PomeLemma 11.17.3.9020E77B2318 Let $Y$ be a complex manifold of dimension $N$, and let $Z\subset Y$ be a closed complex submanifold of dimension $D\geq 1$. If the least smooth convexity index of $Z$ is $\kappa \leq D$, then every smooth convexity index of $Y$ is at least $\kappa $.
Lemma 11.17.39020E77B2318
Proof. Suppose that $Y$ is $p$-convex for an integer $1\leq p<\kappa $. Then a smooth exhaustion function $\psi $ has at least $N-p+1$ positive Levi eigenvalues outside a compact set $K\subset Y$. Its restriction to the closed submanifold $Z$ is an exhaustion function, since its sublevel sets are closed subsets of the compact sublevel sets of $\psi $, and $K\cap Z$ is compact.
At a point of $Z\setminus K$, choose a complex subspace $P\subset T^{1,0}Y$ of dimension $N-p+1$ on which the Levi form of $\psi $ is positive definite. Then
\[ \dim _{\mathbb C}(P\cap T^{1,0}Z)\geq (N-p+1)+D-N=D-p+1. \]
The Levi form of $\psi |_ Z$ is the restriction of the Levi form of $\psi $, since the inclusion is holomorphic, so it has at least $D-p+1$ positive eigenvalues. Since $p<\kappa \leq D$, this shows that $Z$ is $p$-convex, which contradicts the definition of $\kappa $. $\square$
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