Return to Answer

The assertion is false, if I understand well it.

First of all some general information on the notion of weak * topology.

One has a (complex) normed space $B$ and its topological dual $B'$ which is, by definition the vector space of linear maps $f: B \to \mathbb{C}$.

This space has a natural structure of normed space as well with the norm $$||f||' := \sup\{|f(x)| \:|\: x \in B \:, ||x||=1\}\:.$$ However we can equipp $B'$ with another seminormed topology induced by $B$ called weak * topology, it is charachterized by the fact that

$$B' \ni f_n \to f \in B'\quad iff \quad f_n(x)-f(x)\to 0 \quad \forall x\in B\:.$$

The Banach-Alaouglu theorem proves that every norm-closed ball in $B'$ $$\{f \in B' \:|\: ||f|| \leq r\}$$ is compact with respect to the weak * topology provided $B$ is complete (i.e. is a Banach space).

Sometimes it happens that two normed spaces $B_1$ and $B_2$ are in topological duality, that is one is (normed-space) isomorphic to the topological dual of the other. If $B_2$ is isomorphic through $F$ to $B_1' = F(B_2)$, the action of the elements of $B_2$ o the elements of $B_1$ is indicated by means of a pairing: $$\langle b_1, b_2\rangle := (F(b_2))(b_1)\:.$$

Let us pass to the discussed assertion, that the space of the mixed states $S(H)$ is compact with respect to some weak * topology on that. The space $S(H)$ is not linear and it is not a closed ball of the normed (Banach) space of the trace class operators $B_1(H)$ equipped with its natural norm topology induced by the notion of trace. Therefore, one should try to prove that $S(H)$ is a compact subset of $B_1(H)$.

From the sketch of proof, it seems that the authors use the fact that $B_1(H)$ is the topological dual space of some other space. All that in order to apply the Banach-Alaouglu theorem as written above.(*)

As is well known, $B_1(H)$ is in fact (isometrically isomorphic to) the topological dual space of $B_\infty(H)$ (the Banach space of compact operators with the natual norm operator topology).

The duality is represented in terms of the trace operation as $B_1(H)$ is a (both sides *) ideal, as the (left-continuous) pairing

$$\langle A, B \rangle := tr(AB)\:,\:\quad A\in B_1(H)\:, \quad B \in B_\infty(H)$$

The Banach-Alaouglu theorem proves that every closed ball in $B_1(H)$ is therefore compact with respect to the weak * topology induced by the topological duality considered above. In particular, according to this weak* topology, $$B_1(H) \ni \rho_n \to \rho \in B_1(H)\quad iff \quad tr((\rho_n-\rho)A)\to 0 \quad \forall A\in B_1(H)$$ To conclude the proof it would be sufficient to prove that $S(H)$ is closed with respect to that topology, since closed subsets of compact sets are compact as well.

But it is not the case as proved here!

On the other hand, since weak * topology is Hausdroff, compact sets are closed. This proves that the assertion, referred to the said weak * topology, is false as $S(H)$ is not closed with respect to that weak * topology.

Maybe the assertion is true using another weak (natural?) topology.

($*$) Another possibility, not followed by the authors, is to see $B_1(H) \subset B(H)$ and viewing $B(H)$ as the dual of $B_1(H)$, thus equipping this latter with the weak* topology of $B(H)$ also known as ultraweak topology.

Even another way is to see $B_1(H)$ as subset of $B_2(H)$ the Hilbert-space space of Hilbert Schmidt operators and to use the weak * topology on $B_2(H)$ induced by the Riesz anti isomorphism.

The assertion is false, if I understand well it.

First of all some general information on the notion of weak * topology.

One has a (complex) normed space $B$ and its topological dual $B'$ which is, by definition the vector space of continuous linear maps $f: B \to \mathbb{C}$.

