11.10 Homogeneous convex regularization879A1C19426E

Proof. Write $t=|z|^2$. Both branches are continuous, positive semidefinite Hermitian quadratic forms in $v$, and the first is positive definite. Their square roots satisfy the triangle inequality and complex homogeneity; their maximum therefore is a complex norm. In particular $Q$ is positive off zero and continuous, also on the zero section.

The first branch is plurisubharmonic on the total space. For the second, the locally uniformly convergent expansion

expresses it as a limit of sums of squared moduli of holomorphic functions. Thus it too is plurisubharmonic, and so is their maximum.

For $0<t<1/4$, subtracting $t|v|^2$ gives

which is convex. At $z=0$, one has $Q(0,v)=|v|^2$. Hence each fixed fiber has a positive strong convexity margin.

To test a margin on a neighborhood of $(0,(1,0))$, fix any $c>0$ and choose a nonzero $z$ arbitrarily close to zero with $t<c$. At $v=(1,0)$ the second branch is strictly larger than the first, so it remains the maximum on a small fiber ball. Along the real line $v=(1,s)$ in this ball,

Its second derivative is negative. Thus no positive $c$ makes $Q-c|v|^2$ convex on the fiber balls in any such total-space neighborhood. A fiber-linear change of smooth frame and its inverse are uniformly bounded on a smaller base neighborhood, so a positive quadratic margin in another frame would give one in this frame. This proves the failure of the local margin. $\square$

Proof. We first describe a gluing step. Let $f$ be a global section that is smooth near a compact set $L_1$, and let $L_2$ be a compact subset of one of the neighborhoods $U$ in the statement. Choose relatively compact open sets $W_3,W_4$ with $L_1\subset W_3$, $f$ smooth near $\overline W_3$, and $L_2\subset W_4\Subset U$. Choose a smooth function $0\leq \rho \leq 1$ equal to one near $\overline W_3$, with compact support in the region where $f$ is smooth. Put $V=W_3\cup W_4$, and choose a smooth function $-1\leq \psi \leq 1$ equal to one near $L_1\cup L_2$ and equal to minus one near $M\setminus V$.

By stability, for sufficiently small $\xi >0$ the function $f_\xi =f+\xi \psi $ is still a global section. Approximate $f_\xi $ on $\overline W_4$ by a smooth section $g$, with uniform error less than $\delta <\xi /2$, and with arbitrarily small derivative errors near the compact set $\overline W_4\cap \operatorname {supp}\rho $, where $f_\xi $ is smooth. On $V$ define

These definitions agree on the overlap, and $h$ is smooth. Where $\rho $ vanishes, $h$ is the section $g$. Where $\rho $ can be nonzero, $h-f_\xi =(1-\rho )(g-f_\xi )$, and stability applied to the smooth section $f_\xi $ makes $h$ a section if the derivative errors of $g-f_\xi $ are sufficiently small. Moreover, $|h-f_\xi |<\delta $.

Define $f_1=\max \{ f,h\} $ on $V$ and $f_1=f$ outside $V$. Near $L_1\cup L_2$ we have $h>f$, and near the boundary of $V$ we have $h<f$. Thus $f_1$ is continuous, it is a section because $\mathcal S$ is closed under maxima, and it is smooth near $L_1\cup L_2$. Also $\| f_1-f\| _{C^0}<\xi +\delta $. Near $L_1$ it equals $f+\xi $, so the previous smoothness is preserved.

Now cover $M$ by finitely many compact sets $L_2$ of this kind and repeat the step, starting with $L_1$ empty and then taking for $L_1$ the union of the previously treated compact sets. At every step choose $\xi $ and $\delta $ so small that the sum of the finitely many uniform errors is below the prescribed tolerance. The final section is smooth everywhere and has the required uniform error. $\square$

Proof. Complex homogeneity gives the identity

For fixed $(s,t)$, the map $(z,v)\mapsto (z-s,T_ tv)$ is holomorphic and invertible, and its restriction to each fiber is linear. On a compact subset of the cone consisting of vectors of norm one, all these maps with $(s,t)$ sufficiently small stay in a fixed compact subset of the original cone, and their differentials and inverses are uniformly bounded there. A positive local lower bound for the Levi form of $Q$, interpreted through quadratic barriers, therefore gives a positive lower bound for every pullback and for their average. The same argument with real quadratic barriers in the fibers preserves local strong convexity. Only fiber segments contained in the local cone neighborhood are used here.

