1 Introduction

Umegaki divergence is a fundamental concept in quantum information theory and quantum computation. It measures the distinguishability of two quantum states and was first introduced in [1] (also see the survey [2] and references therein). Various generalizations of Umegaki divergences have been introduced and extensively studied in the past few decades. Petz defined the quasi quantum divergence via relative modular operator [3]. Wilde [4] introduced the optimized quantum f-divergence that further generalized the definition by Petz. Another notable generalization is sandwiched \(\alpha \)-Rényi divergence, introduced independently by Wilde et al. [5] and Müller-Lennert et al. [6]. Very recently Hirche and Tomamichel [7] introduced a new family of quantum f-divergence by using the quantum version of hockey-stick divergence.

A crucial property of Umegaki divergence is monotonicity under the actions of quantum operations (also referred to as data processing inequality in the literature), stating that the distinguishability of two quantum states does not increase after undergoing a quantum channel. Monotonicity plays an important role in the study of quantum channel capacity, quantum machine learning, quantum hypothesis testing, etc. The aforementioned generalizations of Umegaki divergence all satisfy the monotonicity under the actions of quantum channels. It is noteworthy that the monotonicity of Umegaki divergence and sandwiched \(\alpha \)-Rényi divergence has been generalized to the positive trace preserving maps in [8] for \(\alpha \in (1,\infty )\) based on [9] and [10] for \(\alpha \in (\frac{1}{2},1)\).

Wilde [4] introduced the optimized quantum f-divergence as a unified definition of various divergences, including Umegaki divergence and sandwiched \(\alpha \)-Rényi divergence. The data processing inequality of optimized quantum f-divergence was proved in [4] by demonstrating invariance under isometries and monotonicity under taking a partial trace. The recoverability and extension of optimized quantum f-divergence to the general von Neumann algebraic setting was accomplished in [11]. In this paper, we prove the monotonicity of optimized quantum f-divergence for positive trace preserving maps satisfying a Schwarz inequality. The main tool is inspired by the well-known Petz recovery map [12]. It remains open whether the optimized quantum f-divergence is monotone under the actions of any positive trace preserving maps.

1.1 Notations

We use \({\mathcal {B}}({\mathcal {H}})\) for the linear space of bounded linear operators defined in the complex Hilbert space \({\mathcal {H}}\). We use \({\mathcal {B}}_{sa}({\mathcal {H}})\subset {\mathcal {B}}({\mathcal {H}})\) for the space of self-adjoint operators. We use \({\mathcal {B}}_{+}( {\mathcal {H}})\subset {\mathcal {B}}({\mathcal {H}})\) for the space of positive definite operators. We use \({{\,\textrm{tr}\,}}\) as the trace on \({\mathcal {B}}({\mathcal {H}})\) and the \(\langle A,B\rangle ={{\,\textrm{tr}\,}}(A^{*}B)\) as the Hilbert–Schmidt inner product.

2 Monotonicity of quantum optimized f-divergence

Let us recall the definition of the optimized quantum f-divergence; see [4] for more explanation and properties.

Definition 2.1

Let f be a function with domain \((0,\infty )\) and range \({\mathbb {R}}\). For positive semi-definite operators \(\rho ,\sigma \in {{\mathcal {B}}({\mathcal {H}}_{S})}\), we define the optimized quantum f-divergence as

$$\begin{aligned} {\tilde{Q}}_{f}(\rho \Vert \sigma )=\sup _{\tau>0,{{{\,\textrm{tr}\,}}}(\tau )\le 1,\epsilon >0}{\tilde{Q}}_{f}\big (\rho \Vert \sigma +\epsilon \Pi _{\sigma }^{\perp };\tau \big ), \end{aligned}$$

where \({\tilde{Q}}_{f}(\rho \Vert \omega ;\tau )\) is defined for positive definite \(\omega ,\tau \in {{\mathcal {B}}_{+}({\mathcal {H}}_{S})}\) as

$$\begin{aligned} {\tilde{Q}}_{f}(\rho \Vert \omega ;\tau )&=\langle \phi ^{\rho }|_{S{\hat{S}}}f\big (\tau _{S}^{-1}\otimes \omega ^{T}_{{\hat{S}}}\big )| \phi ^{\rho }\rangle _{S{\hat{S}}}\hspace{.1cm},\nonumber \\ | \phi ^{\rho }\rangle _{S{\hat{S}}}&=\big (\rho _{S}^{\frac{1}{2}}\otimes I_{{\hat{S}}}\big )| \Gamma \rangle _{S{\hat{S}}}\hspace{.1cm}. \end{aligned}$$
(2.1)

In the above, \(\Pi _{\sigma }^{\perp }\) denotes the projection onto the kernel of \(\sigma \), \({\mathcal {H}}_{S}\) is an auxiliary Hilbert space isomorphic to \({\mathcal {H}}_{S}\),

$$\begin{aligned} | \Gamma \rangle _{S{\hat{S}}}=\sum _{i=1}^{|S|}| i\rangle _{S}| i\rangle _{{\hat{S}}}, \end{aligned}$$

for orthonormal bases \(\{| i\rangle _{S}\}\) and \(\{| i\rangle _{{\hat{S}}}\}\), and the T superscript indicates transpose with respect to the basis \(\{| i\rangle _{{\hat{S}}}\}\).