This space has a natural structure of normed space as well with the norm $$||f||' := \sup\{|f(x)| \:|\: x \in B \:, ||x||=1\}\:.$$ However we can equip $B'$ with another seminormed topology induced by $B$ called weak * topology, it is characterized by the fact that

$$B' \ni f_n \to f \in B'\quad iff \quad f_n(x)-f(x)\to 0 \quad \forall x\in B\:.$$

The Banach-Alaoglu theorem proves that every norm-closed ball in $B'$ $$\{f \in B' \:|\: ||f|| \leq r\}$$ is compact with respect to the weak * topology provided $B$ is complete (i.e. is a Banach space).

Sometimes it happens that two normed spaces $B_1$ and $B_2$ are in topological duality, that is one is (normed-space) isomorphic to the topological dual of the other. If $B_2$ is isomorphic through $F$ to $B_1' = F(B_2)$, the action of the elements of $B_2$ o the elements of $B_1$ is indicated by means of a pairing: $$\langle b_1, b_2\rangle := (F(b_2))(b_1)\:.$$

Let us pass to the discussed assertion, that the space of the mixed states $S(H)$ is compact with respect to some weak * topology on that. The space $S(H)$ is not linear and it is not a closed ball of the normed (Banach) space of the trace class operators $B_1(H)$ equipped with its natural norm topology induced by the notion of trace. Therefore, one should try to prove that $S(H)$ is a compact subset of $B_1(H)$.

From the sketch of proof, it seems that the authors use the fact that $B_1(H)$ is the topological dual space of some other space. All that in order to apply the Banach-Alaoglu theorem as written above.(*)

As is well known, $B_1(H)$ is in fact (isometrically isomorphic to) the topological dual space of $B_\infty(H)$ (the Banach space of compact operators with the natual norm operator topology).

The duality is represented in terms of the trace operation as $B_1(H)$ is a (both sides *) ideal, as the (left-continuous) pairing

$$\langle A, B \rangle := tr(AB)\:,\:\quad A\in B_1(H)\:, \quad B \in B_\infty(H)$$

The Banach-Alaoglu theorem proves that every closed ball in $B_1(H)$ is therefore compact with respect to the weak * topology induced by the topological duality considered above. In particular, according to this weak* topology, $$B_1(H) \ni \rho_n \to \rho \in B_1(H)\quad iff \quad tr((\rho_n-\rho)A)\to 0 \quad \forall A\in B_1(H)$$ To conclude the proof it would be sufficient to prove that $S(H)$ is closed with respect to that topology, since closed subsets of compact sets are compact as well.

But it is not the case as proved here!

On the other hand, since weak * topology is Hausdorff, compact sets are closed. This proves that the assertion, referred to the said weak * topology, is false as $S(H)$ is not closed with respect to that weak * topology.

Maybe the assertion is true using another weak (natural?) topology.

($*$) Another possibility, not followed by the authors, is to see $B_1(H) \subset B(H)$ and viewing $B(H)$ as the dual of $B_1(H)$, thus equipping this latter with the weak* topology of $B(H)$ also known as ultraweak topology.

Even another way is to see $B_1(H)$ as subset of $B_2(H)$ the Hilbert-space space of Hilbert Schmidt operators and to use the weak * topology on $B_2(H)$ induced by the Riesz anti isomorphism.

The assertion is false, if I understand well it.

First of all some general information on the notion of weak * topology.

One has a (complex) normed space $B$ and its topological dual $B'$ which is, by definition the vector space of linear maps $f: B \to \mathbb{C}$.

This space has a natural structure of normed space as well with the norm $$||f||' := \sup\{|f(x)| \:|\: x \in B \:, ||x||=1\}\:.$$ However we can equipp $B'$ with another seminormed topology induced by $B$ called weak * topology, it is charachterized by the fact that

$$B' \ni f_n \to f \in B'\quad iff \quad f_n(x)-f(x)\to 0 \quad \forall x\in B\:.$$

The Banach-Alaouglu theorem proves that every norm-closed ball in $B'$ $$\{f \in B' \:|\: ||f|| \leq r\}$$ is compact with respect to the weak * topology provided $B$ is complete (i.e. is a Banach space).