Ordinary convolution makes $q_\delta $ smooth, and the formula for $Q_\delta $ makes it smooth on the cone and complex homogeneous. Uniform convergence, derivative convergence on compact regions where $Q$ is smooth, and positivity for small $\delta $ follow from the corresponding properties of convolution and from the positive minimum of $Q$ on compact sets of vectors of norm one. $\square$

Proof. Fix a smooth Hermitian squared norm $G$ on $F$, and let $M=\mathbb P(F)$ be the bundle of complex lines in $F$. The quotient $m=Q/G$ is a positive continuous function on the compact manifold $M$. Define a sheaf $\mathcal S$ on $M$ by requiring that the corresponding function $Gm$ on the cone be positive, strictly plurisubharmonic, and locally strongly convex in the fibers.

This sheaf is closed under maxima. Indeed, multiplication by $G>0$ commutes with maxima. Near a point of the cone, subtract a common positive quadratic barrier from both functions; their maximum remains plurisubharmonic, and on sufficiently small fiber balls it remains convex. Positivity is also preserved.

For stability on a compact subset $K$ of $M$, restrict to vectors of norm one lying over $K$, and to a fixed compact annulus around them. Finitely many bundle charts provide positive common lower bounds for both barriers. If $u$ is a smooth function on $M$ that is sufficiently small in $C^2$ near $K$, the Levi form and the real fiber Hessian of $Gu$ are smaller than half these bounds on that annulus. The perturbed function is therefore still positive and has both strict margins. Homogeneity transports the local conditions to the rest of the cone over $K$. This proves the stability required in lemma 11.10.2.

On a sufficiently small projective coordinate chart, lemma 11.10.3 gives local approximation, including derivative control on compact subsets where $m$ is smooth; dividing its regularizations by $G$ gives local smooth sections of $\mathcal S$. By lemma 11.10.2, there are smooth positive global sections $m_ j\to m$ uniformly. Put $Q_ j=Gm_ j$ off the zero section and $Q_ j=0$ on the zero section. Homogeneity gives continuity at zero, uniformly over the compact base, as well as the asserted convergence on a fixed unit sphere bundle.

It remains to see that $p_ j$ is a norm. Local convexity in the fibers away from zero implies convexity on the whole fiber. Indeed, the restriction of $Q_ j$ to a real line avoiding zero is locally convex and hence convex, while on a real line through zero, homogeneity gives $Q_ j(tv)=t^2Q_ j(v)$ for real $t$, which is convex including at zero. Thus the set $B_ x=\{ v:Q_ j(v)\leq 1\} $ is convex, balanced and absorbing, and its gauge is $\sqrt{Q_ j}$. If $v,w$ are nonzero, convexity applied to $v/p_ j(v)$ and $w/p_ j(w)$, with weights proportional to $p_ j(v)$ and $p_ j(w)$, gives the triangle inequality; the cases with a zero vector follow directly. Positivity, smoothness off zero and complex homogeneity give the remaining properties of a smooth complex norm. $\square$

Proof. For $\tau >0$, set $Q_\tau =Q+\tau H$; it is positive, complex homogeneous and strictly plurisubharmonic. On a compact annulus of vectors of norm close to one, local strong convexity of $Q$ in the fibers gives a positive uniform margin in finitely many frames, while the real fiber Hessian of $H$ is uniformly bounded there. Choose $\tau $ so small that the possible negative part of $\tau $ times this Hessian consumes less than half the margin. Homogeneity then gives local strong convexity of $Q_\tau $ in the fibers everywhere off zero. Apply theorem 11.10.4 to $Q_\tau $ and take one of its smooth regularizations; its square root is the required smooth complex norm. $\square$

Proof. For the first assertion, theorem 11.10.4 would produce a smooth complex norm with plurisubharmonic square, contrary to theorem 11.9.4. For the second, use proposition 11.10.5 and the same obstruction. $\square$