Let's consider $\theta_n$ a class of approximations with the following properties:

- all functions $\phi \in \theta_n$ are defined on a symmetric interval [-a, a];

- if $\phi(t)\in\theta_n$, then $\phi(-t)\in\theta_n$;

Consider $d\lambda t=\omega(t) \; dt$, where $\omega$ is an even function.

Prove that if f is an even function on $[-a, a]$, its least squares approximation $\phi\in\theta_n$ is even.

The least squares approximation of a function $f$ is a function $\phi\in\theta_n$ such as:

$||f-\phi||\le||f-\phi_n||$, for every $\phi_n\in\theta_n$.

Also: $||u||=(\int_R|u(t)|^2d\lambda t)^{1/2}$

My attempt of a solution started by trying to prove that the approximation is a linear combination of even functions. So we consider an orthogonal basis $\lbrace\pi_j\rbrace_{j=1}^n$ for the given class of approximations. In this case, the least squares approximation is:

$\phi=\sum_{j=1}^nc_j*\pi_j$, where: $c_j=(\pi_j,f)/(\pi_j,\pi_j)$

(u,v) is the dot product of functions u and v. More precisely:

$(u,v)=\int_{-a}^a u(t)v(t)\omega(t) \; dt$

Using this formula and the hypothesis, it can be proven that if $\pi_j$ is an odd function, then $c_j=0$, because $(\pi_j,f)$ is the integral of an odd function on a symmetric interval.

This way, we eliminate from the approximation $\phi$ all the odd basis functions. So now $\phi$ is a linear combinations of even functions and functions that are neither even, nor odd. The even functions are good, because a linear combination of even function is also a even function. But what happens when the basis function $\pi_j$ is neither even, nor odd? I tried decomposing $\pi_j$ in combinations of even and odd functions, but this approach does not seem to lead anywhere...

Any help would be appreciated.

`$\theta_n$`

of functions, but the later explanation tells me that`$\theta_n$`

is supposed to be a vector space, since linear combinations are considered as candidates. In that case, this is not a research-level question, so it's not appropriate for MO. If, on the other hand, the least-squares solution is to be sought in a truly arbitrary class`$\theta_n$`

, then the solution need not be unique and there are easy examples where no solution is even. I'm voting to close. – Andreas Blass May 20 '11 at 14:43