Solution
= Solution
{c}
\b[False.] Take $V=\{x^2+y^2+z^2<1\}$, the open <unit ball>, and $u(x,y,z)=x$. This is a nonzero <harmonic function>. On the <unit sphere>, the outward <unit normal vector> is $\mathbf n=(x,y,z)$, so
$$
\partial_nu=\mathbf n\cdot(1,0,0)=x=u.
$$
Thus the prescribed <Robin boundary condition> holds. Unlike the preceding case, <Green's identities> now give $\int_V|\nabla u|^2=\int_{\partial V}u^2$, which permits nonzero solutions of the <differential equation>.