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Michael Hardy
Michael Hardy
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Q1: No. You gave a proof above: $\overline{\langle f,\bar g\rangle} = Tr(f\otimes g)$$\overline{\langle f,\bar g\rangle} = \operatorname{Tr}(f\otimes g)$.

Q2: No, if $Tr$$\operatorname{Tr}$ is supposed to be continuous. Namely, $\mathcal S(\mathbb R^k)\subset L^2(\mathbb R^k)$ continuosly, and $L^(\mathbb R^k)\subset \mathcal S'(\mathbb R^k)$$L^2(\mathbb R^k)\subset \mathcal S'(\mathbb R^k)$ countinuous and dense. So any continuous trace would immediately lead to a contradiction to Q1.

Q1: No. You gave a proof above: $\overline{\langle f,\bar g\rangle} = Tr(f\otimes g)$.

Q2: No, if $Tr$ is supposed to be continuous. Namely, $\mathcal S(\mathbb R^k)\subset L^2(\mathbb R^k)$ continuosly, and $L^(\mathbb R^k)\subset \mathcal S'(\mathbb R^k)$ countinuous and dense. So any continuous trace would immediately lead to a contradiction to Q1.

Q1: No. You gave a proof above: $\overline{\langle f,\bar g\rangle} = \operatorname{Tr}(f\otimes g)$.

Q2: No, if $\operatorname{Tr}$ is supposed to be continuous. Namely, $\mathcal S(\mathbb R^k)\subset L^2(\mathbb R^k)$ continuosly, and $L^2(\mathbb R^k)\subset \mathcal S'(\mathbb R^k)$ countinuous and dense. So any continuous trace would immediately lead to a contradiction to Q1.

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Peter Michor
Peter Michor
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Q1: No. You gave a proof above: $\overline{\langle f,\bar g\rangle} = Tr(f\otimes g)$.

Q2: No, if $Tr$ is supposed to be continuous. Namely, $\mathcal S(\mathbb R^k)\subset L^2(\mathbb R^k)$ continuosly, and $L^(\mathbb R^k)\subset \mathcal S'(\mathbb R^k)$ countinuous and dense. So any continuous trace would immediately lead to a contradiction to Q1.