Fix $k\in\mathbb{N}$ and assume $f(x)$ is a real polynomial of degree $n$ such that we have the normalization
$$\int\_{-1}^1f(x)^2\,(1-x)^kdx=1.$$
I am interested in the optimal size of the sum of the coefficients of $f(x)$. To this end, I ask:



> 
> **Question 1:** It appears to me that the maximum value 
>  $$\max\_f\vert f(1)\vert$$
>  is of order $\binom{n+k+1}{k+1}$. Is this true?
> 
> 
> 


Further,



> 
> **Question 2:** Can we find a tight bound so that
>  $$\max\_{f}\vert f(1)\vert\leq c\_n\binom{n+k+1}{k+1}\,?$$
> 
> 
> 


Remark 1: Of course, Question 2 implies Question 1.


Remark 2: I would still be interested in discussions for specific $k$'s.


**Postscript.** MTyson made serious progress although this requires a bit of tweaking on the "loose upper bound". However, his argument suggest that $c\_n$ is actually $c\_k$ which is interesting.



Bound on sum of coefficients of polynomials w.r.t a weighted integral
Consider the $(k,0)$ Jacobi polynomials $P\_n$, which are orthogonal with respect to the weight $(1-x)^k$ on $[-1,1]$. They have squared norm $c\_m=\langle P\_m,P\_m\rangle=\frac{2^{k+1}}{2m+k+1}$ and $P\_m(1)={m+k\choose k}$.


Expand $f$ as the sum $\sum\_{m=0}^n a\_m P\_m$. The constraint is that $\sum\_{m=0}^n a\_m^2c\_m=1$, and we would like to maximize $\sum\_{m=0}^n a\_m{m+k\choose k}$.


We can apply the method of Lagrange multipliers, which amounts to solving
$$
{m+k\choose k}=\frac{\partial}{\partial a\_m}\sum\_{i=0}^n a\_i{i+k\choose k}=\frac{\partial}{\partial a\_m}\sum\_{i=0}^n a\_i^2c\_i=2\lambda c\_ma\_m
$$
subject to the quadratic constraint. That is, $a\_m=\frac{1}{2\lambda c\_m}{m+k\choose k}$ and
$$
1=\sum\_{m=0}^n a\_m^2c\_m=\sum\_{m=0}^n \frac{{m+k\choose k}^2}{4\lambda^2 c\_m},
$$
which gives
$$
\lambda=\frac{1}{2}\sqrt{\sum\_{m=0}^n \frac{{m+k\choose k}^2}{c\_m}}.
$$
Therefore the maximum of $|f(1)|$ is
$$
\sum\_{m=0}^n{m+k\choose k}a\_m=\sum\_{m=0}^n \frac{{m+k\choose k}^2}{2\lambda c\_m}=\sqrt{\sum\_{m=0}^n \frac{{m+k\choose k}^2}{c\_m}}=\sqrt{\sum\_{m=0}^n \frac{{m+k\choose k}^2(2m+k+1)}{2^{k+1}}}.
$$
The terms of the sum are monotonically increasing with $m$, so a loose upper bound is given by
$$
|f(1)|\le\sqrt{\frac{n{n+k\choose k}^2(2n+k+1)}{2^{k+1}}}\le\sqrt{\frac{{n+k\choose k}^2(n+k+1)^2}{2^k}}=\frac{k+1}{2^{k/2}}{n+k+1\choose k+1}.
$$


