Let $D \subset R^m$ be a domain in a $m$-dimensional Euclidean space, $P \in intD$, and $t > 0$ so small that the sphere of radius $t$ centered at the point $P$ sits in $intD$. Let $\phi : D \rightarrow R$ be a smooth function. Let $v(P,t) = \frac{1}{(m-2)!}\int_0^t(t^2-r^2)^{\frac{m-3}{2}}rQ(P,r)dr$, where $Q(P,r) = \frac{1}{\omega_m}\int...\int\phi(x_1 + \beta_1r, ..., x_m + \beta_mr)d\omega_m$, integral being over the unit sphere in $R^m$ centered at the point $P \in intD$, $x_1, ..., x_m$ coordinates of the point $P$, $\omega_m = \frac{2\pi^\frac{m}{2}}{\Gamma(\frac{m}{2})}$ the area of the unit sphere, $d\omega_m$ the surface element of the unit sphere. Levitan B.M. in the paper ["On expansion in eigenfunctions of the Laplace operator", Mat. Sb. (N.S.), 35(77):2 (1954), 267–316] claims (see his formula (2.1.5)) that $v(P,t) = O(\int_0^t(t^2-r^2)^{\frac{m-3}{2}}rdr)$. Can somebody prove this claim or explain why it is true? Thank you in advance.
