Prove that space of Laurent series is not Hilbert

Question

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.

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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?

Answer 1

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.

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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.