The following set

$$ \left\{ \psi_n(x) \right\}_{n\in \mathbb{Z}} = \left\{ \frac{1}{\sqrt{T}}e^{i2\pi \frac{n}{T} x} \right\}_{n \in \mathbb{Z}} $$

Is a complete set in $\mathcal{L}^2(\left[ -T/2 + x_0,T/2 + x_0 \right])$ for all $x_0$, and it is a generalization of the fourier basis in $[-\pi,\pi]$, obtained by raparametrizaton. The set of spherical harmonics

$$ \left\{Y_{l,m}(\theta,\phi)\right\}_{l,m} = \left\{f_{lm}P_l^m(\cos\theta)e^{im\phi} \right\}_{l,m} $$

is also complete in $\mathcal{L}^2(\left[0,2\pi \right]\times\left[0,\pi\right])$ as far as I know. I wonder if there's any reparametrization of that would make the same set complete in different region, just like the one above.

Is it possible?

  • 1
    $\begingroup$ A linear transform of the variables $(\theta, \phi)$ would make $\{Y_{l,m}\}$ complete on any compact, rectangular region of $\mathbb{R} \times \mathbb{R}$. Are you wanting something more complicated that this? $\endgroup$ Feb 22 '18 at 16:44
  • $\begingroup$ No, linear transform I think it's fine, could you write a sketch of the proof? So I can work out the details by myself. I assume though any bijection could work fine for completeness. $\endgroup$ Feb 22 '18 at 16:46

Suppose you want $\{Y_{l,m}(\theta,\phi)\}_{l,m}$ to be complete in $\mathcal{L}^2\left([a,b]\times[c,d]\right)$ for $[a,b]\times[c,d] \subset \mathbb{R}^2$, instead of just on $[0,2\pi] \times[0,\pi]$. Let $x,y:\mathbb{R} \rightarrow \mathbb{R}$ be linear functions so that $x(a) = 0$, $x(b) = 2\pi$, $y(c) = 0$, and $y(d) = \pi$. Then the family of functions $\widetilde{Y}_{l,m}(u,v) = Y_{l,m}(x(u), y(v))$ is the family you want.

One should say something about square integrability here. I'll confine myself to the observation that the Jacobian of the transform between $u,v$ coordinates and $\theta, \phi$ coordinates is constant. To argue completeness, observe that any function in $u,v$ coordinates has an image along $x,y$ in $\theta, \phi$ coordinates.

  • $\begingroup$ We are considering as dot product in this case $$\int_\Omega f(\theta,\phi)g^*(\theta,\phi) d\Omega$$, hence it is strictly dependent from the measure $d\Omega$ (solid angle). By a changing the parameterization shouldn't the measure change as well? I don't think orthonormality would be preserved otherwise. $\endgroup$ Feb 23 '18 at 15:14
  • $\begingroup$ The measure changes, but we can easily determine the change since the Jacobian is constant. So we rescale all $Y$ by $1/\sqrt{J}$, where $J$ is the Jacobian, and we recover orthonormality. $\endgroup$ Feb 23 '18 at 15:31
  • $\begingroup$ But we would preserve orthogonality, right? $\endgroup$ Feb 23 '18 at 16:58
  • $\begingroup$ @user8469759 : Yes. (That's the ...normality.) Also, when you multiplicatively scale zero, you still get zero. $\endgroup$ Feb 24 '18 at 0:50

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