Show there exists a sequence of polynomials which converge uniformly to a continuous $f$.

Question

Let $f : [0,1] \to \mathbb{R}$ be a continuous function that vanishes at $x = 1.$ Show that there exists a sequence of polynomials, each vanishing at $x = 1$, which converges to $f$ uniformly on $[0,1].$

It feels like a Stone-Weierstrass question, but after looking over the Stone-Weierstrass proof several times I am not sure if it truly applies.

My second thought is Arzela-Ascoli - namely that the family of functions you consider are the collection of polynomials, call this collection $P$, that vanish at $x = 1.$ My problem then becomes that this family is not uniformly bound over the interval, and I do not know if we can assert that a sequence of these polynomials converges to a specific $f,$ I think Arzela-Ascoli only proves that a uniformly convergent sequence in $P$ exists.

Anyone have any insight? Thanks in advance.

Answer 1

Another trick may be the following one: take a sequence of polynomials $q_1(x),q_2(x),\ldots$ that uniformly approximate $g(x)=\frac{f(x)}{1-x}$ over $(0,1)$ then consider $p_n(x)=(1-x)q_n(x)$. It is very easy to show that the sequence $p_1(x),p_2(x),\ldots$ meets the given constraints.

Answer 2

Take the Stone Weierstrass polynomials $\{p_n\}$, and modify them. Let $p_n(1) = a_n$ and $q_n(x) = p_n(x) - a_n$. Since $a_n = p_n(1) \rightarrow f(1) = 0$, $\{q_n\}$ are convergent to $f$

Answer 3

As already noted, if $p_n$ is any sequence of polynomials converging uniformly to $f$ then $q_n=p_n-p_n(1)$ works. Another way to look at this is the version of S-W for $C_0(X)$, where $X$ is locally compact Hausdorff; you have $f\in C_0([0,1))$, and that version of S-W says that the polynomials vanishing at $1$ are dense in $C_0([0,1))$.