# $U$ unitary: $\mathbb{T}\ne\sigma(U)$. Prove $\forall\varepsilon>0$ there exists a polynomial $p(z)$ such that $\|U^{-1}-p(U)\|<\varepsilon.$

Let $$U$$ be a unitary operator: $$\mathbb{T}=\{\lambda:|\lambda|=1\}\setminus\sigma(U)\ne\varnothing$$ (the spectrum does not cover the whole circle). Prove that $$\forall\varepsilon>0$$ there exists a polynomial $$p(z)=\sum\limits_{i=0}^Nc_iz^i$$ such that $$\|U^{-1}-p(U)\|<\varepsilon.$$

What can I say is that $$\exists\lambda\in\mathbb{T}: U-\lambda I$$ is invertible. It feels like functional calculus for unitary operators must be useful.

Since the spectrum is not the whole circle, there is at least one point missing. Consider the complex plane minus the ray (i.e. half-line) that starts from the origin and passes through some missing point: this is called a slit plane. The spectrum of $$U$$ is therefore contained in a slit plane. By complex analysis, we know that, on a slit plane, we can define a holomorphic logarithm, i.e. there exists a holomorphic function $$L(z)$$ defined on our slit plance such that $$e^{L(z)}=z$$ for all $$z$$ in the slit plane. In particular, $$e^{L(z)}=z$$ for all $$z\in\sigma(U)$$.
Consider the function $$f(z)=e^{-z}$$ and let $$h=f\circ L$$, so $$h$$ is a holomorphic function defined on $$\sigma(U)$$. In particular, $$h$$ is continuous so we can define using continuous functional calculus the element $$h(U)$$. Since $$zh(z)=ze^{-L(z)}=zz^{-1}=1$$, we have that $$h(U)=U^{-1}$$. But holomorphic functions have power series representations, so, if $$h(z)=\sum_{n=1}^\infty c_nz^n$$ (the convergence is uniform over compact subsets of the domain), then the polynomials $$p_N(z)=\sum_{n=1}^Nc_nz^n$$ covnerge uniformly to $$h$$ on $$\sigma(U)$$, which gives you the desired approximation.
• By replacing $u$ by $\lambda u$ for some modulus-one scalar $\lambda$, you can assume that the point that is missing from the spectrum is $-1$. Then you can simply use the principal branch of the logarithm. Jun 22, 2021 at 12:27
Note that $$\mathbb{C}-\sigma(U)$$ is connected. Use Mergelyan’s theorem, $$1/z$$ can be uniform approached by analytic polynomials on $$\sigma(U)$$.