Sometimes it happens that two normed spaces $B_1$ and $B_2$ are in topological duality, that is one is (normed-space) isomorphic to the topological dual of the other. If $B_2$ is isomorphic through $F$ to $B_1' = F(B_2)$, the action of the elements of $B_2$ o the elements of $B_1$ is indicated by menasd of a pairing: $$\langle b_1, b_2\rangle := (F(b_2))(b_1)\:.$$

Let us pass to the discussed assertion, that the space of the mixed states $S(H)$ is compact with respect to some weak * topology on that. The space $S(H)$ is not linear and it is not a closed ball of the normed (Banach) space of the trace class operators $B_1(H)$ equipped with its natural norm topology induced by the notion of trace. Therefore $S(H)$ should be viewed as a (compact) subset of $B_1(H)$.

From the sketch of proof, it seems that the authors use the fact that $B_1(H)$ is the topological dual space of some other space. All that in order to apply the Banach-Alaouglu theorem as written above.(*)

As is well known, $B_1(H)$ is in fact (isometrically isomorphic to) the topological dual space of $B_\infty(H)$ (the Banach space of compact operators with the natual norm operator topology).

The duality is represented in terms of the trace operation as $B_1(H)$ is a (both sides *) ideal, as the (left-continuous) pairing

$$\langle A, B \rangle := tr(AB)\:,\:\quad A\in B_1(H)\:, \quad B \in B_\infty(H)$$

The Banach-Alaouglu theorem proves that every closed ball in $B_1(H)$ is therefore compact with respect to the weak * topology induced by the topological duality considered above. In particular, according to this weak* topology, $$B_1(H) \ni \rho_n \to \rho \in B_1(H)\quad iff \quad tr((\rho_n-\rho)A)\to 0 \quad \forall A\in B_1(H)$$ To conclude the proof it would be sufficient to prove that $S(H)$ is closed with respect to that topology, since closed subsets of compact sets are compact as well.

But it is not the case as proved here!

On the other hand, since weak * topology is Hausdroff, compact sets are closed. This proves that the assertion, referred to the said weak * topology, is false as $S(H)$ is not closed with respect to that weak * topology.

Maybe the assertion is true using another weak (natural?) topology.

($*$) Another possibility, not followed by the authors, is to see $B_1(H) \subset B(H)$ and viewing $B(H)$ as the dual of $B_1(H)$, thus equipping this latter with the weak* topology of $B(H)$ also known as ultraweak topology.

Even another way is to see $B_1(H)$ as subset of $B_2(H)$ the Hilbert-space space of Hilbert Schmidt operators and to use the weak * topology on $B_2(H)$ induced by the Riesz anti isomorphism.

The assertion is false, if I understand well it.

First of all some general information on the notion of weak * topology.

One has a (complex) normed space $B$ and its topological dual $B'$ which is, by definition the vector space of linear maps $f: B \to \mathbb{C}$.

This space has a natural structure of normed space as well with the norm $$||f||' := \sup\{|f(x)| \:|\: x \in B \:, ||x||=1\}\:.$$ However we can equipp $B'$ with another seminormed topology induced by $B$ called weak * topology, it is charachterized by the fact that

$$B' \ni f_n \to f \in B'\quad iff \quad f_n(x)-f(x)\to 0 \quad \forall x\in B\:.$$

The Banach-Alaouglu theorem proves that every norm-closed ball in $B'$ $$\{f \in B' \:|\: ||f|| \leq r\}$$ is compact with respect to the weak * topology provided $B$ is complete (i.e. is a Banach space).

