involving $\frac{\partial}{\partial w..^\ell}$.

The derivatives of $L$ it remains to check are $\frac{\partial^2}{\partial {b.^\ell}^2}$ and $\frac{\partial}{\partial *}\frac{\partial}{\partial w..^\ell}$. Again we use the expression \[fentry\] $$\sum w..^\ell \sigma \left( \sum w..^{\ell-1} ... \sigma \left( \sum w..^2 \sigma \left( \sum w..^1 z. + b.^1 \right) + b.^2 \right) + ... + b.^{\ell-1} \right) + b.^\ell.$$

First, we compute $$\frac{\partial^2}{\partial {b.^\ell}^2} L(p).$$

The derivative $\frac{\partial^2}{\partial {b.^\ell}^2} L(p) = 2n$, in particular is positive and nonzero.

Next, we compute $$\frac{\partial}{\partial *}\frac{\partial}{\partial w..^\ell} L(p).$$

Well, $$\begin{aligned}
 \label{twoderivativesw}
\frac{\partial}{\partial *} \frac{\partial}{\partial w..^\ell} L(p) = 2 \sum_{\alpha=1}^n \frac{\partial}{\partial *} f_p(x_\alpha) \frac{\partial}{\partial w..^\ell} f_p(x_\alpha) +(f_p(x_\alpha) - y_\alpha) \frac{\partial}{\partial *} \frac{\partial}{\partial w..^\ell} f_p(x_\alpha).\end{aligned}$$

We may check that at any $p \in C$, $$\frac{\partial}{\partial w..^\ell} f_p(x) = 0.$$ So the first term of \[twoderivativesw\] vanishes.

For the second term, we consider two cases.

Case 1: $* = w..^i$.

The coordinates of the derivative $$\frac{\partial}{\partial w..^i} \frac{\partial}{\partial w..^\ell} f_p(z)$$ have the form $$\begin{aligned}
& \sigma \left( \sum w..^{i - 1} \sigma \left( ... \right) + b.^{i-1} \right) \\
&  \cdot w..^{i+1} \sigma' \left( \sum w..^i \sigma \left( ... \right) + b.^i  \right) \\
& \cdot w..^{i+2} \cdot \sigma' \left( \sum w..^{i+1} \sigma \left( ... \right) + b.^{i+1}  \right) \\ 
& \hspace{.3in}\vdots 