Vtyjkyuk

Problem:

Does $L^2([0,1])$ equipped with the inner product $$langle f,g rangle = int_0^1 overlinef(x)g(x) , dx$$ have an orthonormal basis of the form $leftf_n(x):=sumlimits_k=0^n a_k x^2k mid n ge 0right$?

Attempt:

I worked this out by hand and then checked with Mathematica. I think that no such basis exists. However, in my proof I showed that the orthonormal system need not even be complete.

Did I make a mistake somewhere?

Here is the Mathematica code:

u[n_, x_] := Sum[(a[k]+I*b[k])*x^(2*k), k,0,n]

f[m_, n_] := Integrate[ComplexExpand[Conjugate[u[m,x]]*u[n,x]], x,0,1]

Solve[f[0,0]==1, f[0,1]==0, f[1,0]==0, f[1,1]==1, a[0],b[0],a[1],b[1]]

In the code above I expanded each coefficient $a_k$ into its real and imaginary parts, and then solved the set of equations

$$langle f_0,f_0 rangle = 1, langle f_0,f_1 rangle = 0, langle f_1,f_0 rangle = 0, langle f_1,f_1 rangle = 1$$

but got an empty solution set.

The functions $1$, $x^2$, $x^4,ldots,x^2n,ldots$ are linearly independent, so the Gram-Schmidt process does yield an orthonormal sequence with
$f_n(x)=sum_i=0^n a_n,ix^2i$, and the $a_n.i$ real. Is this system complete?

The system is complete iff the polynomials in $x^2$ are dense in
$L^2[0,1]$. By the Stone-Weierstrass theorem, continuous functions on $[0,1]$
can be uniformly approximated by polynomials in $x^2$, and so also approximated
in $L^2[0,1]$ by polynomials in $x^2$. As the continuous functions
are dense in $L^2[0,1]$ then so are the polynomials in $x^2$. The orthonormal
sequence is complete.