slowness of growth of polynomials How to show that for any $d\in\mathbb{N}$ and $\delta>0$ there exists $\varepsilon=\varepsilon(d,\delta)>0$ such that is $f$ is a polynomial of degree $d$, and
$$|f(t)|\leq 1,\quad t\in[0,1],$$ then $$|f(t)|<1+\delta,\quad t\in[0,1+\varepsilon].$$
 (This is an exercise from "Introduction to Ratner Theorems..." by D.W. Morris). It is immediate for $d=0,1,2$, but I am probably lacking some trick to approach the general case. Any help is greatly appreciated.
 A: Claim: For every $d\in\mathbb{N}$, there exists $M_d>0$, such that for $f(x)=\sum_{i=0}^da_ix^i$, if $|f(t)|\le 1$ when $t\in[0,1]$, then $|a_i|\le M_d$, $i=0,\dots d$.

Proof of Claim:
Given $f(x)=\sum_{i=0}^da_ix^i$ and given $b_0,\dots,b_d\in[-1,1]$, consider the following system of linear equations with unknowns $a_0,\dots,a_d$:
$$f\big(\frac{i}{d}\big)=b_i,\quad i=0,\dots,d.\tag{1}$$
Note that the coefficient matrix of $(1)$ is a non-singular Vandermonde matrix whose entries are independent of $b_0,\dots,b_d\in[-1,1]$. It follows that $(1)$ has a unique solution $(a_0,a_1,\dots,a_d)$ and there exists  $M_d>0$, only dependent on $d$, such that $|a_i|\le M_d$, $i=0,\dots d$, which proves the claim. $\quad\square$

The statement follows from the claim immediately. If $f(x)=\sum_{i=0}^da_ix^i$ satisfies that $|f(t)|\le 1$ when $t\in[0,1]$, then $|a_i|\le M_d$, $i=0,\dots d$, and for every $\delta>0$, there exists $\varepsilon=\min(1,\frac{\delta}{2^dM_d})$, such that when $t\in[1,1+\epsilon]$, 
$$|f(t)|\le |f(1)|+|f(t)-f(1)|\le 1+\sum_{i=0}^d|a_i|(t^i-1)\le M_d\varepsilon\sum_{i=1}^d 2^{i-1}\le 1+2^dM_d\varepsilon\le 1+\delta.$$

Edit: An alternative way(indirect but a little more general) to prove the claim is to use the fact that all norms on finite-dimensional vector spaces are equivalent. The claim follows from comparing the two norms of $\mathbb{R}^{d+1}$ below:
$$\|(a_0,\dots,a_d)\|:=\max_{0\le i\le d}|a_i|~,\quad \|(a_0,\dots,a_d)\|':=\max_{0\le t\le 1}|\sum_{i=0}^d a_it^i|.$$
A: Clearly $f $ is uniformly continuous on closed intervals, So given $ \delta >0 $ there exists $ \epsilon'(d,\delta)>0 $ such that if $ |y-t| \leq \epsilon' $ then for all such $y$ we have 
$$ |f(y)-f(t)|< \delta $$
Take $ \epsilon = \min\{ \epsilon',1\} $, then for all $ t \in [1,1+\epsilon] $ we have $ 0 \leq 1-\epsilon \leq t-\epsilon \leq 1 $ and thus $ t-\epsilon \in [0,1] $. So we can choose $y \in [t-\epsilon,1]\subseteq [0,1] $ which clearly satisfies $ |y-t|\leq \epsilon $. Hence we have for all $ t \in [1,1+\epsilon] $
$$ |f(t)|\leq |f(y)|+|f(y)-f(t)| < 1 + \delta $$
For $ t\in [0,1] $ we trivially have $ |f(t)| \leq 1 < 1 + \delta $. So $ |f(t)| < 1+ \delta $ for all $ t \in [0,1+\epsilon ] $
