Proof. By the Weierstrass approximation theorem, there exist polynomials $(p_n)_{n \in {\mathbb {N}}}$ that converge uniformly to f on $[0, r]$ .

Let $\varepsilon> 0$ . By uniform convergence of $(p_n)_{n \in {\mathbb {N}}}$ to f, there exists $N \in {\mathbb {N}}$ such that, for any $n \geqslant N$ , we have $\|p_n - f\|_{[0,r]} < {\varepsilon }/{3}$ , and, in particular, $\|p_N - f\|_{[0,r]} < {\varepsilon }/{3}$ . Since $p_N$ is a polynomial, there exists $N^{\prime } \geqslant N$ such that, for any $n \geqslant N^{\prime }$ ,

$$ \begin{align*}\|p_N (x_n) - P_N (x)\|_{\mathcal{A}} < \frac{\varepsilon}{3}.\end{align*} $$

Let $n \geqslant N'$ . By functional calculus,

$$ \begin{align*} \|f(x_n) - p_N(x_n)\|_{\mathcal{A}} =\|f - p_N\|_{\sigma(x_n)}\leqslant \|f-p_N\|_{[0,r]}<\frac{\varepsilon}{3} \end{align*} $$

since $\sigma (x_n) \subseteq [0,r]$ . Similarly,

$$ \begin{align*} \|f(x) - p_N(x)\|_{\mathcal{A}}<\frac{\varepsilon}{3} \end{align*} $$

since $\sigma (x)\subseteq [0,r]$ . By the triangle inequality for the norm,

$$ \begin{align*} \|f(x_n) - f(x)\|_{\mathcal{A}} & \leqslant \|f(x_n) - p_N(x_n)\|_{\mathcal{A}} + \|p_N(x_n) - p(x)\|_{\mathcal{A}} + \|p_N(x) - f(x)\|_{\mathcal{A}} \\ & < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon. \end{align*} $$

Thus, for any $\varepsilon> 0$ , we can find $N^{\prime }$ such that $\|f(x_n) - f(x)\|_{\mathcal {A}} \leqslant \varepsilon $ for any $n \geqslant N^{\prime }$ . Thus, $(f(x_n))_{n \in {\mathbb {N}}}$ converges to $f(x)$ in the $C^*$ -norm.

Proof. Note that $|x| = \sqrt {x^* x}$ and $|x_n| = \sqrt {x_n^* x_n}$ for every $n \in {\mathbb {N}}$ . For any $n \in {\mathbb {N}}$ , let $a_n = x_n^* x_n$ and $a = x^* x$ . Since multiplication and the adjoint are continuous in the $C^*$ -norm, $(a_n)_{n \in {\mathbb {N}}}$ converges to a in the $C^*$ -norm. For any $n \in {\mathbb {N}}$ , $|x| = \sqrt {a}$ and $|x_n| = \sqrt {a_n}$ . Therefore, by Lemma 2.1 and continuity of the square root, we know that $(|x_n|)_{n \in {\mathbb {N}}}$ converges to $|x|$ in the $C^*$ -norm.

Proof. If $(x_n)_{n \in {\mathbb {N}}}$ converges to x in the $C^*$ -norm, then, since $g(x) = \sqrt {x}$ is continuous, $(\sqrt {x_n})_{n \in {\mathbb {N}}}$ converges to $\sqrt {x}$ in the $C^*$ -norm by Lemma 2.1.

Since x is positive and fixed, $(\sqrt {x_n}\cdot \sqrt {x})_{n \in {\mathbb {N}}}$ converges to $\sqrt {x} \cdot \sqrt {x}$ in the $C^*$ -norm. By definition, $\sqrt {x} \cdot \sqrt {x} = x$ , so $(\sqrt {x_n}\cdot \sqrt {x})_{n \in {\mathbb {N}}}$ converges in the $C^*$ -norm to x. By Lemma 2.2, this implies that $(|\sqrt {x_n}\cdot \sqrt {x}|)_{n \in {\mathbb {N}}}$ converges in the $C^*$ -norm to $|x|=x$ .

