Consider f as a continuous function on [a,b]. How can I show that there exists a sequence {$P_n$} of polynomials such that $P_n \rightarrow f$ uniformlyon [a,b] and such that $P_n(a) = f(a)$ for all n ?
My effort on this question so far was to find Weierstrass's Approx. Thm which says that every continuous function on a closed interval[a,b] can be uniformly approximated by polynomials on [a,b].
The questions is then how to show $P_n(a) = f(a)$.
Thanks for your help.
Without loss of generality, we can assume that $a=0$ and $b=1$. If this is not already the case, then we can use the change of variable $u = \frac{x-a}{b-a}$. Then $u$ ranges over $[0,1]$ as $x$ ranges over the original $[a,b]$.
Now look at the proof of the Weierstrass's Approximation Theorem using Bernstein polynomials on $[0,1]$. The polynomials $P_n$ constructed in that proof are:
$$ P_n(u) = \sum_{k = 0}^{n} f(k/n) {n \choose k} u^k (1 - u)^{n - k} $$
Bernstein shows that these converge uniformly to the given function $f$. They all have the property that $P_n(0) = f(0)$. Actually, $P_n(1) = f(1)$ too, if that's of interest.