We set $D = (d_{ij})_{(i,j) \in \llbracket 1,n\rrbracket^2} = (\sqrt{m_{ij}})_{(i,j) \in \llbracket 1,n\rrbracket^2} \in \mathcal{M}_n(\mathbb{R})$ and $M_c = \left((d_{ij} + c\xi_i^j)^2\right)$ with $c > 0$. Let $X^* \in \mathcal{A}$ be such that $\alpha(X^*) = \sup_{X \in \mathcal{A}} \alpha(X) > 0$, and denote $\alpha^* = \alpha(X^*)$. Show that:
${}^t X^* \Psi(M_{\alpha^*}) X^* = 0$,
$\Psi(M_{\alpha^*})$ has non-negative eigenvalues,
for all $c > \alpha^*$ and for all non-zero vector $X \in \mathcal{H}$, ${}^t X \Psi(M_c) X > 0$.
Conclude that $c^* = \alpha^*$.
We set $D = (d_{ij})_{(i,j) \in \llbracket 1,n\rrbracket^2} = (\sqrt{m_{ij}})_{(i,j) \in \llbracket 1,n\rrbracket^2} \in \mathcal{M}_n(\mathbb{R})$ and $M_c = \left((d_{ij} + c\xi_i^j)^2\right)$ with $c > 0$. Let $X^* \in \mathcal{A}$ be such that $\alpha(X^*) = \sup_{X \in \mathcal{A}} \alpha(X) > 0$, and denote $\alpha^* = \alpha(X^*)$.
Show that:
\begin{itemize}
\item ${}^t X^* \Psi(M_{\alpha^*}) X^* = 0$,
\item $\Psi(M_{\alpha^*})$ has non-negative eigenvalues,
\item for all $c > \alpha^*$ and for all non-zero vector $X \in \mathcal{H}$, ${}^t X \Psi(M_c) X > 0$.
\end{itemize}
Conclude that $c^* = \alpha^*$.