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The generalized fidelity for quantum states that may be sub-normalized is given by (Defn 3.12)

$$F_{*}(\rho, \tau):=\left(\operatorname{Tr}|\sqrt{\rho} \sqrt{\tau}|+\sqrt{(1-\operatorname{Tr} \rho)(1-\operatorname{Tr} \tau)}\right)^{2},$$

where $|A| = \sqrt{A^\dagger A}$.

Is this generalized fidelity also bounded between $0$ and $1$? How can one see this?

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Things state very clear in your link, I've just quoted the main thing and added some reasonings.

Your link says:

Uhlmann's theorem (Theorem 3.10) adapted to the generalized fidelity states that $$ \begin{aligned} F_{*}(\rho, \tau) &=\max _{\varphi, \vartheta} F_{*}(\varphi, \vartheta)=\max _{\vartheta} F_{*}(\phi, \vartheta), \quad \text { where } \\ \sqrt{F_{*}(\varphi, \vartheta)} &=|\langle\varphi \mid \vartheta\rangle|+\sqrt{(1-\operatorname{Tr} \varphi)(1-\operatorname{Tr} \vartheta)} \end{aligned}\tag{1} $$ and $\varphi$ and $\vartheta$ range over all purifications of $\rho$ and $\tau$, respectively, and $\phi$ is a fixed purification of $\rho$. Moreover, using the operators $\hat{\rho}$ and $\hat{\tau}$ defined in the preceding section, we can write $$ F_{*}(\rho, \tau)=F_{*}(\hat{\rho}, \hat{\tau})=(\operatorname{Tr}|\sqrt{\hat{\rho}} \sqrt{\hat{\tau}}|)^{2}\tag{2} $$

So from eq.(2), we can see that it's obviously bounded by 0 and 1.

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