I'm trying to determine whether the following is true:
Let $$f:[0,1] \xrightarrow[]{}\mathbb{R}$$ be continuous and $0\leq x_1<…<x_m\leq1$. Do there exist polynomials, $p_n$ such that $p_n(x_j)=f(x_j)$ for each $j=1,…,m$ and $p_n$ converges to $f$ on $[0,1]$.
So on the one hand we have the Weierstrass approximation theorem telling us that if we do not have our interpolating assumption then there is convergence. A thought I had was to consider $f(x)=|x-\frac{1}{2}|$ then we are looking for polynomials $\bar{p_n}$
such that $||\space(x-\frac{1}{2})\bar{p_n}-|x-\frac{1}{2}|\space||_\infty\xrightarrow[]{}0$ so in a sense we seek $\bar{p_n}$ converging uniformly to a discontinuous function which is not possible. But this is just an idea I have not managed to get to work.
So I would be interested to know if the statement is true and if my proposed counter example can be made to work.
Best Answer
Let $p_i(x)=\frac {\prod_{j \neq i} (x-x_j)} {\prod_{j \neq i} (x_i-x_j)}$ and $P_n(x)=q_n(x)+\sum\limits_{i=1}^{m}p_i(x)[f(x_i)-q_n(x_i)]$ where $(q_n)$ is any sequence of polynomials converging uniformly to $f$. This sequence $(P_n)$ meets your requirements.