Assume a sequence of Bernstein polynomials $\{p_N(t)\}_{N=1}^\infty$ uniformly converges to a continuous function $f(t)$ on $[0,t_f]$. Is it true that the Bernstein coefficients of $p_N(t)$ (e.g., $p_{iN}$) converge to $f(it_f/N)$?

Essentially, this would be the converse of the Bernstein approximation theorem, which states that if $f(t)$ is continuous, then the Bernstein polynomial $$g_N(t) = \sum_{I=0}^N f\left(i \frac{t_f}{N}\right) b_{i,N}(t)$$ uniformly converges to $f(t)$.

This is what I have so far: Let $g_N(t)$ be a Bernstein polynomial approximation of $f(t)$, namely $$g_N(t) = \sum_{I=0}^N f\left(i \frac{t_f}{N}\right) b_{i,N}(t),$$ where $b_{i,N}(t)$ is the Bernstein basis. Then, $$\lim_{N \to \infty} g_N(t) = f(t) \, .$$ On the other hand, by assumption we have $$\lim_{N \to \infty} p_N(t) = f(t) \, .$$ Thus, combining the equations above we get $$\lim_{N \to \infty} (g_N(t) - p_N(t)) = 0\, , $$ which implies $$\lim_{N \to \infty} \sum_{i=0}^N \left(f\left(i \frac{t_f}{N}\right) - p_{iN}\right)b_{i,N}(t) = 0\, .$$ Finally, the result above implies $$\lim_{N \to \infty} \left(f\left(i \frac{t_f}{N}\right) - p_{iN}\right)=0 \, , \quad \forall i \in \{0,\ldots , N\} \, ,$$ which proves the result.

However, I am not quite sure about the last implication. It should be true given the positiveness and partition of unity property of the Bernstein basis, but I am unable to prove it.


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