Some basic properties of parabola in the context of the second-order Hermite interpolation equation.
There is a one-dimensional dataset with n points: x = (x_0, \dots, x_{n-1}) and y = (y_0, \dots, y_{n-1}), such that x_{i} > x_{i-1} for any i. We introduce the following labels:
h_{i-1} = x_i - x_{i-1}
\delta_{i-1} = \frac{y_i - y_{i-1}}{x_i - x_{i-1}}
h is always positive, while \delta can have any value.
On each segment, we approximate the unknown function with a parabola. On the [i-1, i] segment:
\begin{equation}f(x) = a_0 + a_1 (x - x_{i-1}) + a_2 (x - x_{i-1})^2\end{equation}
The first and the second derivative of the parabola are:
\begin{equation}f'(x) = a_1 + 2 a_2 (x - x_{i-1})\end{equation}
\begin{equation}f''(x) = 2 a_2 \end{equation}
The Hermite interpolation theorem provides us with a tool to determine the a coefficients on the [i-1, i] segment if we know y_{i-1}, y_i and one of the first derivative in the end points, either y'_{i-1} or y'_i. Let's assume that we know ("know" in the sense "we can compute") y'_{i-1}. Then there are three equations that the parabola must obey:
\begin{equation}f(x_{i-1}) = y_{i-1} = a_0\end{equation}
\begin{equation}f'(x_{i-1}) = y'_{i-1} = a_1\end{equation}
\begin{equation}f(x_{i}) = y_i = a_0 + a_1 h_{i-1} + a_2 h_{i-1}^2\end{equation}
The solution is then trivial:
\begin{equation}a_0 = y_{i-1}\end{equation}
\begin{equation}a_1 = y'_{i-1}\end{equation}
\begin{equation}a_2 = \frac{y_{i} - y_{i-1} - y'_{i-1} h_{i-1}}{h_{i-1}^2}\end{equation}
Once the a coefficients are known, the parabola is fully determined. That also means that the derivative in the point i is fixed:
f'(x_{i}) = y'_i = a_1 + 2 a_2 (x_i - x_{i-1}) = y'_{i-1} + 2 \frac{y_{i} - y_{i-1} - y'_{i-1} h_{i-1}}{h_{i-1}^2} h_{i-1}
y'_i = \frac{2(y_{i} - y_{i-1}) - y'_{i-1} h_{i-1}}{h_{i-1}}
\begin{equation}y'_i = 2\delta_{i-1} - y'_{i-1}\end{equation}
Local extremum
The extremum of this parabola is located at x_m:
f'(x_m) = 0 = a_1 + 2 a_2 (x_m - x_{i-1})
\begin{equation}x_m = x_{i-1} - \frac{a_1}{2a_2}\end{equation}
Monotonicity
The function f(x) is monotone on the [i-1, i] segment if it does not have an extremum on it, i.e. if
x_m < x_{i-1}\;\;\;\;\;\;\;\;\;\;\mathrm{or}\;\;\;\;\;\;\;\;\;\; x_m > x_i
The first condition readily translates into:
x_{i-1} - \frac{a_1}{2a_2} < x_{i-1}
\frac{a_1}{2a_2} > 0
or, when we substitute a_1 and a_2:
\frac{y'_{i-1} h_{i-1}^2}{2(y_{i} - y_{i-1} - y'_{i-1} h_{i-1})} > 0
\frac{y'_{i-1} h_{i-1}}{2(\delta_{i-1} - y'_{i-1})} > 0
As by definition h_{i-1} > 0
\frac{y'_{i-1}}{2(\delta_{i-1} - y'_{i-1})} > 0
If y'_{i-1} > 0 then \delta_{i-1} - y'_{i-1} > 0 and \delta_{i-1} > y'_{i-1}. If y'_{i-1} < 0, then \delta_{i-1} < y'_{i-1}. Therefore, the first condition (x_m < x_{i-1}) reduces to:
\begin{equation}\boxed{|\delta_{i-1}| > |y'_{i-1}|}\end{equation}
The second condition leads to:
x_{i-1} - \frac{a_1}{2a_2} > x_{i}
-x_{i} + x_{i-1} - \frac{a_1}{2a_2} > 0
h_{i-1} + \frac{a_1}{2a_2} < 0
h_{i-1} + \frac{y'_{i-1} h_{i-1}}{2(\delta_{i-1} - y'_{i-1})} < 0
\frac{2\delta_{i-1} - y'_{i-1}}{2(\delta_{i-1} - y'_{i-1})} < 0
There are two cases. First, if 2\delta_{i-1} - y'_{i-1} > 0, i.e. 2\delta_{i-1} > y'_{i-1}. then it must be \delta_{i-1} - y'_{i-1}< 0, i.e.\delta_{i-1} < y'_{i-1}. Therefore, the first case leads to 2\delta_{i-1} > y'_{i-1} > \delta_{i-1}. Secondly, if 2\delta_{i-1} - y'_{i-1} < 0, i.e. 2\delta_{i-1} < y'_{i-1}, it must be \delta_{i-1} - y'_{i-1}> 0, i.e. \delta_{i-1} > y'_{i-1}. So the second case leades to \delta_{i-1} > y'_{i-1}> 2\delta_{i-1}. The first case is possible only when both \delta_{i-1} and y'_{i-1} are positive and the second case is possible only when both are negative. Therefore, together, the two cases reduce the second condition (x_m > x_{i}) to:
\begin{equation}\boxed{|2\delta_{i-1}| > |y'_{i-1}| > |\delta_{i-1}|}\end{equation}
Together, the two conditions combine into
\begin{equation}\boxed{|2\delta_{i-1}| > |y'_{i-1}|}\end{equation}
A much faster way to reach the same result is to realize that the the derivatives y'_{i-1} and y'_{i} must be of the same sign if the function is monotone on the [i-1, i] segment, i.e.
y'_{i-1} \cdot (2\delta_{i-1} - y'_{i-1}) > 0
which obviously leads to |2\delta_{i-1}| > |y'_{i-1}|.
Therefore, for a Hermite parabola defined by y_{i-1}, y_i and y'_{i-1} to be monotone on the interval (i-1, i), the derivative y'_{i-1} must be of the same sign as the linear slope \delta_{i-1} and smaller in magnitude than 2\delta_{i-1}.
Convexity
For the function f to be convex on (i-1, i) (if parabola is convex on a segment, it's convex everywhere) the condition is f''(x) > 0. In the case of our parabola:
f''(x) = 2a_2 > 0.
\frac{y_{i} - y_{i-1} - y'_{i-1} h_{i-1}}{h_{i-1}^2} > 0
or
y_{i} - y_{i-1} - y'_{i-1} h_{i-1} > 0
\begin{equation}\delta_{i-1} > y'_{i-1}\end{equation}
This result is obvious \delta_{i-1} = y'_{i-1} is the straight line going through the end-points of the segment. For the parabola to be concave, it must hold f''(x) < 0, i.e.
\begin{equation}\delta_{i-1} < y'_{i-1}\end{equation}
