Find a sequence of functions ${p_n}$ which converges uniformly to $f$ on $[0,1]$ with $p_n(0)=p_n(1)=0.$

Question

Suppose $f$ is a continuous function on $[0,1]$ such that $f$ is $0$ at both the endpoints and I have to prove that there is a sequence of polynomials coverging uniformly to $f$ and having the property $p_n(0)=p_n(1)=0$.How to approach this problem from Weierstrass polynomial approximation?

Answer 1

There is a method to have $p_n(a)=f(a)$ and $p_n(b)=f(b)$ for every $n \in \Bbb{N}$ for a continuous function $f:[a,b] \to \Bbb{R}$

Let $f:[a,b] \to \Bbb{R}$.

Exist polynomials $q_n \to f$ uniformly on $[a,b]$.Take:

$$p_n(x)=q_n(x)+f(a)-q_n(a)+\frac{x-a}{b-a}(f(b)-q_n(b)-f(a)+q_n(a))$$

I leave it to you to verify that $p_n \to f$ uniformly on $[a,b]$