Wilde [4] found an equivalent formulation of (2.1) for invertible \(\sigma \):

$$\begin{aligned} {\tilde{Q}}_{f}(\rho \Vert \sigma ;\tau )=\left\langle \rho ^{\frac{1}{2}}, f(\Delta (\sigma ,\tau ))\big (\rho ^{\frac{1}{2}}\big )\right\rangle , \end{aligned}$$

where \(\Delta (\sigma ,\tau )(X):=\sigma X \tau ^{-1}\) is the relative modular operator. For invertible \(\sigma \), the optimized quantum f-divergence can be rewritten as

$$\begin{aligned} {\tilde{Q}}_{f}(\rho \Vert \sigma )=\sup _{\tau >0,{{{\,\textrm{tr}\,}}}(\tau )\le 1}{\tilde{Q}}_{f}(\rho \Vert \sigma {;\,}\tau )\hspace{.1cm}. \end{aligned}$$
(2.2)

Recall that a function \(f: J\subset {\mathbb {R}}\rightarrow {\mathbb {R}}\) is said to be operator anti-monotone if \(A\ge B\) for any \(A,B\in {{\mathcal {B}}_{sa}({\mathcal {H}})}\) with spectra in J implies \(f(A)\le f(B)\). Now, we are ready to state our main theorem.

Theorem 2.2

Let \(\Phi : {B({\mathcal {H}}_{A})\rightarrow B({\mathcal {H}}_{B})}\) be a positive trace preserving linear map satisfying the Schwarz inequality

$$\begin{aligned} \Phi ^{*}(X)\Phi ^{*}(\tau )^{-1}\Phi ^{*}(X^{*})\le \Phi ^{*}(X\tau ^{-1}X^{*}) \end{aligned}$$
(2.3)

for any \(X\in {{\mathcal {B}}({\mathcal {H}}_{B})}\) and \(\tau \in {{\mathcal {B}}_{+}({\mathcal {H}}_{B})}\). Let \(f:(0,\infty )\rightarrow {\mathbb {R}}\) be operator anti-monotone. Let \(\rho \in {{\mathcal {B}}({\mathcal {H}}_{A})}, \sigma \in {{\mathcal {B}}_{+}({\mathcal {H}}_{A})}\) be two quantum states. If \(\Phi (\rho )>0\) and \(\Phi (\sigma )>0\), then we have

$$\begin{aligned} {\tilde{Q}}_{f}(\rho \Vert \sigma )\ge {\tilde{Q}}_{f}(\Phi (\rho )\Vert \Phi (\sigma )). \end{aligned}$$

Proof

We first prove monotonicity for invertible \(\rho \). Then, we modify the proof to show the monotonicity for \(\rho \) which is not necessarily invertible.

Case 1: \(\rho \) is invertible. As mentioned in the introduction, the main tool is the Petz recovery map. We actually only need the partial isometry: \(V_{\rho }:{{\mathcal {B}}({\mathcal {H}}_{B})\rightarrow {\mathcal {B}}({\mathcal {H}}_{A})}\) by

$$\begin{aligned} V_{\rho }(X):=\Phi ^{*}\bigg (X\Phi (\rho )^{-\frac{1}{2}}\bigg )\rho ^{\frac{1}{2}}. \end{aligned}$$

Then, \(V_{\rho }(\Phi (\rho )^{\frac{1}{2}})=\rho ^{\frac{1}{2}}\). For any \(0<\omega \) with \({{\,\textrm{tr}\,}}(\omega )\le 1\), let

$$\begin{aligned} \tau :=\rho ^{\frac{1}{2}}\Phi ^{*}\bigg (\Phi (\rho )^{-\frac{1}{2}}\omega \Phi (\rho )^{-\frac{1}{2}}\bigg )\rho ^{\frac{1}{2}}. \end{aligned}$$
(2.4)

