Possible existence of weight function $\rho (t)$

Question

Consider $L^2[-\pi,\pi]$. We define an inner product on this space by

$$\langle f,g\rangle=\int_{-\pi}^{\pi} f(t)\overline {g(t)} \, dt \quad\to(1)$$
Suppose if we introduce a weight function $\rho(t)$ with the property that $\rho (t)\ge 0 , \quad\forall t $ and that $\rho$ is continuous in $[-\pi,\pi]$, so that $(1)$ becomes $$\langle f,g\rangle=\int_{-\pi}^{\pi}f(t)\overline {g(t)}\rho(t) \, dt$$

then what are the most suitable $\rho(t)$'s which preserves the inner product structure. Some of the possibilities that I know are the Legendre, Laguere and Hermite polynomials and also $1$. I want to know whether there are other possibilities or not and the associated differential equations. Thanks

Answer 1

In order for $\left<\cdot,\cdot\right>:L^2[-a,a]\times L^2[-a,a] \to \mathbb{R}$ to be a valid inner product we need:

• $\rho$ have to be integrable (obviously).
• $\rho$ needs to be real in order to have the conjugate symmetry $\left<f,g\right> = \overline{\left<g,f\right>}$.
• Linearity of $\left<\cdot,\cdot\right>$ follows from linearity of the integral (for all $\rho$) so this gives no further restriction.

The final inner-product condition is positive-definiteness:

• The condition $\left<f,f\right> \geq 0$ requires $\rho \geq 0$.
• The condition $\left<f,f\right>=0\implies f = 0$ a.e. requires that $\rho$ cannot be zero on a whole interval (i.e. that $A = \{x : \rho(x) =0\}$ has measure zero). This can perhaps be relaxed (see BigM's comments below the question).

If the simple conditions above (which are not very restictive at all) are satisfied then $\left<\cdot,\cdot\right>$ is indeed a valid inner-product.