PeterDonis said:
I still don't understand your notation. The argument ##x## of the wave function is a variable. The eigenvalue ##x_0## (or ##a##, or ##y##, or whatever symbol other than just ##x## you want to use) is a constant. The quantity ##\langle x \vert \psi(t) \rangle##, assuming you mean ##\langle x \vert## and ##\vert \psi(t) \rangle## to denote single vectors (i.e., we have chosen a particular eigenvector of position and a particular label ##t## that identifies a single Hilbert space vector), is a number, not a function; it's the inner product of two vectors. So ##\psi(t, x)##, as you've written it, is the value of the wave function for a particular choice of arguments, not the wave function itself.
In the latter part of your post, you write ##u_y(x)##, which looks like a function, but you equate it to ##\langle x \vert y \rangle##, which is again a number, an inner product of two vectors. Unless you mean ##\langle x \vert## to denote a set of vectors, the set of all eigenvectors of position; that would make ##u_y(x)## a function that maps eigenvectors of position to numbers (the number for each eigenvector being its inner product with the specific eigenvector ##\vert y \rangle##. But in that case, once again, the argument ##x## of ##u_y(x)## is a variable; it's not any single eigenvalue, it ranges over all possible eigenvalues.
I'm a bit confused about this debate, because I use standard Dirac notation all the way. I think, we have to really define everything clearly to make progress. I work in the Schrödinger picture of time evolution which is usually introduced first in the introductory QM lecture (which at least in my case at the TU Darmstadt in the mid-1990ies started with the representation-free Dirac approach right away, but we had also a more conventional treatment in terms of wave mechanics in the experimental-physics course lecture before). So let's start with the formalism (the "kinematical part only"):
(a) A (pure) state of a quantum system is described by a ray in Hilbert space, i.e., by a normalized vector ##|\psi(t) \rangle \in \mathcal{H}## modulo a phase factor (where ##\mathcal{H}## is the up to equivalence unique separable Hilbert space, i.e., it can be represented by, e.g., the Hilbert space of square-summable complex sequences, ##\ell^2## or the Hilbert space of square-integrable functions ##\mathrm{L}^2##, corresponding to the representation of the abstract formalism in terms of "matrix mechanics" or "wave mechanics" respectively). Equivalenty, and sometimes more conveniently, the pure state is represented by the corresponding Statistical Operator ##\hat{\rho}_{\psi}(t)=|\psi(t) \rangle \langle \psi(t) |##.
(b) The observables ##A,B,C,\ldots## of the quantum system are described by essential self-adjoint operators ##\hat{A}##, ##\hat{B}##, ##\hat{C},\ldots##. These operators have a complete set of (generalized) eigenvectors with real (generalized) orthonomal eigenvectors (a complete orthonormalized system, CONS, or (generalized basis).
(c) Given the system is prepared in the pure state ##\hat{\rho}_{\psi}(t)## and if ##|a,\beta \rangle## are a CONS of eigenvectors of ##\hat{A}## (where ##\beta## is a label to count the degeneracy of the eigenvalue ##a## of ##\hat{A}##, then the probability to find the value ##a##, when the observable ##A## is measured is given by Born's rule:
$$P_{\psi}(a)=\sum_{\beta} \langle a,\beta|\hat{\rho}_{\psi}(t) a,\beta \rangle=\sum_{\beta} |\langle a,\beta |\Psi(t) \rangle|^2.$$
Here I assumed that ##\beta## runs over a discrete set. If there are continuous parts or if the range of ##\beta## is entirely continuous, then one has to integrate instead.
Now since the set ##|a,\beta \rangle## is a CONS, you can uniquely represent everything in the "##A## representation", i.e., by the vector components
$$\psi(t,a,\beta)=\langle a,\beta |\psi(t) \rangle.$$
One should note that in the Dirac notation you often introduce also co-vectors ("bras") in addition to the vectors ("kets"). In the context of CONSs related to generalized eigenvectors of the self-adjoint operators these live in a larger space than the dual space of ##\mathcal{H}## (note that ##\mathcal{H}^*=\mathcal{H}## due to the scalar product), i.e., in the dual space of the domain (which is the same as the co-domain) of the operator, i.e., ##\langle a,\beta|## is in general a distribution valued linear form on ##\mathcal{H}##).