Then \(\tau \) is invertible since \(\Phi ^{*}\) is unital. Now, we compute the trace of \(\tau \):

$$\begin{aligned} {{\,\textrm{tr}\,}}(\tau )&= {{\,\textrm{tr}\,}}\bigg (\rho \Phi ^{*}\bigg (\Phi (\rho )^{-\frac{1}{2}}\omega \Phi (\rho )^{-\frac{1}{2}}\bigg )\bigg )\\&= {{\,\textrm{tr}\,}}\bigg (\Phi (\rho )\Phi (\rho )^{-\frac{1}{2}}\omega \Phi (\rho )^{-\frac{1}{2}}\bigg )={{\,\textrm{tr}\,}}(\omega )\le 1. \end{aligned}$$

We claim:

$$\begin{aligned} V_{\rho }^{*}\Delta (\sigma ,\tau ) V_{\rho }&\le \Delta (\Phi (\sigma ),\omega )\hspace{.1cm}, \end{aligned}$$
(2.5)
$$\begin{aligned} f(V_{\rho }^{*}\Delta (\sigma ,\tau ) V_{\rho })&\ge f(\Delta (\Phi (\sigma ),\omega )). \end{aligned}$$
(2.6)

Indeed, for any \(X\in {{\mathcal {B}}({\mathcal {H}}_{B})}\),

$$\begin{aligned}&\langle X, V_{\rho }^{*}\Delta (\sigma , \tau ) V_{\rho } (X)\rangle \nonumber \\&\quad = \langle V_{\rho }(X),\Delta (\sigma ,\tau )V_{\rho }(X) \rangle \nonumber \\&\quad = {{\,\textrm{tr}\,}}\bigg (\rho ^{\frac{1}{2}}\Phi ^{*}\big (\Phi (\rho )^{-\frac{1}{2}}X^{*}\big )\sigma \Phi ^{*}\big (X\Phi (\rho )^{-\frac{1}{2}}\big )\rho ^{\frac{1}{2}}\rho ^{-\frac{1}{2}}\Phi ^{*}\big (\Phi (\rho )^{-\frac{1}{2}}\omega \Phi (\rho )^{-\frac{1}{2}}\big )^{-1}\rho ^{-\frac{1}{2}}\bigg )\nonumber \\&\quad = {{\,\textrm{tr}\,}}\bigg (\sigma \Phi ^{*}\big (X\Phi (\rho )^{-\frac{1}{2}}\big )\Phi ^{*}\big (\Phi (\rho )^{-\frac{1}{2}}\omega \Phi (\rho )^{-\frac{1}{2}}\big )^{-1}\Phi ^{*}\big (\Phi (\rho )^{-\frac{1}{2}}X^{*}\big )\bigg )\nonumber \\&\quad \le {{\,\textrm{tr}\,}}(\sigma \Phi ^{*}(X\omega ^{-1} X^{*}))\nonumber \\&\quad = {{{\,\textrm{tr}\,}}(\Phi (\sigma )X\omega ^{-1}X^{*})}\nonumber \\&\quad = \langle X, \Delta (\Phi (\sigma ),\omega )(X) \rangle \hspace{.1cm}, \end{aligned}$$
(2.7)

where the inequality (2.7) follows from the Schwarz inequality (2.3). The inequality (2.6) is an immediate application of (2.5) since f is operator anti-monotone. Now, we show

$$\begin{aligned} {\tilde{Q}}_{f}(\rho \Vert \sigma ;\tau )\ge {\tilde{Q}}_{f}(\Phi (\rho )\Vert \Phi (\sigma ) {;\,}\omega )\hspace{.1cm}. \end{aligned}$$
(2.8)

For any \(\omega >0\) with \({{\,\textrm{tr}\,}}(\omega )\le 1\) and \(\tau \) defined in (2.4), we have:

$$\begin{aligned} {\tilde{Q}}_{f}(\rho \Vert \sigma ;\tau )&= \left\langle V_{\rho }\big (\Phi (\rho )^{\frac{1}{2}}\big ), f(\Delta (\sigma ,\tau )) \big ( V_{\rho }(\Phi (\rho )^{\frac{1}{2}}\big ) \big )\right\rangle \end{aligned}$$
(2.9)
$$\begin{aligned}&= \left\langle \Phi (\rho )^{\frac{1}{2}}, (V_{\rho }^{*}\circ f(\Delta (\sigma ,\tau )) \circ V_{\rho }) \big (\Phi (\rho )^{\frac{1}{2}}\big ) \right\rangle \nonumber \\&\ge \left\langle \Phi (\rho )^{\frac{1}{2}}, f(V_{\rho }^{*}\Delta (\sigma ,\tau )V_{\rho }) \big (\Phi (\rho )^{\frac{1}{2}}\big ) \right\rangle \end{aligned}$$
(2.10)
$$\begin{aligned}&\ge \left\langle \Phi (\rho )^{\frac{1}{2}}, f(\Delta (\Phi (\sigma ),\omega )) \big (\Phi (\rho )^{\frac{1}{2}}\big ) \right\rangle \nonumber \\&= {\tilde{Q}}_{f}(\Phi (\rho )\Vert \Phi (\sigma ){;\,}\omega ) \hspace{.1cm}. \end{aligned}$$
(2.11)

