Prove that $\mathbb P(\|\nabla f(x)\|< Bn) < KB^2$ for a random spherical harmonic $f$ I have a random spherical harmonic of degree $n$ on the sphere $S^2$, i.e.
$$ f = \sum_{k=-n}^{n} \xi_k Y_k$$
with $\xi_k \sim N\left(0, \dfrac{1}{2n+1}\right)$ being independent Gaussians and $\{Y_k\}$ a $L_2$-orthonormal basis of spherical harmonics of degree $n$. (The variances were chosen so to have $\mathbb{E}\| f \|_{L^2} = 1$.)
I want to prove that, given a positive constant $B$ and a point $x \in S^2$, then
$$\mathbb P(\|\nabla f(x)\| < Bn) < KB^2$$
for some positive constant $K$. I thought about showing that the density of the random vector $\nabla f(x)$ is bounded, but I don't know how to compute it.
This inequality is stated without proof in the second to last inequality in page 9 of this article.
 A: $\def\d{\mathrm{d}}\def\peq{\mathrel{\phantom{=}}}$Lemma: If $ξ \sim N(0, D)$ is an $m$-dimensional normal random vector with $D > 0$ and $A$ is an $m × m$ positive definite matrix, then there exists a constant $c > 0$ such that$$
P(ξ^T Aξ < M^2) < c\min(M, M^m). \quad \forall M > 0
$$
Proof: Since $A$ is positive definite, then there exists an $m × m$ invertible real matrix $P$ such that $A = P^T P$. Note that $ξ \sim N(0, D) \Rightarrow η = Pξ \sim N(0, PDP^T)$, and $D_1 = PDP^T > 0$. Because the density
function of $η$, i.e.$$
f_η(x) = \frac{1}{(2π)^{\frac{m}{2}} |D_1|} \exp\left( -\frac{1}{2} x^T D_1^{-1} x \right),
$$
is continuous at $x = 0$, then\begin{align*}
&\peq P(ξ^T Aξ < M^2) = P(η^T η < M^2)\\
&= \int\limits_{|x| < M} f_η(x) \,\d x \sim f_η(0) \int\limits_{|x| < M} \d x \sim c_0 M^m. \quad (M → 0^+)
\end{align*}
Thus there exists $M_0 > 0$ such that $P(ξ^T Aξ < M^2) < (c_0 + 1) M^m$ for $0 < M \leqslant M_0$. Since for $M > M_0$, $P(ξ^T Aξ < M^2) \leqslant 1 < \dfrac{M^m}{M_0^m}$, then taking $c_1 = \max\left(c_0 + 1, \dfrac{1}{M_0^m} \right)$,$$
P(ξ^T Aξ < M^2) < c_1 M^m. \quad \forall M > 0
$$
Also, suppose $η = (η_1, \cdots, η_m)^T$. Since $η_1 \sim N(0, σ_1^2)$ for some $σ_1 > 0$, then\begin{align*}
&\peq P(ξ^T Aξ < M^2) = P(η^T η < M^2) \leqslant P(η_1^2 < M^2)\\
&= \frac{1}{\sqrt{2π} σ_1} \int_{-M}^M \exp\left( -\frac{x^2}{2σ_1^2} \right) \,\d x \leqslant \frac{2M}{\sqrt{2π} σ_1}. \quad \forall M > 0
\end{align*}
Thus it suffices to take $c = \max\left( c_1, \dfrac{2}{\sqrt{2π} σ_1} \right)$.
Now back to the question. Define $ξ = (ξ_{-n}, \cdots, ξ_n)^T$, $Y = (Y_{-n}, \cdots, Y_n)^T$, then$$
|∇f(x)|^2 = \left| \sum_{k = -n}^n ξ_k ∇Y_k(x) \right|^2 = ξ^T ∇Y(x) (∇Y(x))^T ξ.
$$
Note that $ξ \sim N\left( 0, \dfrac{I_{2n + 1}}{2n + 1} \right)$. For $n \geqslant 1$, taking $m = 2n + 1$ and $A = ∇Y(x) (∇Y(x))^T$ in the lemma, there exists a constant $c > 0$ such that for any $M > 0$,\begin{align*}
&\peq P(|∇f(x)| < nB) = P(|∇f(x)|^2 < n^2 B^2)\\
&< c\min(nB, n^{2n + 1} B^{2n + 1}) \leqslant cn^{2n + 1} · B^2. \quad \forall B > 0
\end{align*}
Thus $K$ can be taken as $cn^{2n + 1}$.
