1, \quad 2q > 2.  \label{eq:qbd}$$ But $${\left\vert h_k(x) \right\vert} \leq C k^{-1/12} \label{eq:hsizebd}$$ so $$\begin{gathered}
\begin{aligned}
\int \vert h_k (x) \vert^{2q} \, dx &= \int \vert h_k (x) \vert^2 \vert h_k (x) \vert^{2(q-1)} \, dx\\
& \leq C^{2(q-1)} k^{-(q-1)/6} \int \vert h_k(x) \vert^2 \, dx
\end{aligned} \label{eq:hintbd}\\
\intertext{and}
{\left\Vert h_k \right\Vert}_{2q} \leq C^{1/p} k^{-1/(12p)}, \quad p \geq 1. \label{eq:hnormbd}\end{gathered}$$ This means that the matrix $W$ satisfies the condition with $\alpha = \frac{1}{12p}$ This observation was crucial in [@MiSi]; it gives a broad class of operators covered by so our claims of this sections are applicable to the operators .

But there are much better estimates of $L^p$ norms of the Hermite functions.

As $n \to \infty$,

$$\begin{aligned}
{\left\Vert h_n \right\Vert}_r &\sim n^{-\frac{1}{2} \left( \frac{1}{2} - \frac{1}{r} \right)}, \quad 1 \leq r < 4 \label{eq:hefcnsmallr}\\
{\left\Vert h_n \right\Vert}_r &\sim n^{- \frac{1}{8} } \log n, \quad r = 4 \label{eq:hefcnmidr}\\
{\left\Vert h_n \right\Vert}_r &\sim n^{-\frac{1}{6} \left( \frac{1}{r} + \frac{1}{2} \right)}, \quad r >  4 \label{eq:hefcnlarger}\end{aligned}$$

\[eq:hefcnvarr\]

See [@Thanga Lemma 1.5.2] for the sketch of the proof and further explanations of these claims.

Therefore, could be improved. If $p > 2$ then by $q < 2$, $2q < 4$ so $$\begin{aligned}
{\left\Vert 