Since $\tau $ is $C^*$ -norm continuous, $(|\sqrt {x_n}\cdot \sqrt {x}|)_{n \in {\mathbb {N}}}$ converging to x in the $C^*$ -norm implies that $(\tau (|\sqrt {x_n}\cdot \sqrt {x}|))_{n \in {\mathbb {N}}}$ converges to $\tau (x)$ . Since $x \in D_{\tau }(A)$ , we have $\tau (x) = 1$ . Hence, $(\tau (|\sqrt {x_n}\cdot \sqrt {x}|))_{n \in {\mathbb {N}}}$ converges to $1$ .

Consider

$$ \begin{align*}(d_{B}^{\tau}(x_n,x))_{n \in {\mathbb{N}}} = \Big(\sqrt{1 - \tau(|\sqrt{x_n}\cdot \sqrt{x}|)}\Big)_{n \in {\mathbb{N}}}.\end{align*} $$

Since $(\tau (|\sqrt {x_n}\cdot \sqrt {x}|))_{n \in {\mathbb {N}}}$ converges to $1$ , it follows that $(d_{B}^{\tau }(x_n,x))_{n \in {\mathbb {N}}} $ converges to $\sqrt {1-1} = 0$ . Hence, $(x_n)_{n \in {\mathbb {N}}}$ converges to x with respect to the Bures metric $d_{B}^{\tau }$ .

Proof. The fact that the topology induced by $d_{C([0,1])}$ is finer is provided by Corollary 2.4. Thus, it remains to show that the topologies are not equal. To accomplish this, we find a sequence in $D_\rho (C([0,1]))$ that converges with respect to $d_B^\rho $ , but does not converge uniformly (that is, with respect to $d_{C([0,1])}$ ).

Let $n \in {\mathbb {N}}$ . Consider

(2.1) $$ \begin{align} f_{n}(x)= \begin{cases} 2nx & \text{if } x \in [0, {1}/{2n}],\\ 1 & \text{if } x \in ({1}/{2n}, 1-{1}/{2n}),\\ 2nx-2n+2 & \text{if } x \in [1-{1}/{2n}, 1], \end{cases} \end{align} $$

defined for all $x \in [0,1]$ (see Figure 1). Note that $f_n \in C([0,1]).$ Define $f:[0,1]\rightarrow {\mathbb {R}}$ by $f(x)=1$ for all $x \in [0,1].$ Note that $f \in D_\rho (C([0,1])).$

Figure 1 $f_1,f_2,f_3$ given by the formula (2.1). (Plotted in GeoGebra).

Next, we check that $f_n\in D_\rho (C([0,1])$ for every $n \in {\mathbb {N}}$ . Let $n\in {\mathbb {N}}$ . First, $f_n\geqslant 0$ by construction. Second,

$$ \begin{align*} \rho(f_n) = \int_{0}^{1} f_n(x) \,dx & = \int_{0}^{{1}/{2n}} 2nx \,dx + \int_{{1}/{2n}}^{1 - {1}/{2n}} 1 \,dx + \int_{1 - {1}/{2n}}^{1} (2nx-2n+2) \,dx \\ &= [nx^2]_0^{{1}/{2n}} + [x]_{{1}/{2n}}^{1 - {1}/{2n}} + [nx^2-2nx+2x]_{1 - {1}/{2n}}^{1} \\ &= \frac{1}{4n} + 1 - \frac{1}{n} + \frac{3}{4n} = 1. \end{align*} $$

Hence, $f_n \in D_\rho (C([0,1])).$

We now prove that the sequence of functions $(f_n)_{n\in {\mathbb {N}}}$ converges to the function $f $ in the Bures metric, $d_B^\rho $ , but fails to converge to f uniformly.

