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Suppose I have two sequences of univariate Bernstein polynomials: $$B_{n}(x)=\sum_{i=0}^n a_{n,i}b_{n,i}(x), \quad x \in [0,1],$$ and $$P_{n}(x)=\sum_{i=0}^n c_{n,i}b_{n,i}(x), \quad x \in [0,1],$$ where $b_{n,i}=\binom{n}{i}x^i (1-x)^{n-i}$.

Suppose it is known that sequences $a_n=(a_{n,0},a_{n,1}, \ldots, a_{n,n})$ and $c_n=(c_{n,0},c_{n,1}, \ldots, c_{n,n})$ are such that $B_{n}(x)$ and $P_{n}(x)$ converge uniformly on $[0,1]$ to the same function as $n \rightarrow \infty$.

I think we should be able to find some traditional norm $\|\cdot\|_{n+1}$ in $\mathbb{R}^{n+1}$ to conclude from here that
$$\|a_n-c_n\|_{n+1} \rightarrow 0 \text{ as } n \rightarrow \infty$$ but not sure how establish that exactly....

If it helps, I am happy to assume that the function in the limit is smooth (even infinitely differentiable).

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    @copper.hat Thanks for the comment. I edited the question a bit. I think finding that norm (the same norm in for each $n$) is a part of the question. – 2017-02-05
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    Of course such a sequence of norms exists; take $\|a_n\| := \|B_n\|_\infty$. The interesting question would then be: is there a "simpler" sequence of norms for which this works? – 2017-02-05
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    @D.Thomine Yes, I agree. The idea is to see whether one can make this conclusion for a more traditional/usual norm (like Manhattan, Euclidean, etc). – 2017-02-05
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    @D.Thomine Edited question to reflect your comment, thanks. – 2017-02-05

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I think that you can't get non-trivial bounds for any $\ell^p$ norm. Fix $p \in [1, \infty]$. Assume that there exists a sequence $(K_n)$ such that, for all $f \in \mathcal{C} ([0,1])$, for all $(a_{n,i})$ and $(c_{n,i})$ such that the associated Bernstein polynomials $A_n$ and $C_n$ converge uniformly to $f$,

$$\lim_{n \to + \infty} K_n \|a_{n,\cdot}-c_{n,\cdot}\|_{\ell^p} = 0.$$

Fix $f$ compactly supported in $(0,1)$. Choose $a_{n,i} := f(i/n)$. Then, let $a_{n,i}^{(0)} := a_{n,i}+a_{n,i+1}$ if $i$ is even, and $0$ otherwise. Similarly, define $a_{n,i}^{(1)} := a_{n,i}+a_{n,i-1}$ if $i$ is odd, and $0$ otherwise.

Since $b_{n,i+1}/b_{n,i}$ converges to $1$ uniformly in $i$ as long as $0 \ll i \ll n$, and since all the $a_{n,i}$ are $0$ for $i$ close to $0$ or $n$ (because $f$ is compactly supported), we get that $A_n^{(0)}$ and $A_n^{(1)}$ both converge uniformly to $f$ (that section would require more details, but I only foresee some tedious $\varepsilon-\delta$ arguments).

By hypothesis,

$$\lim_{n \to + \infty} K_n \|a_{n,i}^{(0)}-a_{n,i}^{(1)}\|_{\ell^p} = 0.$$

But, since $a_{n,i}^{(0)}$ and $a_{n,i}^{(1)}$ have disjoint support but take the same values, using the expression for the $\ell^p$ norm,

$$\|a_{n,i}^{(0)}-a_{n,i}^{(1)}\|_{\ell^p} = 2^{\frac{1}{p}} \|a_{n,i}^{(0)}\|_{\ell^p} \sim 2^{1+\frac{1}{p}} \|a_{n,i}\|_{\ell^p}.$$

Hence, $$\lim_{n \to + \infty} K_n \|a_{n,i}\|_{\ell^p} = 0$$. This sequence of norm is not strong enough to distinguish the uniform convergence of Bernstein polynomials.

The problem is that, as $n$ grow large, polynomials $b_{n,i}$ with close values of $i$ get very correlated (and somewhat interchangeable). This clashes with the expression for the $\ell^p$ norm, which doesn't distinguish between indices which are close and indices farther away.