Now let's clarify this with the example we are discussing here, namely the motion of a scalar non-relativistic particles restricted to one direction in space. As for any real-world physical system, you have to make a model to construct the Hilbert space, and the most straight-forward one is to use "canonical quantization". In the classical case the phase space is given by ##x## (position) and ##p## (momentum) of the particle. Now these observables become self-adjoint operators on a Hilbert space, ##\hat{x}## and ##\hat{p}##, and the entire algebra of observables is defined by the Lie-algebra of commutators. The only non-trivial commutator is
$$[\hat{x},\hat{p}]=\mathrm{i} \hat{1}.$$
The most difficult part to construct the Hilbert space such that you can calculate real things is to figure out the spectrum of the operators and how these operators can be realized. Here, this works as follows. The heuristics is that the momentum is generating spatial translations in the classical case, and that's also the case in quantum theory, as is demonstrated as follows. Consider the operator-valued function
$$\hat{X}(\xi)=\exp(\mathrm{i} \xi \hat{p}) \hat{x} \exp(-\mathrm{i} \xi \hat{p}), \quad \xi \in \mathbb{R}.$$
Then
$$\mathrm{d}_{\xi} \hat{X} = \mathrm{i} \exp(\mathrm{i} \xi \hat{p}) [\hat{p},\hat{x}] \exp(-\mathrm{i} \xi \hat{p})=\hat{1}.$$
In the last step we've used the commutator relation. Since further obviously ##\hat{X}(0)=\hat{x}## we thus find
$$\hat{X}(\xi)=\hat{x}+\xi \hat{1}.$$
Now since ##\hat{x}## is self-adjoint it as a (generalized) eigenvector ##|x_0 \rangle## with a real eigenvalue ##x_0##. Then let's consider ##\exp(-\mathrm{i} \xi \hat{p} |x_0 \rangle##. From the above it's suggestive to think that this is also an eigenvector of the position operator thus we calculate
$$\hat{x} \exp(-\mathrm{i} \xi \hat{p}) |x_0 \rangle=\exp(-\mathrm{i} \xi \hat{p}) \hat{X}(\xi) |x_0 \rangle= \exp(-\mathrm{i} \xi \hat{p}) (\hat{x}+\xi \hat{1}) |x_0 \rangle = (x_0+\xi) \exp(-\mathrm{i} \xi \hat{p})|x_0 \rangle.$$
This means that ##\exp(-\mathrm{i} \xi \hat{p} |x_0 \rangle## is position eigenvector with eigenvalue ##x_0+\xi##, which implies that the spectrum of ##\hat{x}## is the entire real line ##\mathbb{R}##. Since we further assume that the representation of the Heisenberg commutator algebra is irreducible (since there's no other independent observable in the classical system that "commutes" with the position observable) the CONS of generalized eigenvectors of the position observable are given by
$$|x \rangle=\exp(-\mathrm{i} x \hat{p}) |0 \rangle, \quad x \in \mathbb{R}.$$
The generalized eigenvectors are normalized "to a ##\delta## distribution",
$$\langle x'|\langle x \rangle=\delta(x-x')$$
for convenience. Further the completeness relation is assumed to hold:
$$\int_{\mathbb{R}} |x \rangle \langle x|=\hat{1}.$$
Now we are ready to work in position representation (aka "wave mechanics"). The pure states are represented in the above specified sense by the "components"
$$\psi(t,x)=\langle x|\psi(t) \rangle,$$
and the state ket itself is given due to the completeness relation by
$$|\psi(t) \rangle=\int_{\mathbb{R}} \mathrm{d} x |x \rangle \psi(x,t).$$
The scalar product is realized by integrals, again due to the completeness relation, by
$$\langle \psi_1|\psi_2 \rangle=\int_{\mathbb{R}} \mathrm{d} x \langle \psi_1|x \rangle \langle x|\psi_2 \rangle=\int_{\mathbb{R}} \mathrm{d} x \psi_1^*(x) \psi_2(x),$$
i.e., we have a map between the abstract Hilbert-space ##\mathcal{H}## and the Hilbert space ##\mathrm{L}^2(\mathbb{R},\mathbb{C})## of complex-valued square-integrable functions (modulo equivalence of functions that only deviate on set of real numbers of Lebesgue-measure 0).
Finally we need the momentum operator. It's uniquely defined when we know how it acts on the generalized position eigenvectors, but that's easy since in our choice of the basis we have
$$|x \rangle = \exp(-\mathrm{i} x \hat{p}) | 0 \rangle \; \Rightarrow \; \hat{p} |x \rangle=\mathrm{i} \mathrm{d}_x |x \rangle.$$
Taking the adjoint of this relation we find
$$\langle x| \hat{p}=-\mathrm{i} \mathrm{d}_x \langle x|.$$
Multiplying with ##|\psi \rangle## we find that
$$\tilde{p} \psi(x)=\langle x|\hat{p} \psi \rangle=-\mathrm{i} \mathrm{d}_x \langle x|\psi \rangle=-\mathrm{i} \psi'(x).$$
Now you can easily find the momentum eigenvectors in position representation by solving the eigenvalue equation
$$\tilde{p} u_p(x)=p u_p (x), \text{where} \quad u_p(x)=\langle x|p \rangle.$$
Of course ##u_p## is a function of position with a real parameter ##p## (or, if you wish a function of two independent variables ##x## and ##p##). This gives (including the "normalization to a ##\delta## distribution"),
$$u_p(x)=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} x p),$$
and you can easily transform from the position to the momentum distribution, via
$$\tilde{\psi}(p)=\langle p|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x \langle p|x \rangle \langle x|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x \frac{1}{\sqrt{2 \pi}} \exp(-\mathrm{i} p x) \psi(x),$$
i.e., you get the momentum-space wave function as Fourier transform of the position-space wave function and vice versa,
$$\psi(x)=\int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi}} \exp(+\mathrm{i} p x) \tilde{\psi}(p).$$
I hope this clarifies the standard notation of the Dirac formalism sufficiently.