Is the set of polynomials of degree $n$, whose coefficients sum to $0$ dense over $C[0,1]$ or $C[-1, 0]$? 
Let $B = \Big{\{} \ \sum\limits_{i=0}^n a_ix^i; a_i \in R, \sum\limits_{i=0}^n a_i = 0 \Big{\}}$. Is $B$ dense over $C[0,1]$? Is it dense over $C[-1, 0]$?

Intuitively, I want to say that $B$ is dense over both. Given $f$, I can find a polynomial $p \in B$ where $p(0) = a_0 = -(a_1 + ... + a_n)$. This implies that $f(0) = a_0 = -(a_1 + ... + a_n)$. I can't personally see this fact restricting $f$ in any way and this argument is for both $C[0,1]$ and $C[-1,0]$. The issue is I don't know if the interval really makes a difference since I've only recently started proving density of polynomial sets and so I don't have a lot of examples to work from.
 A: I believe you have mis-stated the question, because for a given fixed $n$, the set of polynomials of some fixed degree $n$ is not dense on $C[0,1]$ (equipped as usual with the supremum norm) even without the restriction that the coefficients add to $0$.
The interesting question you may have meant to ask is whether the set of all polynomials of finite degree $n \in \Bbb{Z}+ \cup \{0\}$ whose coefficients add to zero is dense in
$C[0,1]$ (equipped as usual with the supremum norm).
Here, if you remove the restriction on the sum of coefficients, the set of all polynomials is dense, because for any continuous function $f$ (which must also be finite since the interval is closed) you can find a polynomial approximation having arbitrarily small maximal error by allowing high degree $n$.  So there will be a set of polynomials $f_n$ that converge uniformly to $f$.
Now apply the restriction that the sum of coefficients is zero.  Since $f_n(1)$ is precisely the sum of coefficients in $f_n$, the sequnce of approximations will not converge uniformly to any function that has value $f(1) \neq 0$.  So $B$ is not dense over $C[0,1]$.
The question for $C[-1,0]$ is more subtle, because nobody said anything about the alternating-signs sum of coefficients (which would be the value at $f_n(-1)$). The way to see that $B$ is dense over $C[-1,0]$ is to note that the Legendre polynomials $P_n$ have sums of coefficients equal to $1$, so if you expand $g(x) = f - \int_{-1}^0 f(x)dx$ in Legendre polynomials up to order $n$, and then add $\frac1n \int_{-1}^0 f(x)dx$ from each of those polynomials you have the desired sequence of polynomials.  This fails if $\int_{-1}^0 f(x)dx=0$, but in that case you can write $f_n$ as the $n$-th degree Legendre polynomial approximation to $h_n(x)$ which is the $g(x)$ you would obtain by approximating $f(x)  - \frac1n$ and still get a sequence that approaches $f(n)$ uniformly.  
So 
$B$ is dense over $C[-1,0]$.
