Finite-size charge multipole remainder

For each component $k$, write $D^2{b}_k(x)[u,v]={\sum }_{i,j=1}^3{a}_{ij}(x)u_i{v}_j$. The coefficients $a_{ij}$ are continuous because $b$ is $C^2$. The ball is compact by [F3], so the finitely many coefficients have a common bound $C_k<\mathrm{\infty }$ there. Since $\mathrm{\mid }{u}_i\mathrm{\mid }\le \mathrm{\mid }u\mathrm{\mid }$ and $\mathrm{\mid }{v}_j\mathrm{\mid }\le \mathrm{\mid }v\mathrm{\mid }$, the sum has magnitude at most $9C_k\mathrm{\mid }u\mathrm{\mid }\mathrm{\mid }v\mathrm{\mid }$, hence the displayed bilinear supremum and $M_k$ are finite. For nonzero $u,v$, apply that supremum to $u\mathrm{/}\mathrm{\mid }u\mathrm{\mid },v\mathrm{/}\mathrm{\mid }v\mathrm{\mid }$ and multiply by $\mathrm{\mid }u\mathrm{\mid }\mathrm{\mid }v\mathrm{\mid }$; when either vector is zero the bilinear expression is zero. This proves the homogeneous bound used below. Since $\mathrm{\mid }y-z\mathrm{\mid }\le \mathrm{\ell }$ on the support, monotonicity and homogeneity in [F2] show that each moment is absolutely integrable: $\mathrm{\mid }Q\mathrm{\mid }\le \mathrm{\parallel }\rho {\mathrm{\parallel }}_1$ and $\mathrm{\mid }{d}_i\mathrm{\mid }\le \mathrm{\ell }\mathrm{\parallel }\rho {\mathrm{\parallel }}_1$. By compactness of the ball and continuity of each $b_k$, [F3] bounds $b_k$ there, so $\rho b_k$ is integrable as well.

Fix $y\in {\bar{B}}_{\mathrm{\ell }}(z)$ and component $k$, and set $\varphi (t)=b_k(z+t(y-z))$ on $[0,1]$. The segment stays in the closed ball, and the chain rule [F4] gives ${\varphi }^{\mathrm{\prime }}(0)=D{b}_k(z)(y-z)$ and ${\varphi }^{\mathrm{\prime }\mathrm{\prime }}(t)=D^2{b}_k(z+t(y-z))[y-z,y-z]$, hence $\mathrm{\mid }{\varphi }^{\mathrm{\prime }\mathrm{\prime }}(t)\mathrm{\mid }\le C:=M_k\mathrm{\mid }y-z{\mathrm{\mid }}^2$. Set $H(t)=(1-t){\varphi }^{\mathrm{\prime }}(t)+\varphi (t)$ and $f(t):=H^{\mathrm{\prime }}(t)=(1-t){\varphi }^{\mathrm{\prime }\mathrm{\prime }}(t)$. This derivative is continuous and Riemann integrable by [F3], so [F4] gives $r_k(y):=b_k(y)-b_k(z)-D{b}_k(z)(y-z)={\int }_0^{1}f(t)dt$. To bound this Riemann integral, let $h_{\pm }(t):=\pm C(1-t)$. The pointwise inequalities $h_{-}\le f\le h_{+}$ imply $L(h_{-},P)\le L(f,P)$ and $U(f,P)\le U(h_{+},P)$ for every partition, by the infimum, supremum and positive weights in [F6]. Taking the corresponding supremum and infimum gives $\int h_{-}\le \int f\le \int h_{+}$. The continuous functions $h_{\pm }$ have primitives $\pm C(t-t^2\mathrm{/}2)$, so [F4] evaluates their integrals as $\pm C\mathrm{/}2$. Consequently $\mathrm{\mid }{r}_k(y)\mathrm{\mid }\le \frac{1}{2}{M}_k\mathrm{\mid }y-z{\mathrm{\mid }}^2$.

The function $r_k$ defined in step 2.1 is continuous on the ball. Its bound there makes $\rho r_k$ absolutely Lebesgue integrable by [F2]. Multiply that component expansion by $\rho (y)$ and integrate over ${\bar{B}}_{\mathrm{\ell }}(z)$. Linearity [F1] gives ${\mathcal{B}}_k=Q{b}_k(z)+D{b}_k(z)d+R_k$, with $R_k={\int }_{{\bar{B}}_{\mathrm{\ell }}(z)}\rho (y)r_k(y)d^{3}y$. By the Lebesgue integral triangle inequality and domination [F2], $\mathrm{\mid }{R}_k\mathrm{\mid }\le \frac{1}{2}{M}_k{\mathrm{\ell }}^2{\int }_{{\bar{B}}_{\mathrm{\ell }}(z)}\mathrm{\mid }\rho \mathrm{\mid }{d}^{3}y\le \frac{1}{2}{M}_k{\mathrm{\ell }}^2\mathrm{\parallel }\rho {\mathrm{\parallel }}_1$.