Bounds for the coefficients for a family of polynomials Let $I=[a,b], a<b$ be an interval and $c>0$. Let $\mathcal P_m$ be the family of all non-negative-valued polynomials (coefficients can still be negative) of degree bounded by the integer $m$. 
Can we find a positive integer $M$ that only depends on $I,m,c$ such that any coefficent of any polynomial $p$ in $\mathcal P_m$ with $p(I)\subset [0,c]$ is bounded by M? (Namely $M$ is the uniform bound for the coeffients of polynomials in this family). 
 A: The answer is yes but I am not sure the method below is optimal or practical in finding $M$ and it doesn't depend on positivity, only on the property that $|P(x)| \le c$ 
(note that translating by a constant you actually can move between the two cases so I don't think that is essential anyway).
The proof is by induction (the interval will be fixed so we drop it from indexing though it will be clear from the proof that the constant depends also on it)- so assume we have $n \ge 1, \mathcal P_{n-1, c}$ the family as in the OP (degree at most $n-1$ and $c>0$ arbitrary) but asking only $|P(x)| \le c, P \in \mathcal P_{n-1, c}, x \in I$ and we can find $M(n-1, c)$ s.t.
$|a| \le  M(n-1, c)$ for any $a$ coefficient of some polynomial in $\mathcal P_{n-1, c}$ and we need to prove the same for $\mathcal P_{n, c}$
(we will find a real positive $M_{n,c}$ bound as if we want an integer we can take $[M]+1$)
In degree $0$ the problem is trivial with $M(0, c)=c$ so we have our base case. 
The fundamental result we use is that if $T_n(x)=\cos n\theta, x=\cos \theta, -1 \le x \le 1, 0 \le \theta \le \pi$ is the Chebyshev polynomial of the first kind and degree $n$ and $P$ is a monic polynomial of degree $n$ on some interval $I=[a,b]$, we have $\max_I |P(x)| \ge 2(\frac{b-a}{4})^n$ with equality iff $P(x)=2(\frac{b-a}{4})^nT_n(\frac{2x-a-b}{b-a})$
This is a classic result and it follows from the equioscillation of the Chebyshev polynomial between its $n$ zeroes (in other words it goes monotonically from $1$ to $-1$ and back like the cosine but now in $n$ equally spaced and explicit intervals in $\theta$, while its leading coefficient is $2^{n-1}$ so if $|P| < 2^{1-n}$ on $[-1,1]$, $P$ monic of degree $n$, $2^{1-n}T_n-P$ would have $n$ zeroes in $[-1,1]$ but degree $n-1$ so contradiction and then we move to $[a,b]$ by translation...)
But now if $d_n=2(\frac{b-a}{4})^n$ we get that if $P$ has degree precisely $n$ and leading coefficient $a_n \ne 0$, while $|P| \le c$ on $I$, we must have $|a_n| \le \frac {c}{d_n}$. But now $a_nx^n$ sends $I$ into an interval $J_{a_n}$ which is uniformly bounded in $|a_n| \le \frac{d_n}{c}$ so there is $p_{n,c} < q_{n,c}$ s.t $a_nx^n(I) \subset [p_{n,c},q_{n,c}]$ for all $a_n$ that can appear as leading coefficients of a degree $n$ polynomial in $\mathcal P_{n, c}$, which means that $|P(x)-a_nx^n| \le c+|p_{n,c}|+|q_{n,c}|=c_1$ for all $P \in \mathcal P_{n, c}, x \in I$ where $a_n$ was the leading coefficient of $P$ 
Hence $P(x)-a_nx^n \in \mathcal P_{n-1, c_1}$ for all $P \in \mathcal P_{n, c}$ of degree $n$ and since $c<c_1$ we also have that polynomials in $\mathcal P_{n, c}$ that have degree $n-1$ or less are in $\mathcal P_{n-1, c_1}$. So we can apply induction and find $M_{n-1,c_1}$ and hence taking $M_{n,c}= \max (M_{n-1,c_1}, \frac{d_n}{c})$ we get a coefficient bound as required for $\mathcal P_{n, c}$!
Note that $p_{n,c},q_{n,c}$ are computable directly in terms of $a,b,n,\frac{d_n}{c}$ and we can actually take $c_1=c+ \max (|p_{n,c}|,|q_{n,c}|)$ so get a computable $M$ if we have a computable $M_1$ in degree $n-1$ and as in degree zero we start at a nice $M=c$ we probably can doably backtrack for nice values of $a,b$, but in general the computations may get ugly fast especially if we want to optimize $M$ and $ab <0$
