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Suppose I have $L^2(\Omega)$ which has two inner products that are both norm-equivalent.

The eigenfunctions of the Laplacian $\Delta$ we know forms an orthonormal basis of $L^2(\Omega)$ -- with respect to which inner product? Both of them? How is that? Is it because they're both norm equivalent?

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When one speaks about the inner product in $L^2(\Omega)$ one always means $\langle f, g \rangle = \int_\Omega f(x)g(x) \, dx$. The eigenfunctions of the Laplacian are orthogonal with respect to this inner product. If there is a second equivalent inner product, then the eigenfunctions will still be orthogonal. However whether they are orthonormal depends on how they are scaled and this may be different even if the inner products are equivalent. – user38355 Jul 11 '13 at 12:36
How do you show that they're still orthogonal? (I wonder if you mean norm-equivalent when you write equivalent inner product) – aere Jul 11 '13 at 12:40
If you have a second inner product, you'll have a second Laplacian. – Neal Jul 11 '13 at 12:58
@Neal Please can you elaborate. – aere Jul 11 '13 at 13:16
@aere Right. I didn't read the question carefully enough and thought you meant that the inner products satisfy an inequality like $c(u,v)_1 \le (u,v)_2 \le C(u,v)_1$. – user38355 Jul 11 '13 at 16:21

In general, when one talks about orthogonality of eigenfunctions of $\Delta$, the inner product $$(u,v)_2=\int uv,\ \forall\ u,v\in L^2 $$ is being assumed.

We can have another inner product with equivalently norm, but in this new inner product the eigenfunctions are not orthogonal, indeed, let $p\in L^\infty$ satisfying $$0<a\leq p$$

Define $$(u,v)_p=\int puv,\ \forall\ u,v\in L^2$$

You can check that $(\cdot,\cdot)_p$ is a inner product in $L^2$ and $|u|=(u,u)_p^{1/2}$ is equivalently to $(u,u)_2^{1/2}$.

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Hmm. Consider a coercive elliptic operator $A$ and a Hilbert space $H$. Then I'm sure I Read somewhere that the eigenfunctions of $A$ form o.n basis with respect to $H$. Here everything is abstract, so I am wondering why the usual $L^2$ inner product is so special. – aere Jul 11 '13 at 13:18
As I have said in the beginning of my answer, "In general" the inner product used is $(u,v)_2$. This does not mean that we can not use another inner product, in fact we can use it and if it satisfies the hypothesis of the Spectral Theorem, then you get another basis, however, as I have showed, such basis does not need to be orthogonal in every inner product you have. – Tomás Jul 11 '13 at 13:22
@aere Actually more special functions are constructed so that they are orthogonal under a weighted $L^2$-inner product. – Shuhao Cao Jul 12 '13 at 22:19
@aere: I believe that $A$ is usually assumed to be formally self-adjoint. Then the inner product is implicit. – Sam Jul 12 '13 at 22:23

Here are some additional examples illustrate Tomás' answer that orthogonality doesn't translate among weighted $L^2$-inner product.

A $W$-weighted $L^2$-inner product is: $$ (u,v)_W := \int_{\Omega} W uv. $$ Notice that the equivalence of the norm is: $$ c \|u\|_{L^2(\Omega)} \leq \|u\|_{W} \leq C\|u\|_{L^2(\Omega)}, $$ by definition this is just for $u$: $$ c\int_{\Omega} u^2 \leq \int_{\Omega} W u^2 \leq C\int_{\Omega} u^2, $$ not for different $u$ and $v$: $$ c\int_{\Omega} uv \stackrel{?}{\leq } \int_{\Omega} W uv\stackrel{?}{\leq } C\int_{\Omega} uv, $$ Hence the norm equivalence tells you nothing about orthogonality.

Example: We can construct the polynomial to our interest by Gram-Schmidt procedure to the polynomial basis set of $L^2(\Omega)$, under $W$-weighted $L^2$-inner product. Then a normalization will give you the orthonormal basis set for $L^2(\Omega)$.

Different weight produces different set of polynomials, and they are only orthogonal under their own weighted inner product that they have been constructed from.

If we take $\Omega = (-1,1)$, famous examples are:

  1. $W = 1 \longrightarrow$ Legendre polynomials.
  2. $W = \sqrt{1-x^2} \longrightarrow$ Chebyshev polynomials.
  3. $W = (1+x)^{\alpha}(1-x)^{\beta}\longrightarrow$ Jacobi polynomials.

Moreover, if the weight $W$ is bounded above and below, then the weighted $L^2$- and unweighted $L^2$-norm are equivalent.

Last remark: Also OP's question reminds me of the eigenfunctions of the Hamiltonian in Schrödinger's equation, for different potential terms, you can have Bessel functions, spherical harmonics, Airy functions, Laguerre functions, ... and the list gones on and on. Different eigenfunctions are orthogonal under different weight.

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