The proof is constructive, which will be done in two steps.

The first step, as the main part, is to find a solution $u$ with the prescribed growth (or decay) rate. We will use the method of separation of variables described as follows.

For a fixed slice $Y^{2n-1}\subset\mathcal{C}^{n}$, let $\{\Lambda_{k}\}_{k=0}^{\infty}$ with $\Lambda_{0}=0$ be the spectrum of $\Delta_{\mathcal{C}^{n}}$ acting on functions. Let $\{\varphi_{k}\}_{k=0}^{\infty}$ be the eigenfunctions satisfying

Given a function $v$ and for any fixed $z\geq 1$, we have the fiberwise $L^{2}$-expansion on $Y^{2n-1}$,

Then we can first construct a formal solution

to (5.218), which holds in the $L^{2}$-sense for each fixed $z\geq 1$. Here the coefficient functions $u_{k}(z)$ are the particular solutions constructed in Lemma 5.15. The main part is to prove that the above series $u(z,\bm{y})$ converges with higher regularity and hence $u(z,\bm{y})$ is a regular solution to (5.218).

To begin with, we will prove that the series $u(z,\bm{y})$ converges in the $C^{0}$-norm and hence gives a $C^{0}$-function. Combining Lemma 5.13, Lemma 5.15 and the eigenfunction estimate in Lemma 3.32, we have

Applying Weyl’s law to the spectrum $\{\Lambda_{k}\}_{k=1}^{\infty}$,

where $C_{0}>0$ depends only on $Y^{2n-1}$ and $k$ is sufficiently large. Let $K_{0}\geq 2n+1$, then

Therefore, $u\in C^{0}(\mathcal{C}^{n})$ and $u$ satisfies the $C^{0}$-asymptotic estimate in (5.219).

Based on the above $C^{0}$-regularity, we will apply the standard elliptic regularity on $(\mathcal{C}^{n},g_{\mathcal{C}^{n}})$ to show that $u\in C^{2}(\mathcal{C}^{n})$ is a regular solution to $\Delta_{g_{\mathcal{C}^{n}}}u=v$. We take the partial sums

of the expansions

It is obvious that,

For every $\bm{x}\equiv(z,\bm{y})\in\mathcal{C}^{n}$, we will apply the elliptic regularity on the ball $B_{2}(\bm{x})\subset\mathcal{C}^{n}$ to obtain the higher regularity of $u$.

As a starter, by the same arguments as the above, we have $\|V_{N}-v\|_{C^{0}(B_{2}(\bm{x}))}\to 0$ as $N\to\infty$. The proof of the higher order convergence is almost verbatim. In fact, we just need to use $\|v\|_{C^{2K_{0}+m}}$ with $m\leq K_{0}$. Since $\Delta_{g_{\mathcal{C}}}U_{N}=V_{N}$, the standard $W^{2,p}$- implies that regularity for every $1<p<\infty$,

By assumption $v\in C^{3K_{0}}(\mathcal{C}^{n})$ for $K_{0}\geq 2n+1$, so it follows that $\|V_{N}\|_{C^{2}(B_{2}(\bm{x}))}\leq C_{\bm{x}}$. Therefore, for every $1<p<\infty$,

Now it suffices to choose $p>2n$, so the Sobolev embedding implies

which implies that $U_{N}\to u$ in the $C^{3}$-norm with respect to $g_{\mathcal{C}^{n}}$. The proof of the first step is done.

We have constructed a solution $u$ satisfying $|u(\bm{x})|\leq C\cdot e^{\eta\cdot z^{\frac{n}{2}}}$. Now we are ready to show that

This can be accomplished by the elliptic $W^{2,p}$-estimate. Since a Calabi space $(\mathcal{C}^{n},g_{\mathcal{C}^{n}})$ is collapsed with bounded curvatures as $z\to+\infty$, so there is some constant $r_{0}>0$ such that for each $\bm{x}\in\mathcal{C}^{n}$ satisfying $z(\bm{x})\geq 1$, the universal cover $(\widetilde{B_{2r_{0}}(\bm{x})},\tilde{\bm{x}})$ is non-collapsing. Now we lift the solution $u$ to this non-collapsing local universal cover, then for any $p>1$, there exists $C_{p}>0$ such that

We can choose any $p>2n$, then Sobolev embedding gives

In particular,

where $\alpha\equiv 1-\frac{2n}{p}$. So the proof of the proposition is done.

∎