Let $n \in {\mathbb {N}}.$ Then

$$ \begin{align*} \rho(\sqrt{f_n}\sqrt{f}) &= \int_{0}^{{1}/{2n}} (\sqrt{2nx}\cdot 1) \,dx + \int_{{1}/{2n}}^{1 - {1}/{2n}} (\sqrt{1}\cdot 1) \,dx \\ &\quad + \int_{1 - {1}/{2n}}^{1} (\sqrt{2nx-2n+2}\cdot 1) \,dx \\ &= \int_{0}^{{1}/{2n}} \sqrt{2nx} \,dx + \int_{{1}/{2n}}^{1 - {1}/{2n}} 1 \,dx + \int_{1 - {1}/{2n}}^{1} \sqrt{2nx-2n+2} \,dx \\ &= \sqrt{2n} \cdot \bigg[\frac{2}{3} x^{{3}/{2}}\bigg]_0^{{1}/{2n}} + [x]_{{1}/{2n}}^{1 - {1}/{2n}} + \sqrt{2} \cdot \bigg[\frac{2}{3n} \cdot (nx-n+1)^{{3}/{2}}\bigg]_{1 - {1}/{2n}}^{1} \\ &= \frac{1}{3n} + 1 - \frac{1}{n} + \frac{2\sqrt{2}}{3n} - \frac{1}{3n} = 1 - \frac{3-2\sqrt{2}}{3n}. \end{align*} $$

Therefore,

$$ \begin{align*}d_{B}^{\rho}(f_n,f) = \sqrt{1 - \rho(\sqrt{f_n}\sqrt{f})} = \sqrt{1 - \bigg(1 - \frac{3-2\sqrt{2}}{3n}\bigg)} = \sqrt{\frac{3-2\sqrt{2}}{3n}}\end{align*} $$

and $\lim _{n \to \infty }d_{B}^{\rho }(f_n,f) =0$ . Therefore, we have shown that $(f_n)_{n \in {\mathbb {N}}}$ converges to $f $ in the $\rho $ -Bures metric.

Finally, suppose, by way of contradiction, that $(f_n)_{n\in {\mathbb {N}}}$ converges uniformly to $f $ . Then, for any $\delta> 0$ , there exists $M \in {\mathbb {N}}$ such that, for any $n> M$ and any $x \in [0,1]$ , we have $|f_n(x) - f| < \delta $ . Therefore, for some $\delta _0 \in (0,1)$ , there exists $M_0 \in {\mathbb {N}}$ such that, for any $n> M_0$ and any $x \in [0,1]$ , we have $|f_n(x) - f| < \delta _0$ . However, let $x = 1$ . Then $|f_n(1) - f(1)| = |2 - 1| = 1> \delta _0$ , which is a contradiction. Hence, our assumption is false.

Proof. For $a\in {\mathcal {A}}_+$ , there exists $b\in {\mathcal {A}}$ such that $a=b^*b$ by [Reference Davidson3, Lemma I.4.3]. Let $c\in {\mathcal {A}}$ . Since $\tau $ is a trace,

$$ \begin{align*} \varphi_a^\tau(c^*c)= \tau(b^*bc^*c)= \tau(cb^*bc^*)= \tau(cb^*(cb^*)^*)\geqslant 0. \end{align*} $$

Thus, $\varphi $ is a positive linear functional, and, in particular, $\varphi _a^\tau (1_{\mathcal {A}})=\|\varphi _a^\tau \|_{\mathrm {op}}$ by [Reference Davidson3, Lemma I.9.5]. Since $a \in D_\tau ({\mathcal {A}})$ ,

$$ \begin{align*} \|\varphi_a^\tau\|_{\mathrm{op}}=\tau(a1_{\mathcal{A}})=\tau(a)=1. \end{align*} $$

Thus, $\varphi _a^\tau $ is a state, and therefore, $\Phi _\tau $ is well defined.

Next, let $a,a'\in {\mathcal {A}}_+$ such that $\Phi _\tau (a)=\Phi _\tau (a')$ . Hence, $\varphi _a^\tau ((a-a')^*)=\varphi ^\tau _{a'}((a-a')^*)$ . Thus, $\tau (a(a-a')^*)=\tau (a'(a-a')^*)$ and so $\tau ((a-a')(a-a')^*)=0$ . Therefore, $a-a'=0$ as $\tau $ is faithful, that is, $a=a'.$ Thus, $\Phi _\tau $ is injective.

