Lemma 6.8.6E20924A743CA

Proof. (1) The function $F$ is smooth with $\Delta F=4\sum _ j\partial _{z_ j}\partial _{\bar z_ j}F=0$, by (5.2.9); so its real and imaginary parts are harmonic and have the ball mean value property (proposition 3.1.5). Thus $F(y)$ is the mean of $F$ over $B(y,\delta )$, and the Cauchy–Schwarz inequality gives (6.8.7).

(2) Homogeneous polynomials $P_ l$, $P_{l'}$ of different degrees are orthogonal in $L^2$ of the unit sphere: $P_ l(\mathrm e^{\mathrm i\theta }\omega )=\mathrm e^{\mathrm il\theta }P_ l(\omega )$, and surface measure is invariant under the orthogonal maps $\omega \mapsto \mathrm e^{\mathrm i\theta }\omega $ (subsection 3.1.1), so $\int _{S^{2n-1}}P_ l\overline{P_{l'}}\, \mathrm dS=\mathrm e^{\mathrm i(l-l')\theta }\int _{S^{2n-1}}P_ l\overline{P_{l'}}\, \mathrm dS$ for every $\theta $. For small $t>0$ the homogeneous expansion $F(x+t\omega )=\sum _ lt^ lP_ l(\omega )$ converges uniformly in $\omega \in S^{2n-1}$, by (5.2.1); integrating term by term, we obtain $\int _{S^{2n-1}}|F(x+t\omega )|^2\, \mathrm dS(\omega )=\sum _{l\geq k}t^{2l}\int _{S^{2n-1}}|P_ l|^2\, \mathrm dS$, which implies the first inequality. The last integral is positive, because $P_ k$ does not vanish identically on the sphere, by homogeneity, and nonempty relatively open subsets of the sphere have positive surface measure. $\square$