Growth rate of derivatives of $\phi_n(t):=\mathbb E_x[f(x_1)f(tx_1 + (1-t^2)^{1/2} x_2)]$, where $x=(x_1,\ldots,x_n)$ is uniform on unit-sphere

Question

Suppose $f:[-1,1] \to \mathbb R$ is function which is $k$ times continuously-differentiable a.e on $(-1,1)$, for some fixed $k \ge 1$. Let $n$ be a large positive integer, define the function $\phi_n:[-1,1] \to \mathbb R$ by $$ \phi_n(t) = E_x[f(x_1)f(tx_1 + (1-t^2)^{1/2} x_2)], $$ where $x=(x_1,x_2,\ldots,x_n)$ uniform on the unit-sphere in $\mathbb R^n$.

Question 1. Is true that $\phi_n$ is $k$ times continuously-differentiable on $(-1,1)$ ?

In the event that the answer of the above question is affirmative, consider the following question.

Question 2. In terms of $k$ and $n$, what is the order of growth the quantity $\alpha_{n,k}:=\sup_{t \in [-1/2,1/2]}|\phi^{(k)}_n(t)|$ ?

My rough guess is that $\alpha_{n,k} \lesssim E_x[|x_1|] = \mathcal O(1/n^{1/2})$.

Related question: 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 $0$..

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Example

Because $f$ is $k$ times continuously-differentiable a.e on $(-1,1)$, we can construct continuous functions $g_0,g_1, \ldots,g_k:[-1,1] \to \mathbb R$ such that $g_k = f^{(k)}$ a.e on $(-1,1)$. In particular, $g_0 = f$ a.e on $(-1,1)$.

For $k=1$, we have $\phi^{(1)}_n(t) = E_x[g_0(x_1)(x_1-t(1-t^2)^{-1/2}x_2)g_1(tx+(1-t^2)^{1/2}x_2)]$, using the boundedness of $g_0$ and $g_1$, one computes

$$ \begin{split} \alpha_{n,1}^2 &\lesssim \sup_{|t| \le 1/2}E|x_1-t(1-t^2)^{-1/2}x_2|^2 = \sup_{|t| \le 1/2}E[x_1^2+t^2(1-t^2)^{-1}x_2^2]\\ &= E[x_1^2 + (1/3)x_2^2] \lesssim 1/n. \end{split} $$

Thus, $\alpha_{n,1} = \mathcal O(1/n^{1/2})$.

Answer 1

We extend the intuition in the case $k=1$ to more general $k$. I'm not 100% sure of all the steps. Comments welcome.

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As, noted before

Because $f$ is $k$ times continuously-differentiable a.e on $(-1,1)$, we can construct continuous functions $g_0,g_1, \ldots,g_k:[-1,1] \to \mathbb R$ such that $f^{(k)}=g_k$ a.e on $(-1,1)$. In particular, $f=g_0$ a.e on $(-1,1)$.

For $x=(x_1,x_2,\ldots,x_n)$ drawn uniformly at random on $S_{n-1}$, consider the (random) function $u_n(t)=u_n(t;x):=tx_1+(1-t^2)^{1/2}x_2$. Using the Faà di Bruno's formula, we can write the $k$th derivative of $\phi_n$ as follows $$ \phi_n^{(k)}(t) = E_x\left[g_0(x_1)\sum \frac{k!}{\prod_{i=1}^k m_i!i!^{m_i}}g_{m_1+\ldots+m_k}(u_n(t))\prod_{i=1}^k (u_n^{(i)}(t))^{m_i}\right], $$ where the sum is over all $k$-tuples $(m_1,\ldots,m_k)$ such that $\sum_{i=1}^k im_i = k$. Thus, again because all the $g_j$'s are bounded (being continuous on the compact interval $[-1,1]$), one has

$$ \alpha_{n,k} \lesssim \sum \frac{k!}{\prod_{i=1}^km_i!i!^{m_i}}E_x\left[\prod_{i=1}^k \sup_{|t| \le 1/2}|u^{(i)}(t)|^{m_i}\right]. \tag{1} $$

Now, for all $t \in [-1/2,1/2]$, we note the following:

• $u_n^{(0)}(t)=u_n(t)$, and so $|u^{(0)}_n(t)|=|u_n(t)| = |tx_1+(1-t^2)^{1/2}| \lesssim \max(|x_1|,|x_2|)$.
• $u_n^{(1)} = x_1-t(1-t^2)^{-1/2}x_2$, and so $|u^{(1)}_n(t)|=|x_1-t(1-t^2)^{-1/2}| \lesssim \max(|x_1|,|x_2|)$.
• for $i \ge 2$, we have $u_n^{(i)}$ is of the form $p_i(t)(1-t^2)^{-(2i-1)}$ where $p_i(t)$ is a polynomial of degree $i-1$ in $t$, and so $|u_n^{(i)}(t)| \lesssim |x_2|$.

Putting things together, we have $\sup_{t \in [-1/2,1/2]}|u_n^{(i)}(t)| \lesssim \max(|x_1|,|x_2|)$ for all $i$. We deduce that

$$ \begin{split} E_x\left[\prod_{i=1}^k \sup_{|t| \le 1/2}|u^{(i)}(t)|^{m_i}\right] &\lesssim E_x[\max(|x_1|,|x_2|)^{m_1+\ldots + m_k}]\\ &\le 2 E_x[|x_1|^{m_1+\ldots + m_k}]\\ &= \mathcal O(n^{-1/2}). \end{split} $$ were in the last line we've used the fact that $m_1+\ldots+m_k \ge 1$ since all the $m_i$'s can't be equal to $0$.

Combining with (1) gives $\alpha_{n,k} = \mathcal O(n^{-1/2})$ as was suspected.