Well this question was asked few years ago, but the answer doesn't satisfy me. Here is a link:
related question: How to Change the Interval of Interpolation from [-1,1] to [a,b] for Chebyshev Nodes
I want to show that for given $n+1$ interpolation points $y_0,y_1,...,y_n$ in [a,b] , $E(y) = |f(y)-p_{n+1}(y)| \le \frac{(b-a)^{n+1}}{2^{(2n+1)} (n+1)!} \cdot max_{y\in[a,b]}|f^{n+1}(y)|$.
Well for $\bar{y}\in[a,b]$ $|f(\bar{y})-p_{n+1}(\bar{y})| = |p_{n+2}(\bar{y})-p_{n+1}(\bar{y})| = |f[y_0,y_1,...,y_n,\bar{y}]| \cdot |\prod_{i=0}^{i=n}(y-y_i)| $. $|f[y_0,y_1,...,y_n,\bar{y}]|$, (Usimg Rolle theorem), is bounded by $max_{y\in[a,b]}|f^{n+1}(y)|$.
So we are left with $|\prod_{i=0}^{i=n}(y-y_i)| $ , cause $y\in[a,b]$ in order to use known results for Chebyshev's interpolation in $[-1,1]$, we define $h(y) = \frac{2y+(b+a)}{b-a}=x\in[-1,1]$, when $h$ is an homeomorphism.
Now, denoting Chebishev polynomials with $T_(n) (h(y))$, we know that $T_{n+1}(x)=2T_n(x)-T_n-1(x)$ , and T_{n+1} is a polynomial of deg $n+1$ with $n$ different roots, which are in our case: $x_0=h(y_0),x_1=h(y_1),...,x_n=h(y_n)$. Thus $T_{n+1}(x) = c\cdot \prod_0^n(x-x_i) , c\in \mathbb{R}$. From the recursion formula, $T_{n+1}(x) = 2^n\cdot x^{n+1} + l.o.t(x^n)$.
From those two conclusions, we learn that $c=2^n$ , so $T_{n+1}(x) = 2^n \cdot \prod_0^n(x-x_i)$
So $\prod_0^n(x-x_i) = \frac{T_{n+1}(x)}{2^n}= \frac{2^n\cdot x^{n+1} + l.o.t(x^n)} {2^n} = x^{n+1} + l.o.t(x^n) =_{x=h(y)=\frac{2y +b+a}{b-a}} = \frac{2^{n+1}y^{n+1}}{(b-a)^{n+1}} + l.o.t$
Unfortunately I don't see how to continue from this point...
