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Consider the symmetric Laurent polynomialIt satisfies $f(1)=1$, so the Alexander polynomial realization theorem gives a knot $K_3$ with ${\Delta }_{K_3}(t)\doteq f(t)$. The polynomial is not a unit of $\mathbb{Z}[t^{\pm 1}]$, whereas

Here functional calculus convergence means that for every $f\in C(\mathbb{T})$,

Embed ${\mathcal{H}}_n$ in $\mathcal{H}$ and write ${\tilde{A}}_n=P_n^{\ast }{A}_n{P}_n$. The hypothesis gives ${\tilde{A}}_n\rightharpoonup A$ in the weak operator topology. Since $A_n$ and $A$ are unitary operators,Thus ${\tilde{A}}_n\to A$ in the strong operator topology. Applying the same argument to the adjoints gives ${\tilde{A}}_n^{\ast }\to A^{\ast }$ strongly.

Products of uniformly bounded strongly convergent operators converge strongly, so for every integer $k$,strongly, with negative $k$ interpreted through adjoints. Therefore convergence holds for every Laurent polynomial. The Stone-Weierstrass theorem says that Laurent polynomials are uniformly dense in $C(\mathbb{T})$. Since the continuous functional calculus is contractive, uniform approximation finishes the proof for every $f\in C(\mathbb{T})$.