Suppose $h$ is a seminorm on $\mathbb{C}^n$, $I=\{X\in \mathbb{C}^n : h(X)<1\}$, $V=\{X\in \mathbb{C}^n: h(X)=0\}$, $U$ is the orthogonal complement of $V$ in $\mathbb{C}^n$ and $I_0 =I\cap U$. Let $$\mathcal{F}=\{s:\mathbb{C}^n\times \mathbb{C}^n\rightarrow \mathbb{C}: s\,\, \text{is a pseudo-Hermitian scalar product such that}\,\,q_s (X)=\sqrt{s(X,X)}\leq h(X)\}$$and let $\mathbb{E}(s)=\{X\in \mathbb{C}^n: q_s (X)<1\}\cap U$. It is easy to see that $V\subset I_0 +V\subset I\subset \mathbb{E}(s)+V$ and $h$ is a norm on $U$.
In a text, it is claimed that since $I_0$ is bounded, there is an $s\in \mathcal{F}$ such that the volume (in the Lebesgue measure on $U$) of $\mathbb{E}(s)$ is bounded.
I could not verify the claim. What is the idea behind this claim?
Thank you.
Thanks to Anne Bauval who gave me the idea in the comments above.
If $P$ is the projection of $\mathbb{C}^n$ onto $U$, it is easy to see from the semi-norm property of $h$ that for any $X\in \mathbb{C}^n$, $h(X)=h(P(X))$. Now, $h$ is a norm on the finite-dimensional space $U$. Hence there is a $c>0$ such that the $c||\cdot||_2\leq h(\cdot)\leq \frac{1}{c}||\cdot||_2$ on $U$. Now we define $s:\mathbb{C}^n\times \mathbb{C}^n\rightarrow \mathbb{C}$ by $s(X,Y)=c^2 \langle P(X),P(Y)\rangle$. It is easy to see that for this $s$, $q_s (X)=q_s (P(X))\leq h(P(X))=h(X)$. Also, with this $s$, $\mathbb{E}(s)\subset \frac{1}{c^2} I_0$.