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Andreas Thom
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There are four facts which clarify things a little bit:

  1. The inclusion $\ell^1 {\mathbb Z} \subset C(S^1)$ preserves the spectrum. (That is Wiener's Theorem)

  2. If $f = \sum_i a_i z^{i}\in \ell^1 {\mathbb Z}$ has non-negative coefficients, then $\|f\| = \|f\|_{C(S^1)}$.

  3. If $f(z)>0$ for all $z \in S^1$, then $f$ has a square-root in $\ell^1 {\mathbb Z}$ by holomorphic functional calculus.

  4. $2 - z+z^{-1}$$2 - z-z^{-1}$ is non-negative, but has no square-root in $\ell^1{\mathbb Z}$.

There are four facts which clarify things a little bit:

  1. The inclusion $\ell^1 {\mathbb Z} \subset C(S^1)$ preserves the spectrum. (That is Wiener's Theorem)

  2. If $f = \sum_i a_i z^{i}\in \ell^1 {\mathbb Z}$ has non-negative coefficients, then $\|f\| = \|f\|_{C(S^1)}$.

  3. If $f(z)>0$ for all $z \in S^1$, then $f$ has a square-root in $\ell^1 {\mathbb Z}$ by holomorphic functional calculus.

  4. $2 - z+z^{-1}$ is non-negative, but has no square-root in $\ell^1{\mathbb Z}$.

There are four facts which clarify things a little bit:

  1. The inclusion $\ell^1 {\mathbb Z} \subset C(S^1)$ preserves the spectrum. (That is Wiener's Theorem)

  2. If $f = \sum_i a_i z^{i}\in \ell^1 {\mathbb Z}$ has non-negative coefficients, then $\|f\| = \|f\|_{C(S^1)}$.

  3. If $f(z)>0$ for all $z \in S^1$, then $f$ has a square-root in $\ell^1 {\mathbb Z}$ by holomorphic functional calculus.

  4. $2 - z-z^{-1}$ is non-negative, but has no square-root in $\ell^1{\mathbb Z}$.

Source Link
Andreas Thom
  • 25.5k
  • 4
  • 82
  • 142

There are four facts which clarify things a little bit:

  1. The inclusion $\ell^1 {\mathbb Z} \subset C(S^1)$ preserves the spectrum. (That is Wiener's Theorem)

  2. If $f = \sum_i a_i z^{i}\in \ell^1 {\mathbb Z}$ has non-negative coefficients, then $\|f\| = \|f\|_{C(S^1)}$.

  3. If $f(z)>0$ for all $z \in S^1$, then $f$ has a square-root in $\ell^1 {\mathbb Z}$ by holomorphic functional calculus.

  4. $2 - z+z^{-1}$ is non-negative, but has no square-root in $\ell^1{\mathbb Z}$.