Suppose first $ f(0)=0 $,
Extend f to [-1,1] by defining $ f(x)=-f(-x) $, then $ f\in C[-1,1] $. By Weirstrass, there exists a sequence of polynomials $ {q_n} $ which converge uniformly to f on[-1,1]. ie given $ \epsilon>0 $, there exists N such that
$ |q_n-f|<\epsilon $ for $ n\geq N $, $ x\in [-1,1] $
Let $ p_n(x)=\frac{q_n(x)-q_n(-x)}{2} $ Then $ p_n $ are odd polynomials and for $ x\in [0,1] $and $ n\geq N $
$ |p_n(x)-f(x)|=|\frac{q_n(x)-q_n(-x)}{2}-f(x)|=\frac{1}{2}|q_n(x)-f(x)-q_n(-x)+f(-x)| \leq \frac{1}{2}({|q_n(x)-f(x)|+|q_n(-x)-f(-x)|}) \leq\frac{1}{2}{(\epsilon+\epsilon)}=\epsilon $
which shows that $ p_n $converge to f uniformlyon [0,1]
Conversely, suppose that such odd polynomials $ {p_n} $ exist, we claim that $ f(0)=0 $
suppose not, let $ f(0)=M>0 $ ( if M<0 consider -f),then for large enough N
$ |p_n(0)-f(0)|<\frac{M}{2} $, ie $ |p_n(0)-M|<\frac{M}{2} $ which means $ p_n(0)>\frac{M}{2} $
But this means the constant term of such polynomials is non-zero which contradicts the fact that they are odd.
