I'm having problems with the second derivative of the function $g(\alpha) = f(\textbf{y}(\alpha))$ (which I will define more precisely below). I tried calculating it myself, could anyone just simply check that my reasoning is correct or not? :) Here goes:
Let $\textbf{x} = (x_1, x_2, ..., x_n)$ and $\textbf{d} = (d_1, d_2, ..., d_n) \in \mathbb{R}^n$. Lets define $\textbf{y}(\alpha) = \textbf{x} + \alpha\textbf{d}$, where $0 \leq \alpha \leq \bar{\alpha}$.
Now lets define the function $g(\alpha) = f(\:\textbf{y}(\alpha)\:)$. Calculate $g''(\alpha)$.
Here is my attempted solution: First I calculate $g'(\alpha)$:
$$g'(\alpha) = \frac{\text{d}}{\text{d}\alpha}\left( \:f(\:\textbf{y}(\alpha)\: )\: \right) = \frac{\text{d}}{\text{d}\alpha}(\:f(y_1,y_2, ..., y_n)\:) = \frac{\text{d}}{\text{d}\alpha}(\: f(x_1 + \alpha d_1, x_2 + \alpha d_2, ..., x_n + \alpha d_n) \:) = \frac{\partial f}{\partial y_1}\cdot\frac{\partial y_1}{\partial \alpha} + \cdots + \frac{\partial f}{\partial y_n}\cdot\frac{\partial y_n}{\partial \alpha} = \;\;\;\;\;\;\;\;\;\;\;\;\;\frac{\partial f}{\partial y_1}\cdot d_1 + \cdots + \frac{\partial f}{\partial y_n}\cdot d_n = \nabla f(\:\textbf{y}(\alpha)\:)\cdot \textbf{d} = h(\textbf{y})$$
Next I calculate $g''(\alpha)$:
$$g''(\alpha) = \frac{\text{d}}{\text{d}\alpha}(\: h(\textbf{y})\:) = \frac{\text{d}}{\text{d}\alpha}(\: h(y_1, ..., y_n) \:) = \frac{\partial h}{\partial y_1}\cdot\frac{\partial y_1}{\partial \alpha} + \cdots + \frac{\partial h}{\partial y_n}\cdot\frac{\partial y_n}{\partial \alpha} = \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{\partial h}{\partial y_1}\cdot d_1 + \cdots + \frac{\partial h}{\partial y_n}\cdot d_n = \nabla h(\: \textbf{y}(\alpha)\:)\cdot \textbf{d}$$
Now:
$$\frac{\partial h}{\partial y_i} = \frac{\partial }{\partial y_i} \left( \frac{\partial f}{\partial y_1}\cdot d_1 + \cdots + \frac{\partial f}{\partial y_n}\cdot d_n \right) = \frac{\partial^2f}{\partial y_1\partial y_i}\cdot d_1 + \cdots + \frac{\partial^2f}{\partial y_n\partial y_i}\cdot d_n$$
So:
$$g''(\alpha) = \frac{\partial^2f}{\partial y_1^2}\cdot d_1^2 + \cdots + \frac{\partial^2f}{\partial y_n\partial y_1}\cdot d_nd_1 + \cdots + \frac{\partial^2f}{\partial y_1\partial y_n}\cdot d_1d_n + \cdots + \frac{\partial^2f}{\partial y_n^2}\cdot d_n^2$$
Or:
$$g''(\alpha) = \textbf{d} \;\nabla^2 f(\:\textbf{y}(\alpha)\:) \:\textbf{d}^T$$
As I said before, could someone verify that I got it right? :) Thank you for any help !