Prove that space of Laurent series is not Hilbert Let $z_0 \in \mathbb{C}, s>0,T(z_0,s):=\{z\in\mathbb{C}:|z-z_0|=s\}$ and let $V=V(z_0,s)$ be a vector space over field $\mathbb{C}$ of all Laurent series that are uniformly and absolutely convergent on $T(z_0,s)$ and:
$$<f,g>:=\frac{1}{2 \pi}\int_0^{2\pi}f(z_0+se^{it})\overline{g(z_0+se^{it})}dt$$ $f,g \in V$
Show that space $V$ with a dot product $<\cdot,\cdot>$ is not a Hilbert space.

So I should show that $V$ is not complete. I other words i need to show that the norm induced by this dot product doesnt make $V$ complete.
The norm is following:
$$\|f\|_V=\Big(\frac{1}{2\pi} \int_0^{2\pi} |f(z_0+se^{it})|^2 dt\Big)^{\frac{1}{2}}$$
Now I just need to find a Cauchy sequence that is not convergent to the element of $V$. But I dont have any idea how can I do that. Any help please?
 A: Two and a half hints:


*

*You can without loss of generality assume $z_0 = 0$ and $s = 1$, the general case differs only in book-keeping.

*Determine the inner product (and hence the norm) in terms of the coefficients of the Laurent series. You should see a connection with a familiar Hilbert space.

*Find what the absolute convergence of the Laurent series means for the coefficients. You should find an identification of $V$ with a familiar Banach space which is known not to be a Hilbert space.


If you have the last, you no longer need to find a Cauchy sequence in $V$ that is not convergent, but then it is easy to explicitly find one.

Since the Laurent series are assumed to be absolutely and uniformly convergent on the circle, we can interchange summation and integration:
\begin{align}
\frac{1}{2\pi}\int_0^{2\pi} f(e^{it})\overline{g(e^{it})}\,dt
&= \frac{1}{2\pi} \int_0^{2\pi}\Biggl(\sum_{n = -\infty}^\infty a_n e^{int}\Biggr)\Biggl(\sum_{k = -\infty}^\infty \overline{b_k} e^{-ikt}\Biggr)\,dt\\
&= \sum_{n,k} a_n \overline{b_k} \underbrace{\frac{1}{2\pi}\int_0^{2\pi} e^{i(n-k)t}\,dt}_{\delta_{n,k}}\\
&= \sum_{n = -\infty}^\infty a_n \overline{b_n},
\end{align}
where
$$f(e^{it}) = \sum_{n = -\infty}^\infty a_n e^{int}\quad\text{and}\quad g(e^{it}) = \sum_{k = -\infty}^\infty b_k e^{ikt}.$$
Thus the inner product corresponds to the inner product on the sequence space $\ell^2(\mathbb{Z})$. The assumption that the series are absolutely and uniformly convergent means that the coefficients of the Laurent series are absolutely summable, or that $V \cong \ell^1(\mathbb{Z})\subset \ell^2(\mathbb{Z})$. Since $\ell^1(\mathbb{Z})$ is a dense proper subspace of $\ell^2(\mathbb{Z})$, it follows that $V$ is not complete.
