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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.

Can you please help me? Any hint is appreciate.

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2 Answers 2

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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.

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    $\begingroup$ 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. $\endgroup$
    – J. De Ro
    Jun 22, 2021 at 12:27
  • $\begingroup$ @QuantumSpace exactly, one can rotate without any harm to the generality:) $\endgroup$ Jun 22, 2021 at 12:28
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Note that $\mathbb{C}-\sigma(U)$ is connected. Use Mergelyan’s theorem, $1/z$ can be uniform approached by analytic polynomials on $\sigma(U)$.

Edit: Thanks for David’s comment, here Mergelyan’s theorem can be replaced by Runge’s theorem, which is much more elementary.

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    $\begingroup$ In fact Runge's theorem is enough here, and of course it's much more elementary $\endgroup$ Jun 22, 2021 at 12:24

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