Sometimes it happens that two normed spaces $B_1$ and $B_2$ are in topological duality, that is one is (normed-space) isomorphic to the topological dual of the other. If $B_2$ is isomorphic through $F$ to $B_1' = F(B_2)$, the action of the elements of $B_2$ o the elements of $B_1$ is indicated by means of a pairing: $$\langle b_1, b_2\rangle := (F(b_2))(b_1)\:.$$

Let us pass to the discussed assertion, that the space of the mixed states $S(H)$ is compact with respect to some weak * topology on that. The space $S(H)$ is not linear and it is not a closed ball of the normed (Banach) space of the trace class operators $B_1(H)$ equipped with its natural norm topology induced by the notion of trace. Therefore, one should try to prove that $S(H)$ is a compact subset of $B_1(H)$.

From the sketch of proof, it seems that the authors use the fact that $B_1(H)$ is the topological dual space of some other space. All that in order to apply the Banach-Alaouglu theorem as written above.(*)

As is well known, $B_1(H)$ is in fact (isometrically isomorphic to) the topological dual space of $B_\infty(H)$ (the Banach space of compact operators with the natual norm operator topology).

The duality is represented in terms of the trace operation as $B_1(H)$ is a (both sides *) ideal, as the (left-continuous) pairing

$$\langle A, B \rangle := tr(AB)\:,\:\quad A\in B_1(H)\:, \quad B \in B_\infty(H)$$

The Banach-Alaouglu theorem proves that every closed ball in $B_1(H)$ is therefore compact with respect to the weak * topology induced by the topological duality considered above. In particular, according to this weak* topology, $$B_1(H) \ni \rho_n \to \rho \in B_1(H)\quad iff \quad tr((\rho_n-\rho)A)\to 0 \quad \forall A\in B_1(H)$$ To conclude the proof it would be sufficient to prove that $S(H)$ is closed with respect to that topology, since closed subsets of compact sets are compact as well.

But it is not the case as proved here!

On the other hand, since weak * topology is Hausdroff, compact sets are closed. This proves that the assertion, referred to the said weak * topology, is false as $S(H)$ is not closed with respect to that weak * topology.

Maybe the assertion is true using another weak (natural?) topology.

($*$) Another possibility, not followed by the authors, is to see $B_1(H) \subset B(H)$ and viewing $B(H)$ as the dual of $B_1(H)$, thus equipping this latter with the weak* topology of $B(H)$ also known as ultraweak topology.

Even another way is to see $B_1(H)$ as subset of $B_2(H)$ the Hilbert-space space of Hilbert Schmidt operators and to use the weak * topology on $B_2(H)$ induced by the Riesz anti isomorphism.

The assertion is false, if I understand well it.

From the sketch of proof, it seems that the authors use the fact that $B_1(H)$ (the Banach space of the trace class operators equipped with its natural norm topology induced by the notion of trace) is the topological dual space of some other space. All that in order to apply the Banach-Alaouglu theorem.(*)

As is well known, $B_1(H)$ is in fact the topological dual space of $B_\infty(H)$ (the Banach space of compact operators with the natual norm operator topology).

The duality is represented in terms of the trace operation as $B_1(H)$ is a (both sides *) ideal, as the (left-continuous) pairing

$$\langle A, B \rangle := tr(AB)\:,\:\quad A\in B_1(H)\:, \quad B \in B_\infty(H)$$

The Banach-Alaouglu theorem proves that every closed ball in $B_1(H)$ is therefore compact with respect to the weak * topology induced by the topological duality considered above. In particular, according to this weak* topology, $$B_1(H) \ni \rho_n \to \rho \in B_1(H)\quad iff \quad tr((\rho_n-\rho)A)\to 0 \quad \forall A\in B_1(H)$$ To conclude the proof it would be sufficient to prove that $S(H)$ is closed with respect to that topology, since closed subsets of compact sets are compact as well.

But it is not the case as proved here!

On the other hand, since weak * topology is Hausdroff, compact sets are closed. This proves that the assertion, referred to the said weak * topology, is false as $S(H)$ is not closed with respect to that weak * topology.

Maybe the assertion is true using another weak (natural?) topology.

