Suppose $f$ is a real-valued differentiable function and $f′\in C([0,1])$ and $f(0) = 0$.
Prove that, $\forall\varepsilon\gt0$ , there is a polynomials $p$ of the form $p(x) = \sum_{n=1}^k a_{2n+1}x^{2n+1}$ s. t. $\|p-f\|+\|p'-f'\|\leqslant\varepsilon$, where $\|g\|=\sup_{x\in[0,1]}|g(x)|\forall g\in C([0,1])$.
I know that there is a continuous function $g$ on $[0, 1]$ s. t. $\|f − g\|\lt\varepsilon$, but I don't know where to go from here.
Note that polynomials in $x^{2}$ form dense set in $C[0,1]$ by Stone - Weierstrass Theorem. Choose a polynomial $q$ in $x^{2}$ such that $\|q-f'\|<\epsilon /2$ and take $p(x)=\int_0^{x} q(t)dt$ so that $\|p'-f'\|=\|q-f'\|<\epsilon /2$. Recall that $f(x)=\int_0^{x} f'(t)dt$. This gives $\|p-f\|<\epsilon /2$. Note that $p$ invovles only odd powers of $x$.