Proof. Let $(a_n)_{n \in {\mathbb {N}}}$ be a sequence in $D_\tau ({\mathcal {A}})$ that converges to $a\in D_\tau ({\mathcal {A}})$ with respect to $d_{\mathcal {A}}$ . Then, for each $b \in {\mathcal {A}}$ ,

$$ \begin{align*} \lim_{n\to \infty} \varphi^ \tau_{a_n}(b)=\lim_{n\to \infty} \tau(a_nb)=\tau(ab)=\varphi_a^\tau(b) \end{align*} $$

since multiplication is continuous and $\tau $ is continuous. Since convergence in $\mathrm {mk}_L$ is equivalent to weak* convergence by definition, the proof is complete.

Proof. We only need to show that $D_\tau ({\mathcal {A}})$ is closed and bounded with respect to the $C^*$ -norm. First, it is closed since $\tau $ is continuous with respect to the $C^*$ -norm.

Since ${\mathcal {A}}$ is finite dimensional, the norms $\|\cdot \|_{\mathcal {A}}$ and $\|\cdot \|_\tau $ are equivalent, where $\|a\|_\tau =\sqrt {\tau (a^*a)}$ for $ a\in {\mathcal {A}}$ . Now, let $a \in D_\tau ({\mathcal {A}})$ and let $\sqrt {a}$ denote its unique positive square root. By the $C^*$ -identity,

$$ \begin{align*} \sqrt{\|a\|_{\mathcal{A}}} = \sqrt{\|(\sqrt{a})^*\sqrt{a}\|_{\mathcal{A}}} & =\|\sqrt{a}\|_{\mathcal{A}} \\ &\leqslant \alpha \|\sqrt{a}\|_\tau =\alpha\sqrt{\tau((\sqrt{a})^*\sqrt{a})}=\alpha\sqrt{\tau(a)}=\alpha, \end{align*} $$

since $\tau (a)=1$ . So $\|a\|_{\mathcal {A}}\leqslant \alpha ^2$ . Hence, $D_\tau ({\mathcal {A}})$ is bounded and thus is compact by the Heine–Borel theorem.

Proof. The proof that $\Phi _\tau $ is continuous from $(D_\tau ({\mathcal {A}}), d_{\mathcal {A}})$ to $(D_\tau ({\mathcal {A}}),d^\tau _L )$ is the same as the proof of Proposition 3.3. Since $(D_\tau ({\mathcal {A}}), d_{\mathcal {A}})$ is compact by Proposition 3.4, $\Phi _\tau $ is a homeomorphism onto its image with respect to the weak* topology on $S({\mathcal {A}})$ .

Since convergence in $\mathrm {mk}_L$ is equivalent to weak* convergence by definition, $ d_{\mathcal {A}} $ and $ d^\tau _L $ are topologically equivalent.

Proof. Consider

$$ \begin{align*} \mathrm{id}_{D_\tau({\mathcal{A}})}: (D_\tau({\mathcal{A}}), d_{\mathcal{A}}) \longrightarrow (D_\tau({\mathcal{A}}), d_B^\tau). \end{align*} $$

By Theorem 2.3, convergence in $d_{\mathcal {A}}$ implies convergence in $d^\tau _B$ . Hence, $ \mathrm {id}_{D_\tau ({\mathcal {A}})}$ is continuous, and since $(D_\tau ({\mathcal {A}}), d_{\mathcal {A}})$ is compact, $ \mathrm {id}_{D_\tau ({\mathcal {A}})}$ is a homeomorphism. Hence, $d_{\mathcal {A}}$ and $d_B^\tau $ are topologically equivalent.

Proof. This follows immediately from Theorems 3.6 and 3.5 and since homeomorphism is an equivalence relation.

