to $b.^\ell$, differently. For $b.^\ell$, there is only one coordinate in which the derivative could be nonzero. In this component, the second factor of (\[onederivative\]) is 1. Thus we are left with $$\frac{\partial L}{\partial b.^\ell} = \sum_{\alpha=1}^n 2 (f_p(x_\alpha) - y_\alpha)$$ This vanishes by our choice of $b^\ell$. This suffices to show that $$\frac{\partial}{\partial b.^\ell} L(p) = 0.$$

Fix any point $p$ in $C$. By Theorem \[starlocus\], $p$ is a critical point. We will now show that every second derivative of $L$ with respect to the parameters $\{w_i, b_j\}$ vanishes, except for $$\frac{\partial^2}{\partial {b.^\ell}^2}= 2n.$$ This suffices to prove the claim that for every point $p$ in $C$, the Hessian of $L$ at $p$ has one positive eigenvalue and the remaining eigenvalues all vanish.

Differentiation with respect to any two parameters $\gamma_1, \gamma_2$ gives $$\begin{aligned}
\label{twoderivatives}
\frac{\partial}{\partial \gamma_1} \frac{\partial}{\partial \gamma_2} L(p) = 2 \sum_{\alpha=1}^n \frac{\partial}{\partial \gamma_1} f_p(x_\alpha) \frac{\partial}{\partial \gamma_2} f_p(x_\alpha) +(f_p(x_\alpha) - y_\alpha) \frac{\partial}{\partial \gamma_1} \frac{\partial}{\partial \gamma_2} f_p(x_\alpha).\end{aligned}$$

By Lemma \[notell\], at any point $p \in C$, any derivative of $f_p(x)$ not involving $\frac{\partial}{\partial w..^\ell}$ vanishes except $\frac{\partial}{\partial {b.^\ell}^2}$. Therefore with the exception of the derivative $\frac{\partial^2}{\partial {b.^\ell}^2}$, both terms of \[twoderivatives\] vanish for any second derivative of $L$ not 