= Solution
An <electron hole> is a missing <electron> in an otherwise almost filled <valence band>. A completely filled band has no net transport current: its velocities cancel over the <Brillouin zone>. Describing a few missing states as mobile excitations is therefore simpler than following almost every occupied <electron> state.
Near an isotropic valence maximum, write the <electron> dispersion as $E_v(\mathbf k)=E_v(0)-\hbar^2k^2/(2m_h)$ with $m_h>0$. The <electron> curvature <mass> is negative there. Removing that <electron> creates an excitation with <momentum> $\hbar\mathbf k_h=-\hbar\mathbf k$ and <energy> measured upward from the filled reference,
$$
E_h(\mathbf k_h)=E_v(0)-E_v(-\mathbf k_h)=\frac{\hbar^2k_h^2}{2m_h}.
$$
Thus
$$
\boxed{m_h^*=-m_{e,\mathrm{valence}}^*>0.}
$$
The sign refers to the positive excitation <energy> of a missing <electron> near a maximum, not to negative vacuum <mass>. A general maximum gives the corresponding positive hole-curvature tensor.<Lecture Notes for Solid State Physics by Steven H. Simon>[§16.2, printed pp. 178–179: missing-electron carriers, opposite charge and <band effective mass>]{ref}
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