I am a PhD student in Physics. Let us consider a vector in an infinite dimensional Hilbert space as


\begin{equation}
|f\rangle\equiv
\begin{bmatrix}
 1 \\
 z \\
 z^2 \\
 \vdots
\end{bmatrix},
\end{equation}


where $z$ is a complex number. The norm of the vector $|f\rangle$ is
\begin{equation}
\left\lVert\langle f|f\rangle\right\rVert=1+|z|^2+|z|^4+\cdots=\frac{1}{1-|z|^2}.
\end{equation}


Hence, the vector $|f\rangle$ is normalizable only when $|z|<1$, and the normalized vector should be $|z\rangle=\sqrt{1-|z|^2}|f\rangle$. The overlap between any two distinct vectors is
\begin{equation}
\langle z\_2|z\_1\rangle = \frac{\sqrt{1-|z\_2|^2}\sqrt{1-|z\_1|^2}}{1-z^\*\_2z\_1},
\end{equation}
which yields
\begin{equation}
|\langle z\_2|z\_1\rangle|^2=\frac{(1-|z\_2|^2)(1-|z\_1|^2)}{|z\_1-z\_2|^2+(1-|z\_2|^2)(1-|z\_1|^2)}\leq 1,
\end{equation}
Hence, any two distinct vectors are non-orthogonal, and the overlap of the two vectors is always small than one. 


Now, my question is: can these vectors form a complete/overcomplete basis in the infinite dimensional Hilbert space?


For any two arbitrary vectors $|\psi\rangle$ and $|\phi\rangle$, the above statement is true when the following relation is valid
\begin{equation}
\langle\psi|\phi\rangle=\frac{1}{\pi}\int \langle\psi|z\rangle\langle z|\phi\rangle g(z)d^2z,
\end{equation}
where $g(z)$ is an unknown integration kernal. If we denote $z=re^{i\phi}$, the integration becomes
\begin{align}
\langle\psi|\phi\rangle&=\frac{1}{2\pi}\sum\_{n,m}\langle\psi|m\rangle\langle n|\phi\rangle\int\_0^1 g(r)r^{n+m+1}dr\int\_0^{2\pi}e^{i(n-m)}d\phi\\
&=\frac{1}{2\pi}\sum\_{n,m}\langle\psi|m\rangle\langle n|\phi\rangle\int\_0^1 g(r)r^{n+m+1}dr\int\_0^{2\pi}e^{i(n-m)}d\phi\\
&=\sum\_{n}\langle\psi|n\rangle\langle n|\phi\rangle\int\_0^1 g(r)r^{2n+1}dr,
\end{align}
where we have selected a kernal function $g(r)$ that is independent of the polar angle $\phi$. Hence, if we can prove that for any positive integer $n$
\begin{equation}
\int\_0^1 g(r)r^{2n+1}dr=1,
\end{equation}
then $|z\rangle$ for $|z|<1$ will form a overcomplete basis in the infinite dimensional Hilbert space. However, I can't find such $g(r)$. Therefore, I doubt that $|z\rangle$ do not form a complete basis.



How to prove or disprove a type of states form an overcomplete basis in the Hilbert space?
As suggested by Yemon Choi, the mapping $\hat{\psi}(z) = \langle \psi | f \rangle$ is an analytic function on the unit disc, a member of the $H^2$ complex Hardy space. Even though $f$ is not a member of the $\ell^2$ Hilbert space when $|z| = 1$, the mapping $\hat{\psi}$ extends to the boundary of the unit disk (in the sense of almost everywhere limit along radii, for example), and by [Parseval's identity](https://en.wikipedia.org/wiki/Parseval%27s_identity),
$$\begin{aligned}\langle \psi | \phi \rangle & = \frac{1}{2 \pi} \int\_0^{2\pi} \hat{\psi}(e^{i t}) \overline{\hat{\phi}(e^{i t})} dt \\ & = \frac{1}{2 \pi} \int\_0^{2\pi} \langle \psi | f\_{e^{it}} \rangle \langle f\_{e^{it}} | \phi \rangle dt .\end{aligned}$$
In other words, your "function" $g(z)$ should be the limit of $\tfrac{1}{2 \pi} (1 - r^2)^{-1/2}$ times the Lebesgue measure on the boundary of the disk of radius $r$ as $r \to 1^-$.


