Bound on sum of coefficients of polynomials w.r.t a weighted integral

Question

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.

Answer 1

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 $$ |f(1)|\le\sqrt{\frac{n{n+k\choose k}^2(2n+k+1)}{2^{k+1}}}\le\sqrt{\frac{{n+k\choose k}^2(n+k+1)^2}{2^k}}=\frac{k+1}{2^{k/2}}{n+k+1\choose k+1}. $$

Answer 2

I noticed that we can really tighten the "loose upper bound" given by MTyson and hence obtain an optimal result answering Question 2. This rests on the identity $$\sum_{m=0}^n\binom{m+k}k^2(2m+k+1)=(k+1)\binom{n+k+1}{k+1}^2$$ which can be proved in different ways. Consequently, we have $$\max_f\vert f(1)\vert=\sqrt{\frac{k+1}{2^{k+1}}}\binom{n+k+1}{k+1}.$$ Observe that $c_k=\sqrt{\frac{k+1}{2^{k+1}}}$ is the sought-after factor.

In the end, I thank MTyson.