Let $C[0,1]$ denote the set of all real-valued continuous functions on $[0,1]$. Consider the normed linear space
$$X=\{f\in C[0,1]:f(\dfrac{1}{2})=0\},$$ with the sup norm. Show that the set $P=\{f\in X:f$ is a polynomial$\}$ is dense in $X$.
My Thought.
For an $f$ in $X$, we know there exists a sequence of polynomials $(p_n)$ converging to $f$ in the sup norm. This implies pointwise-convergence, so $p_n(\dfrac{1}{2})\to f(\dfrac{1}{2})=0$. Hence, defining a sequence $(q_n)$ as $q_n(x)=p_n(x)-p_n(\dfrac{1}{2})$, we get a sequence of polynomials in $P$ converging uniformly to $f$, as required.