Complete sequence in $L^2[0,1]$

Question

The Weierstrass Approximation Theorem implies that $ \left\{x^{k}\right\}_{k \geq 0} $ is complete in $L^2[0,1]$, I'm wondering if $ \left\{x^{2 k}\right\}_{k \geq 0} $ still complete. Is there a proof that shows that this sequence is also complete?

Answer 1

Any continuous function on $[0,1]$ can be approximated uniformly (hence also in $L^{2}$ norm) by polynomials in $x^{2}$ by Stone - Weierstrass Theorem. Hence the given family is complete.

Answer 2

Choose $f$ and $\epsilon>0$ and let $g(x) = f(\sqrt{x})$. Find a polynomial $p$ such that $\|g-p\|_\infty < \epsilon$. Then $\sup_{x\in [0,1]} |f(x)-p(x^2)| < \epsilon$.

Hence the span of the functions $x \mapsto x^{2k}$ is dense in $C[0,1]$ and hence in $L^2[0,1]$.