($*$) Another possibility, not followed by the authors, is to see $B_1(H) \subset B(H)$ and viewing $B(H)$ as the dual of $B_1(H)$, thus equipping this latter with the weak* topology of $B(H)$ also known as ultraweak topology.

Even another way is to see $B_1(H)$ as subset of $B_2(H)$ the Hilbert-space space of Hilbert Schmidt operators and to use the weak * topology on $B_2(H)$ induced by the Riesz anti isomorphism.

The assertion is false, if I understand well it.

First of all some general information on the notion of weak * topology.

One has a (complex) normed space $B$ and its topological dual $B'$ which is, by definition the vector space of linear maps $f: B \to \mathbb{C}$.

This space has a natural structure of normed space as well with the norm $$||f||' := \sup\{|f(x)| \:|\: x \in B \:, ||x||=1\}\:.$$ However we can equipp $B'$ with another seminormed topology induced by $B$ called weak * topology, it is charachterized by the fact that

$$B' \ni f_n \to f \in B'\quad iff \quad f_n(x)-f(x)\to 0 \quad \forall x\in B\:.$$

The Banach-Alaouglu theorem proves that every norm-closed ball in $B'$ $$\{f \in B' \:|\: ||f|| \leq r\}$$ is compact with respect to the weak * topology provided $B$ is complete (i.e. is a Banach space).

Sometimes it happens that two normed spaces $B_1$ and $B_2$ are in topological duality, that is one is (normed-space) isomorphic to the topological dual of the other. If $B_2$ is isomorphic through $F$ to $B_1' = F(B_2)$, the action of the elements of $B_2$ o the elements of $B_1$ is indicated by menasd of a pairing: $$\langle b_1, b_2\rangle := (F(b_2))(b_1)\:.$$

Let us pass to the discussed assertion, that the space of the mixed states $S(H)$ is compact with respect to some weak * topology on that. The space $S(H)$ is not linear and it is not a closed ball of the normed (Banach) space of the trace class operators $B_1(H)$ equipped with its natural norm topology induced by the notion of trace. Therefore $S(H)$ should be viewed as a (compact) subset of $B_1(H)$.

From the sketch of proof, it seems that the authors use the fact that $B_1(H)$ is the topological dual space of some other space. All that in order to apply the Banach-Alaouglu theorem as written above.(*)

As is well known, $B_1(H)$ is in fact (isometrically isomorphic to) the topological dual space of $B_\infty(H)$ (the Banach space of compact operators with the natual norm operator topology).

The duality is represented in terms of the trace operation as $B_1(H)$ is a (both sides *) ideal, as the (left-continuous) pairing

$$\langle A, B \rangle := tr(AB)\:,\:\quad A\in B_1(H)\:, \quad B \in B_\infty(H)$$

The Banach-Alaouglu theorem proves that every closed ball in $B_1(H)$ is therefore compact with respect to the weak * topology induced by the topological duality considered above. In particular, according to this weak* topology, $$B_1(H) \ni \rho_n \to \rho \in B_1(H)\quad iff \quad tr((\rho_n-\rho)A)\to 0 \quad \forall A\in B_1(H)$$ To conclude the proof it would be sufficient to prove that $S(H)$ is closed with respect to that topology, since closed subsets of compact sets are compact as well.

But it is not the case as proved here!

On the other hand, since weak * topology is Hausdroff, compact sets are closed. This proves that the assertion, referred to the said weak * topology, is false as $S(H)$ is not closed with respect to that weak * topology.

Maybe the assertion is true using another weak (natural?) topology.

($*$) Another possibility, not followed by the authors, is to see $B_1(H) \subset B(H)$ and viewing $B(H)$ as the dual of $B_1(H)$, thus equipping this latter with the weak* topology of $B(H)$ also known as ultraweak topology.

Even another way is to see $B_1(H)$ as subset of $B_2(H)$ the Hilbert-space space of Hilbert Schmidt operators and to use the weak * topology on $B_2(H)$ induced by the Riesz anti isomorphism.