= Solution
Apply the <chain rule> to $f_0=f_0(E)$ with $E=H_0(\mathbf J)$:
$$
\mathbf k\cdot\partial_{\mathbf J}f_0=f_0'(E)\mathbf k\cdot\boldsymbol\Omega.
$$
On the <resonance> selected by the <Dirac delta function>, $\mathbf k\cdot\boldsymbol\Omega=k_\phi\Omega_{\mathrm p}$. Thus
$$
\boxed{\mathcal T=\pi(2\pi)^3\Omega_{\mathrm p}\sum_{\mathbf k}k_\phi^2\int|\Psi_{\mathbf k}|^2\bigl[-f_0'(E)\bigr]
\delta\bigl(\mathbf k\cdot\boldsymbol\Omega-k_\phi\Omega_{\mathrm p}\bigr)\,d^3J\geq0.}
$$
Every weight is nonnegative, and $-f_0'(E)>0$. Strict positivity additionally requires a populated nonaxisymmetric <resonance> with nonzero coupling on a set of positive surface measure. \b[The unconditional statement is nonnegativity, rather than positive definiteness.] As a concrete counterexample to strict positivity of this formula, retain one nonzero conjugate harmonic pair with $k_\phi\ne0$ and choose its smooth coefficient $\Psi_{\mathbf k}$ supported in an open region $|\Delta_{\mathbf k}|>d>0$. The perturbation is nonzero but has zero resonant coupling, so $\mathcal T=0$ even when $f_0'(E)<0$. The even simpler zero-perturbation case also gives zero. Whether a particular physical <stellar bar> has nonzero resonant overlap is additional information.
For a coupled <stellar bar> and halo, the halo gains positive <angular momentum> and exerts the opposite <torque> on the <stellar bar>. If the halo <mass> is $M_h$ and the rigid <stellar bar> has constant positive <moment of inertia> $I_b$, then
$$
\boxed{I_b\frac{d\Omega_{\mathrm p}}{dt}=-M_h\mathcal T\leq0.}
$$
This <bar-halo angular-momentum transfer> is a form of <dynamical friction>: a resonantly coupled <stellar bar> tends to slow down while the <dark matter halo> spins up. Actual long-term evolution changes both the <stellar bar> pattern and the halo <distribution function>, so the fixed-pattern response must then be updated. If the <stellar bar> changes its <moment of inertia>, the general relation is $d(I_b\Omega_{\mathrm p})/dt=-M_h\mathcal T$; positive <moment of inertia> alone would not fix the sign of $d\Omega_{\mathrm p}/dt$ without that rigidity assumption.
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