$\mathbb{R}^{\mathbb{N}}$ Hilbert space with uncountable orthonormal basis? Let's say we have a Hilbert space of sequences $u_n\in\mathbb{R}^{\mathbb{N}}$, equipped with some inner product $(\cdot,\cdot)$. Let's say there is a Hermitian operator $M$ acting on this space, with a continuous spectrum of eigenvalues $\lambda\in\mathbb{R}$. Because $M$ is Hermitian, its eigenvectors are automatically orthogonal to each other $(a^{\lambda},a^{\lambda'})=0$ for $\lambda\neq\lambda'$. The eigenvectors are assumed to be in the Hilbert space (i.e. have finite norm).
Owing to the existence of this Hermitian operator $M$, this Hilbert space has an uncountable orthonormal basis. I don't see any obstacles for such a scenario to occur, but it seems counter-intuitive for a sequence $u_n$ to have finite norm while also being orthogonal to an uncountable number of sequences.
Does this scenario make sense? Or is there something about $M$ and the Hilbert space which makes it impossible?
Edit: The reason I ask this question is that I seem to have come across a scenario in which this does in fact occur. The Hilbert space in question has the unusual inner product
$$(u,v)=\sum_{n=0}^{\infty}\frac{(2n+1)!!}{(2n)!!}u_nv_n.$$
And the Hermitian operator is
$$M_{nm}=(4n+3)\delta_{nm}+2n\delta_{n-1m}+(2n+3)\delta_{n+1m}.$$
You can check that it is Hermitian with respect to this inner product by noting that $\frac{(2n+1)!!}{(2n)!!}M_{nm}$ is symmetric. This operator can be written in the form $M=A+A^{\dagger}+3$, where
$$A_{nm}=2n\delta_{nm}+(2n+3)\delta_{n+1m}.$$
These operators obey the commutator
$$[A,A^{\dagger}]=2M.$$
This means in particular that
$$e^{cA}Me^{-cA}=e^{2c}M$$
Which implies that if $v$ is an eigenvector of $M$ with eigenvalue $\lambda$, then $e^{-cA}v$ is an eigenvector with eigenvalue $e^{2c}\lambda$, for any constant $c$. Thus providing a continuous spectrum of eigenvalues.
To find an eigenvector $v$, we may solve the recursive relation associated with the equation $Mv=\lambda v$, and find that there are normalizeable solutions for real $\lambda<0$, with asymptotic behavior
$$v_n\sim(-1)^n\frac{1}{\sqrt{n}}e^{-\sqrt{2n|\lambda|}},\quad n\to\infty$$
Hermiticity of $M$
Since this has been questioned a couple of times I will show that $M$ is Hermitian with respect to this inner product. As I mentioned it is equivalent to showing that $\frac{(2n+1)!!}{(2n)!!}M_{nm}$ is symmetric.
$$\frac{(2n+1)!!}{(2n)!!}M_{nm}=\frac{(2n+1)!!}{(2n)!!}(4n+3)\delta_{nm}+\frac{(2n+1)!!}{(2n-2)!!}\delta_{n-1m}+\frac{(2n+3)!!}{(2n)!!}\delta_{n+1m}$$
This is symmetric since
$$\frac{(2n+1)!!}{(2n-2)!!}\delta_{n-1m}=\frac{(2m+3)!!}{(2m)!!}\delta_{m+1n}.$$
Reviewing the definition of a Hermitian operator, it presumes it is a bounded operator, which we have found that this operator is not (assuming the rest of my calculations are correct). Perhaps the statement about orthogonality of eigenvectors is not true when the "Hermitian" operator is unbounded?
 A: An example of such a Hilbert space is $\ell^2=\{ x_n| \sum x_n^2<\infty\}$. This is a complete inner product space with inner product $\langle x_n,y_n\rangle=\sum x_ny_n$. We can automatically see that the sequences with a one in a single entry create an orthonormal basis for $\ell^2$, i.e. $B=(e^i_n)$ where $e_n^i=\delta_{ni}$. This implies that there cannot be an uncountable basis for such a vector space and hence such a hermitian operator cannot exist. I don't believe we can construct another sequence Hilbert space since the norm must be compatible with the inner product.
A: Assuming you've defined your Hilbert space to be the space of sequences with finite norm with respect to your inner product, by rescaling appropriately, the Hilbert space of sequences you describe is isomorphic to $\ell^2(\mathbb{N})$, which is separable (admits a countable orthonormal basis), and hence which has the property that any set of orthonormal vectors in it must be at most countable. It follows that a bounded self-adjoint operator on $\ell^2(\mathbb{N})$ can have at most countably many distinct eigenvalues.
So your argument must break down somewhere, although I'm not sure where. Your calculations imply that the spectrum of $M$ is unbounded, so either those calculations are wrong or $M$ is unbounded. But I don't think even an unbounded self-adjoint operator on $\ell^2(\mathbb{N})$ can have uncountably many distinct eigenvalues. (I'm not even sure what an unbounded self-adjoint operator is, really.) So maybe $M$ is not self-adjoint, but I haven't checked this.
