= Solution
The <Weak Harnack inequality> is an estimate for a nonnegative <weak supersolution>. Write $F=(f^1,\ldots,f^n)$, let $B_{2R}(x_0)\subset\subset\Omega$, and assume $u\geq0$ almost everywhere on this ball. There are $p>0$ and $C<\infty$, depending only on the dimension, the <uniform ellipticity> bounds, $q$, and the scaled norms of the lower-order coefficients, such that
$$
\boxed{\left(\frac{1}{|B_R(x_0)|}\int_{B_R(x_0)}u^p\right)^{1/p}\leq C\left(\operatorname*{ess\,inf}_{B_R(x_0)}u+R^{1-n/q}\|F\|_{L^q(B_{2R})}+R^{2-2n/q}\|g\|_{L^{q/2}(B_{2R})}\right).}
$$
Here $\langle v\rangle_E=|E|^{-1}\int_Ev$ is the <integral average>. The <weak supersolution> inequality uses the sign convention $Lu\leq\operatorname{div}F+g$. For instance, the coefficient dependence can be expressed using
$$
R^{1-n/q}(\|b\|_{L^q(B_{2R})}+\|c\|_{L^q(B_{2R})}),\qquad R^{2-2n/q}\|d\|_{L^{q/2}(B_{2R})}.
$$
Thus one may use the same $p,C$ on all sufficiently small balls when the global coefficient norms and <uniform ellipticity> bounds are fixed. The nonnegativity condition is essential to this formulation; a signed <weak supersolution> may first be shifted, with the resulting change in its forcing included. As usual, supremum and infimum statements for functions in a <Sobolev space> mean <essential suprema> and <essential infima>. The standard multidimensional statement uses $n\geq2$, so $q/2>1$. In dimension one the analogous formulation requires $q\geq2$; the printed restriction $q>n$ alone would allow $q/2<1$, which does not suffice. To see the obstruction, smooth the nonnegative capped function $\min\{1,|x|/\varepsilon\}$ by a <mollifier> of width $\tau$. Its positive second derivative is a bump of mass $2/\varepsilon$ and has $L^{q/2}$ size of order $\varepsilon^{-1}\tau^{2/q-1}$. For fixed $\varepsilon$ and $1<q<2$, this size and the minimum both tend to zero as $\tau\downarrow0$, whereas the average on a fixed larger interval stays positive. Thus the <Weak Harnack inequality> cannot have the displayed uniform forcing bound in that range. Shrinking $\varepsilon$ and choosing $\tau$ still smaller likewise defeats a uniform Hölder estimate based on that norm alone.
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