Orthogonal Complements of polynomials in $L^2[0,1]$ I have two very difficult questions in my home work in function analysis, in which I have two calculate the complements of the following sets, in $L^2[0,1]$:


*

*All polynomials in the variable $x^2$

*All polynomials with coefficients summing up to $0$.


My thoughts so far:


*

*I thought that it would be easier to use $V^\perp=\overline V^\perp $.

*Since the polynomials with variable $x$ are dense in $C[0,1]$, I thought that given $f\in C[0,1]$, I'd take $f\circ \sqrt x$ (which is only in $C[0,1]$), and take a series $f_n$ of polynomials in $x$ that converges in $L^\infty$ to $f\circ \sqrt x$, and therefore $f_n\circ x^2$ is  polynomials in $x^2$ that should converge (am I right? why?) to $f\circ \sqrt x \circ x^2=f$. Since it converges in $L^\infty$, it converges to the same limit in $L^2$, and we have that the closure includes $C[0,1]$, and since $C[0,1]$ is dense we get that $V^\perp=\overline V^\perp={L^2[0,1]}^\perp=\{0\}$. Am I right?
And I have absolutely no idea about 2.
Appreciate your help..
 A: *

*In the first case, let $M$ be the linear subspace of $L^{2}[0,1]$ spanned by polynomials in $x^{2}$. The closure $\overline{M}$ of $M$ in $L^{2}[0,1]$ includes all of the $\cos(n\pi x)$ functions for $n=0,1,2,3,\cdots$. This set of functions is a complete orthonormal basis of $L^{2}[0,1]$. To see this, suppose $f \in L^{2}[0,1]$, and extend $f$ to an even function $f_{e}$ on $[-1,1]$. The functions $\{ \frac{1}{\sqrt{2}}e^{in\pi x}\}_{n=-\infty}^{\infty}$ form a complete orthonormal basis of $L^{2}[-1,1]$. So the following converges in $L^{2}[-1,1]$:
$$
\begin{align}
                 f_{e} & = \sum_{n=-\infty}^{\infty}\frac{1}{2}\int_{-1}^{1}f_{e}(t)e^{-in\pi t}\,dt\,e^{in\pi x} \\
          & = \sum_{n=-\infty}^{\infty}\frac{1}{2}\int_{0}^{1}f(t)(e^{in\pi t}+e^{-in\pi t})\,dt\;e^{in\pi x} \\
       & = \sum_{n=-\infty}^{\infty}\int_{0}^{1}f(t)\cos(n\pi t)\,dt\,e^{inx} \\
     & = \int_{0}^{1}f(t)\,dt+\sum_{n=1}^{\infty}2\int_{0}^{1}f(t)\cos(n\pi t)\,dt\,\cos(n\pi x).
\end{align}
$$
The above converges in $L^{2}[-1,1]$ to $f_{e}$. So this series also converges in $L^{2}[0,1]$ to $f$. Therefore $\overline{M}=L^{2}[0,1]$ and $M^{\perp}=\{0\}$.

*Let $M$ be the subspace of polynomials whose coefficients sum to $0$. Then
$$
        x^{k}-x^{n} \in M,\;\;\; n =1,2,3,\cdot
$$
Furthermore, $L^{2}-\lim_{n} (x^{k}-x^{n})=x^{k}$. Therefore, all polynomials are in the closure $\overline{M}$. The polynomials are dense in $L^{2}[0,1]$. So $M^{\perp}=\{0\}$.
