Theorem 15.35.10092A98EE953D

Proof. Put $w_ j=2^{-j}$ and $M_ j=\max _{X_0}F_ j$.

Step 1: a smooth baseline. Fix $C^ j$ norms $\| \cdot \| _{C^ j}$ on $X_0$, computed in a fixed finite atlas, so that they increase with $j$. Choose $0<\varepsilon _ j\leq 1/2$ with

Then $E=\sum _ jw_ j\varepsilon _ jF_ j$ converges in every $C^ l$ norm, so $E$ is smooth on $X_0$, and $0<E\leq b_0/4$.

Step 2: bumps escaping to the pole. Choose a smooth function $\chi :\mathbb R\to [0,1]$, supported in $[0,1]$, positive on $(0,1)$ and flat at $0$ and $1$. Starting with $T_0=0$, choose inductively $T_ j\geq 2$ with $T_ j>T_{j-1}+2$ and

This is possible because, by lemma 15.35.8 with $\tau =1/j$, the factor $\mathrm e^{(L^\alpha -1)/j}W(L)$ is uniformly large on sufficiently distant intervals. On the annulus $\{ T_ j\leq L\leq T_ j+1\} $ put $\beta _ j=\chi (L-T_ j)B_{1/j}/Z_ j$, extend $\beta _ j$ by zero, and set $g_ j=\varepsilon _ j+(1-\varepsilon _ j)\beta _ j$. Since $\chi $ is flat at the endpoints, $\beta _ j$ is smooth; it vanishes near $o$ and near the hyperplane $\{ L=1\} $. Its integral against $\kappa ^ k$ is one, so $g_ j$ is a smooth strictly positive probability density with respect to $\kappa ^ k$. Moreover,

Step 3: the scalars $c_ H(f_ j)$. By theorem 14.2.1 on $Y$, there is a smooth $\phi _ j$ with $\kappa +\mathrm{dd}^{\mathrm{c}}\phi _ j>0$ and $(\kappa +\mathrm{dd}^{\mathrm{c}}\phi _ j)^ k=g_ j\kappa ^ k$; the scalar is one because both sides have mass one. On $X_0$, the same theorem gives a smooth $v_ j$ with $h_0+\mathrm{dd}^{\mathrm{c}}v_ j>0$ and $(h_0+\mathrm{dd}^{\mathrm{c}}v_ j)^ m=c_{h_0}(F_ j)F_ j\, \mathrm dV_0$. Since only one term of the binomial expansion survives for dimensional reasons,

Uniqueness of the scalar gives $c_ H(f_ j)=C c_{h_0}(F_ j)\leq 2^{-2j}$.

Step 4: properties of $f$. By Tonelli's theorem, $\int _ Xf\, \mathrm dV=\sum _ jw_ j=1$. The density of $f\, \mathrm{d}V$ with respect to $\mu $ is

The annuli are disjoint and escape to the pole, so only finitely many bumps meet any compact subset of $Y\setminus \{ o\} $; hence $F$, and therefore $f$, is smooth off the pole fiber. On the pole fiber all bumps vanish and $F=E$, so $f$ is finite everywhere. Also $f\geq w_1f_1\geq \frac12\min _ Xf_1>0$. If $\mathrm{MA}^{\mathrm{ext}}_ H(u)\geq af\, \mathrm{d}V$ with $a>0$, then $\mathrm{MA}^{\mathrm{ext}}_ H(u)\geq aw_ jf_ j\, \mathrm{d}V$, and lemma 15.27.1 gives

This is impossible, so $b_ H(f)=0$ by theorem 15.28.4.

Step 5: exponential subsolutions. Fix $\lambda >0$, and choose $J$ with $1/j\leq \lambda $ for $j\geq J$. Since $L^\alpha -1\geq 0$, we have $B_{1/j}\leq B_\lambda $ for these $j$. As the bumps have disjoint supports, Step 2 bounds their total contribution $\sum _{j\geq J}w_ j(1-\varepsilon _ j)F_ j\beta _ j$ by $B_\lambda /2$. The baseline satisfies $E\leq b_0/4\leq B_\lambda /4$, because $B_\lambda \geq q_\alpha \geq b_0$. The remaining finitely many bumps are bounded by the constant

and $K_\lambda \leq (K_\lambda /b_0)B_\lambda $. Hence $F\leq A_\lambda B_\lambda $ with $A_\lambda =3/4+K_\lambda /b_0$, and theorem 15.35.9 gives the exponential subsolution $u_\alpha +\lambda ^{-1}\log (CV_0/A_\lambda )$ and the greatest solution. Finally, a bounded solution $u$ would satisfy $\mathrm{MA}^{\mathrm{ext}}_ H(u)\geq \mathrm{e}^{\lambda \inf _ Xu}f\, \mathrm{d}V$, a positive scaled subsolution, contradicting $b_ H(f)=0$. Hence every solution is unbounded. $\square$