Let $f \in \mathcal{C}^4(\mathbb{R})$ be a real valued function, let $t_0< t_1< t_2$ be real numbers and $p$ is a polynomial of degree $≤3$, which satisfies $p(t_j)=f(t_j) , \ j= 0,1,2$ and $p'(t_1) =f'(t_1)$.
a) Prove that for $t \in [t_0,t_2]$, $$|f(t)−p(t)| \leq \frac{1}{24} \underset{\xi \in [t_0,t_2]}{\max}|f^{(4)}(\xi)||(t−t_0)(t−t_1)^2(t−t_2)|$$
b) Furthermore, is there a function $f \in \mathcal{C}^4(\mathbb{R})$, $f \not\in \mathcal{P_3}$ and a corresponding polynomial $p \in \mathcal{P_3}$ and a $t \in (t_0,t_2), \ t \neq t_1$, for which you have equality in the above inequality?
Here is my attempt at proving a).
We wish to prove that the error of the interpolation polynomial satisfies: \begin{align} |f(t) - p(t)| \; \leq \frac{1}{24} \; \ \underset{t \in [t_0,t_2]}{\max} \ |f^{(4)} (\xi)| \ |(t-t_0)(t-t_1)^2(t-t_2)| \end{align}
We fix a value $x \in [t_0,t_2]$ and introduce the following auxiliary function \begin{align} F(t) = e(t) - \underbrace{\frac{f(x)-p(x)}{(x-t_0)(x-t_1)^2(x-t_2)}}_{c} \underbrace{(t-t_0)(t-t_1)^2(t-t_2)}_{w(t)} \end{align} where $e(t) = f(x) - p(x)$. We note that $c$ does not depend on $t$, but on a fixed value of $x$ which makes $c$ a constant.
At the points $t_i, \ i=0,1,2$, we have $F(t_i) = e(t_i) - cw(t_i) = 0-c \cdot 0 = 0$. We have a double zero from the $(t-t_1)^2$ term in $c$. Furthermore, $F$ has a zero at the fixed value $x \in [t_0,t_2]$ since \begin{align*} F(x) &= e(x) -\frac{f(x)-p(x)}{(x-t_0)(x-t_1)^2(x-t_2)} (x-t_0)(x-t_1)^2(x-t_2)\\ &= e(x) - (f(x) - p(x))\\ &= 0 \end{align*} Thus, $F(t)$ has five zeroes in $[t_0,t_2]$.
By Rolle's theorem, $F'(t)$ will have four zeroes: one in the interval $[t_0,t_1]$ and one in the interval $[t_1,t_2]$, and yet another one since $t_1$ is a double root. Furthermore, we have the extra condition that $f'(t_1) = p'(t_1)$, we have that $f'(t_1) - p'(t_1) = e'(t_1) = 0$. Applying Rolle's theorem again to $F''(t)$, we get three zeroes, $F^{(3)}(t)$ has two zeroes and finally we see that $F^{(4)}(t)$ has one zero, which we denote $xi$, i.e. $F^{(4)}(\xi) = 0$.
As we are differentiating, we pay close attention to $w(t)$, which is a polynomial of degree 4. Differentiating, we get \begin{align*} w'(t) &= 4t^3 + \dots\\ w''(t) &= 12t^2 + \dots\\ w^{(3)}(t) &= 24t + \dots\\ w^{(4)}(t) &= 24 \\ \end{align*}
So we have $$F^{(4)}(\xi) = e^{(4)}(\xi) - 24c$$ and so $$ c \ = \ \frac{e^{(4)}(\xi)}{24} \; \left( = \frac{f(x)-p(x)}{(x-t_0)(x-t_1)^2(x-t_2)} \right) $$
Thus \begin{align*} F^{(4)}(\xi) &= \frac{e^{(4)}(\xi)}{24} = \frac{f(x)-p(x)}{(x-t_0)(x-t_1)^2(x-t_2)}\\ \iff f(x)-p(x) &= \frac{e^{(4)}(\xi)}{24} \cdot (x-t_0)(x-t_1)^2(x-t_2) \\ & \leq \frac{1}{24} \underset{\xi \in [t_0,t_2]}{\max}|f^{(4)}(\xi)||(t-t_0)(t−t_1)^2(t−t_2)|\\ \end{align*}
I'm not entirely convinced I've done this correctly, so I'm happy to accept some hints. Also, I have no idea how to approach b), so please share any tips!