= Solution
Let $G=L^{-1}$ and use the bound from part (b),
$$
|Gh|_{2,\alpha;B}\leq M|h|_{0,\alpha;B}.
$$
On the closed ball $\mathcal B_\varepsilon\subset C_0^{2,\alpha}(B)$ define
$$
T(u)=G(f-cu-g(u)).
$$
The Hölder product estimate and part (i) give
$$
|T(u)|_{2,\alpha}
\leq M\big(|f|_{0,\alpha}+C|c|_{0,\alpha}\varepsilon+\varepsilon^2\big),
$$
and
$$
|T(u)-T(v)|_{2,\alpha}
\leq M\big(C|c|_{0,\alpha}+2\varepsilon\big)|u-v|_{2,\alpha}.
$$
Choose $\varepsilon_0$ so that $2M\varepsilon_0<1/2$, then choose $\delta_0$ so that the remaining terms make $T$ preserve $\mathcal B_{\varepsilon_0}$ and have contraction constant less than one. The <Banach fixed-point theorem> gives a unique $u$ in that ball satisfying
$$
Lu+g(u)+cu=f,
\qquad u|_{\partial B}=0.
$$
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