A finite harmonic Taylor series and its Cauchy bound

Work under [F2]. Write $u(x)=x_1^2-x_2^2$ for $x\in {\mathbb{R}}^n$. Its coordinate partials are ${\mathrm{\partial }}_{1}u=2x_1$, ${\mathrm{\partial }}_{2}u=-2x_2$ and ${\mathrm{\partial }}_{i}u=0$ for $i\ge 3$, so the second coordinate partials are ${\mathrm{\partial }}_1{\mathrm{\partial }}_{1}u=2$, ${\mathrm{\partial }}_2{\mathrm{\partial }}_{2}u=-2$ and all others vanish; hence $\mathrm{\Delta }u=2-2=0$, and $u$ is harmonic on every Euclidean ball.

Expand $u$ about $a$: writing $h=x-a$, $u(a+h)=(a_1+h_1{)}^2-(a_2+h_2{)}^2=u(a)+2(a_1{h}_1-a_2{h}_2)+(h_1^2-h_2^2)$, and there is no term of degree three or higher. Hence $D^{\alpha }u(a)=0$ for $\mathrm{\mid }\alpha \mathrm{\mid }\ge 3$, the Taylor series terminates at degree two, and it equals $u$ at every point (the finite sum is the expansion above), in agreement with the general real-analytic representation of [F1].

Factorial Cauchy bound. For $r>0$ let $M:={\sup }_{B_{2r}(a)}\mathrm{\mid }u\mathrm{\mid }<\mathrm{\infty }$; the polynomial is harmonic on all of ${\mathbb{R}}^n$ by step 1.1, so [F1] gives $\mathrm{\mid }{D}^{\alpha }u(a)\mathrm{\mid }\le M{C}_n^{\mathrm{\mid }\alpha \mathrm{\mid }}\mathrm{\mid }\alpha \mathrm{\mid }!r^{-\mathrm{\mid }\alpha \mathrm{\mid }}$ for every multi-index. For $\mathrm{\mid }\alpha \mathrm{\mid }\ge 3$ the derivative is actually zero. The finite expansion of step 2.1 checks the normalization $D^{\alpha }u(a)\mathrm{/}\alpha !$ directly: its $h_1^2$ coefficient is $1=D^{(2,0,\dots )}u(a)\mathrm{/}2!$.