Consider a partition of the domain $\Omega=[a,b]$ $$ a=x_{1}<x_{2}<\cdots < x_{N}=b $$ with mesh size $h=\max\{x_{i+1}-x_{i}:i=1,...,N-1\}$. Let V be an inner product space, with inner product given by \begin{align*} \langle u,v \rangle = \int_{\Omega} uv \end{align*}
Here's my attempt:
Let $P_{h}(f)$ be the orthogonal projection of $f$ over $V_{h}$ (The space of piecewise continuous functions). Since $P_{h}(f)$ is the orthogonal projection, then it satisfies \begin{align*} ||f-P_{h}(f)|| \leq || f-p(x)|| \ \ \forall p \in V_{h} \end{align*} In particular this inequality holds for the interpolation polynomial $I_{f}(x)$ of $f$, i.e \begin{align*} ||f-P_{h}(f)|| &\leq || f-I_{f}|| \\[2mm] & = \Bigg|\Bigg| \frac{f^{2}(\xi_{x})}{2}\pi_{2}(x) \Bigg|\Bigg| \\[2mm] & \leq \Bigg|\Bigg| \frac{M_{2}}{2}\pi_{2}(x) \Bigg|\Bigg| \end{align*} where $\pi_{2}(x)= \prod_{i=1}^{N}(x-x_{i})$ y $M_{2} = \max_{t\in \Omega} |f^{(2)}(t)|$.
But I'm stuck, I don't know how to continue. Any hints?
actually you're on last step. as you mentioned if $I_i$ is linear interpolation of $f$ on interval $[x_i,x_{i+1}]$, then we have : $$ |f-I_i| \leq \frac{f^{(2)}(\xi_i)}{8} (x_{i+1}-x_i)^2 \qquad \xi_i \in [x_i,x_{i+1}] $$ so for piecewise interpolation $I_f$ of $f$ on $[a,b]$ we have : $$ |f-I_f| \leq \frac{1}{8} \max_{i=0,\dots,N-1}\{f^{(2)}(\xi_i) .(x_{i+1}-x_i)^2\} \leq\frac{M}{8} h^2 $$ so $C = \frac{M}{8} = \frac{1}{8} \max_{t\in \Omega} |f^{(2)}(t)|$ works. (assuming $f\in C^2(\Omega)$)