Approximation of continuous functions vanishing at a point with polynomials vanishing at the same point

Question

Let $C[0, 1]$ denote the set of all real-valued continuous functions on $[0, 1]$. Consider the normed linear space
$$ X=\{f\in C[0,1]| f(\frac{1}{2})=0\} $$ with the sup-norm $$||f|| = \sup\{|f(t)| : t ∈ [0, 1]\}.$$ Show that the set $$P=\{f\in X | f \text{ is a polynomial}\}$$ is dense in $X$.

I was trying to approach this problem using Weierstrass' approximation theorem but by doing so, for each $f\in X$, I can only find polynomials with roots in a neighborhood of $\frac{1}{2}$.

Answer 1

Hint: Let $p$ be a polynomial such that $|p(x)-f(x)| <\epsilon/2$ for all $x$. We can write $p(x)$ in the form $ \sum\limits_{k=0}^{N}a_k(x-\frac 1 2)^{k}$. Let $q(x)= \sum\limits_{k=1}^{N}a_k(x-\frac 1 2)^{k}$. $q$ is a polynomial vanishing at $\frac 1 2$. Verify that $|q(x)-f(x)| <\epsilon$ for all $x$.

Answer 2

Hint: Every polynomial with root in a neighborhood of $\frac12$ can be approximated by a polynomial with root at $\frac12$. Just shift it a little bit in $x$ or $y$-direction.