# Does convergence of Bernstein polynomials imply convergence of its coefficients?

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.