{split}
$$
This completes the proof of **Claim 2.** $\quad\quad\quad\Box$