I am reading the book, Computational Differential Equations by Erikson, Estep, Hansbo and Johnson. My problem setting is as follows -
Given the nodal points $\zeta_0, \zeta_1 \in [a, b]$, and the Lagrange polynomials, $$\lambda_0(x) = \frac{x - \zeta_1}{\zeta_0 - \zeta_1}, \lambda_1(x) = \frac{x - \zeta_0}{\zeta_1 - \zeta_0},$$ the Lagrange interpolant of a function $f(x)$ is $$\Pi f(x) = \lambda_0f(\zeta_0) + \lambda_1f(\zeta_1) \; \; (*)$$ The error is derived to be $$|\Pi f(x) -f(x)| \leq \frac{1}{2}|(x-\zeta_0)(x - \zeta_1)| \underset{\zeta \in [a, b]}{\mathrm{max}} |f''(\zeta)|$$ In contrast to the usual proof that invokes Rolle's theorem repeatedly to derive the error estimate, here the proof for $\zeta_0 < x < \zeta_1$ proceeds as follows:
Expand $f(\zeta_0), f(\zeta_1)$ around $x$ in Taylor's series: $$f(\zeta_0) = f(x) + f'(x) (\zeta_0 -x) + f''(c_0) (\zeta_0 -x)^2/2$$ $$f(\zeta_1) = f(x) + f'(x) (\zeta_1 -x) + f''(c_1) (\zeta_1 -x)^2/2$$
Plugging these expansions into (*), and using $$\lambda_0(x) + \lambda_1(x) = 1, \;\;(\zeta_0 -x)\lambda_0 + (\zeta_1 - x) \lambda_1 = 0 \;\; (**),$$
$$\Pi f(x) - f(x) = 1/2 [f''(c_0) (\zeta_0 -x)^2\lambda_0(x) + f''(c_1) (\zeta_1 -x)^2\lambda_1(x) ] \; \; (**)$$
Now taking absolute value and simplifying the inequalities, we use $|x - \zeta_0| = x - \zeta_0$ but $|x - \zeta_1| = \zeta_1 - x$ due to the position of x between $\zeta_0, \zeta_1$ to get $$|\Pi f(x) -f(x)| \leq \frac{1}{2}|(x-\zeta_0)(x - \zeta_1)| \underset{\zeta \in [a, b]}{\mathrm{max}} |f''(\zeta)|$$.
My question starts here: Now this is clear, but the authors then leave as exercise (labelled as "hard") the proof of the analogous result $$|\Pi f(x) -f(x)| \leq \frac{1}{6}|(x-\zeta_0)(x - \zeta_1)(x - \zeta_2)| \underset{\zeta \in [a, b]}{\mathrm{max}} |f'''(\zeta)|$$
for the case of quadratic Lagrange basis functions $\lambda_0(x), \lambda_1(x), \lambda_2(x)$ with nodal points $\zeta_0, \zeta_1, \zeta_2$.
If I assume that $\zeta_0 < x < \zeta_1 < \zeta_2$, and try expanding in Taylor's series around $x$, and extend the identities (**) to eliminate $f'(x), f''(x)$, I end up with
$$|\Pi f(x) -f(x)| \leq \frac{1}{6}|(x-\zeta_0)(x - \zeta_1)(x - \zeta_2)| \underset{\zeta \in [a, b]}{\mathrm{max}} |f'''(\zeta)| \; g(x) $$ where $$g(x) = \left[ \dfrac{(x-\zeta_0)^2}{(\zeta_1-\zeta_0)(\zeta_2 - \zeta_0)} + \dfrac{(x-\zeta_1)^2}{(\zeta_1-\zeta_0)(\zeta_2 - \zeta_1)} + \dfrac{(x-\zeta_2)^2}{(\zeta_2 -\zeta_0)(\zeta_2 - \zeta_1)} \right]$$
The last expression $g(x)$ is definitely greater than 1 for $\zeta_0 < x < \zeta_1$!
I think there is some clever trick that I probably need to introduce in choosing what to expand around what point, but I am not able to get it.
Could somebody suggest how to proceed and derive the estimate in this case ?