Proof. First, let $x \in D_\tau ({\mathbb {C}}^2)$ and let $x'\in {\mathbb {C}}^2$ . Then

$$ \begin{align*} \Phi_\tau(x)(x')=\phi^\tau_x(x')=\tau(xx')=x_1x^{\prime}_1+x_2x_2'. \end{align*} $$

Now, assume that $y \in D_\tau ({\mathbb {C}}^2)$ . Since $\tau (x)=1=\tau (y)$ , we have $x_1+x_2=1=y_1+y_2$ and so $x_1-y_1=y_2-x_2$ . Hence, if $L^B(x')\leqslant 1$ ,

$$ \begin{align*} |\varphi^\tau_x(x')-\varphi^\tau_y(x')|& =|x_1x_1'+x_2x_2'-y_1x_1'-y_2x_2'|\\ & = |(x_1-y_1)x_1'+(x_2-y_2)x_2'|\\ &= |(x_1-y_1)x_1'-(y_2-x_2)x_2'|\\ & =|(x_1-y_1)x_1'-(x_1-y_1)x_2'|\\ & = |(x_1-y_1)(x_1'-x_2')|\\ & =|x_1-y_1|\cdot |x_1'-x_2'|\\ & \leqslant |x_1-y_1|. \end{align*} $$

Now, consider $x'=(1,0)$ . Then

$$ \begin{align*} |\varphi^\tau_x(x')-\varphi^\tau_y(x')|=|x_1-y_1|. \end{align*} $$

Thus,

$$ \begin{align*} d_{L^B}^\tau(x,y)=\mathrm{sup} \{ |\varphi^\tau_x(x')-\varphi^\tau_y(x')| : L^B(x')\leqslant 1\}=|x_1-y_1|. \end{align*} $$

We also note that

$$ \begin{align*} \|x-y\|_{{\mathbb{C}}^2}=\max\{ |x_1-y_1|, |x_2-y_2|\}=\max\{|x_1-y_1|, |x_1-y_1|\}=|x_1-y_1|, \end{align*} $$

as desired.

Proof. Corollary 3.7 provides topological equivalence.

Consider $x=(1,0)\in D_\tau ({\mathbb {C}}^2)$ . Let $y=(y_1,y_2)\in D_\tau ({\mathbb {C}}^2)$ . By Theorem 4.1,

$$ \begin{align*}d_{L^B}^\tau(x,y)=|1-y_1|.\end{align*} $$

Also,

$$ \begin{align*} d_B^{\tau}(x,y)=\sqrt{1-\tau(|\sqrt{x}\sqrt{y}|)}=\sqrt{1-\sqrt{y_1}}.\end{align*} $$

Now, consider the ratio

$$ \begin{align*}\frac{d_B^\tau(x,y)}{d_{L^B}^\tau(x,y)}=\frac{\sqrt{1-\sqrt{y_1}}}{|1-y_1|}.\end{align*} $$

We have $\lim _{y_1\to 1^-}\sqrt {1-\sqrt {y_1}}=0$ and $\lim _{y_1\to 1^-}1-y_1=0$ . Further,

$$ \begin{align*}\frac{d}{dy_1}\Big(\sqrt{1-\sqrt{y_1}}\Big)=\frac{-1}{4\sqrt{y_1}\sqrt{1-\sqrt{y_1}}}, \quad \frac{d}{dy_1}(1-y_1)=-1\end{align*} $$

and

$$ \begin{align*} \lim_{y_1\to1^-} \frac{\frac{-1}{4\sqrt{y_1}\sqrt{1-\sqrt{y_1}}}}{-1}=\lim_{y_1\to1^-}\frac{1}{4\sqrt{y_1}\sqrt{1-\sqrt{y_1}}}=\infty. \end{align*} $$

Thus, by L’Hopital’s rule,

$$ \begin{align*} \lim_{y_1\to 1^-}\frac{d_B^\tau(x,y)}{d_{L^B}^\tau(x,y)}=\infty. \end{align*} $$

Hence, the set

$$ \begin{align*} \bigg\{ \frac{d_B^\tau(x,y)}{d_{L^B}^\tau(x,y)}: x,y \in D_\tau({\mathbb{C}}^2), x \neq y\bigg\} \end{align*} $$

is unbounded, and so the metrics $d_B^\tau $ and $ d_{L^B}^\tau $ are not metric equivalent.