Pome7A772B0D8822Lemma 11.17.1.7A772B0D8822 Fix $\delta ,B>0$. On a finite-dimensional Hermitian vector space, put
for covectors $\xi $ and vectors $c$ with $|c|\leq B$. There is $\gamma >0$, depending only on $\delta $ and $B$, such that $s_ c^2-\gamma |\xi |^2$ is real convex for every such $c$.
Proof. For $0<\tau \leq 1$, replace $s_ c$ by the norm
which is smooth off zero. Its Lipschitz constant is at most $L=\sqrt{B^2+1}+\delta $, uniformly in $c$ and $\tau $. All derivatives below are real derivatives on the dual space. At $\xi \ne 0$, set $u=\xi /|\xi |$ and decompose a vector as $w=\alpha u+b$ with $b$ orthogonal to $u$ for the real inner product. Write $s=s_{c,\tau }$ and $A=\mathrm ds_\xi (w)$. Since the first summand of $s$ is convex and $s\geq \delta |\xi |$,
Homogeneity gives $\mathrm ds_\xi (u)=s(\xi )/|\xi |\geq \delta $. Since $|\mathrm ds_\xi (b)|\leq L|b|$, we have
Take $\gamma =\delta ^4/(\delta ^2+2L^2)$. Then $\gamma \leq \delta ^2/2$ and $\gamma (1+2L^2/\delta ^2)=\delta ^2$, so $D^2(s^2)(w,w)\geq 2\gamma |w|^2$.
The homogeneous function $s^2-\gamma |\xi |^2$ of degree two extends as a $C^1$ function at zero with differential zero. Its restriction to every real line is convex, including at zero, because its second derivative is nonnegative away from zero and its first derivatives match there. Thus it is convex on the whole space. Finally,
and the locally uniform limit as $\tau \downarrow 0$ proves the assertion with the same $\gamma $. $\square$
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