The equality (2.9) follows from the observation that \(V_{\rho }(\Phi (\rho )^{\frac{1}{2}})=\rho ^{\frac{1}{2}}\). The inequality (2.11) follows from (2.6). The inequality (2.10) is a direct application of the operator Jensen inequality [13] for operator convex functions. Here, we use the facts that \(f:(0,\infty )\rightarrow {\mathbb {R}}\) is operator concave if \(f:(0,\infty )\rightarrow {\mathbb {R}}\) is operator monotone [14, Theorem III.2] (also see the proof of [15, Theorem V.2.5]) and that f is operator convex if and only if \(-f\) is operator concave. Then, we have the monotonicity:

$$\begin{aligned} {\tilde{Q}}_{f}(\rho \Vert \sigma )\ge&\sup _{\tau \text { defined by} (2.4)}{\tilde{Q}}_{f}(\rho \Vert \sigma ;\tau ) \end{aligned}$$
(2.12)
$$\begin{aligned} \ge&\sup _{\omega >0,{{{\,\textrm{tr}\,}}}(\omega )\le 1} {\tilde{Q}}_{f}(\Phi (\rho )\Vert \Phi (\sigma );\omega ) \end{aligned}$$
(2.13)
$$\begin{aligned}&= {\tilde{Q}}_{f}(\Phi (\rho )\Vert \Phi (\sigma ))\hspace{.1cm}, \end{aligned}$$
(2.14)

where we use the equivalent definition (2.2) in (2.12) and (2.14). The inequality (2.13) follows from (2.8).

Case 2: \(\rho \) is not necessarily invertible. We use \(\rho ^{-\frac{1}{2}}\) to denote the square-root inverse on the supports of \(\rho \). Let \(V_{\rho }:{{\mathcal {B}}({\mathcal {H}}_{B})\rightarrow {\mathcal {B}}({\mathcal {H}}_{A})}\) be defined as in Case 1: \(V_{\rho }(X)=\Phi ^{*}\left( X\Phi (\rho )^{-\frac{1}{2}}\right) \rho ^{\frac{1}{2}}\hspace{.1cm}.\) For any \(0<\omega \) with \({{\,\textrm{tr}\,}}(\omega )\le 1\), let

$$\begin{aligned} \tau _{\delta }:=(1-\delta )\rho ^{\frac{1}{2}}\Phi ^{*}\left( \Phi (\rho )^{-\frac{1}{2}}\omega \Phi (\rho )^{-\frac{1}{2}}\right) \rho ^{\frac{1}{2}}+\delta {\tilde{\omega }}, \end{aligned}$$
(2.15)

where \(\delta \in (0,1)\) and \({\tilde{\omega }}\in {{\mathcal {B}}_{+}({\mathcal {H}}_{A})}\) is any invertible quantum state and \({{\,\textrm{tr}\,}}(\tau _{\delta })\le 1\). For simplicity, we use the following abbreviation

$$\begin{aligned} \bar{\omega }:=\Phi ^{*}\left( \Phi (\rho )^{-\frac{1}{2}}\omega \Phi (\rho )^{-\frac{1}{2}}\right) \hspace{.1cm}. \end{aligned}$$

It is obvious that \(\bar{\omega }\) is also invertible since \(\omega \) and \(\Phi (\rho )\) are invertible and \(\Phi ^{*}\) is unital. We first claim that

$$\begin{aligned} \left\| \tau _{\delta }^{-\frac{1}{2}}\rho ^{\frac{1}{2}}\bar{\omega }\rho ^{\frac{1}{2}}\tau _{\delta }^{-\frac{1}{2}}\right\| _{\infty }\le \frac{1}{1-\delta }. \end{aligned}$$
(2.16)

Indeed, we have

$$\begin{aligned}&\left\| \tau _{\delta }^{-\frac{1}{2}}\rho ^{\frac{1}{2}}\bar{\omega }\rho ^{\frac{1}{2}}\tau _{\delta }^{-\frac{1}{2}}\right\| _{\infty }\\&\quad = \left\| \left( (1-\delta )\rho ^{\frac{1}{2}}\bar{\omega }\rho ^{\frac{1}{2}}+\delta {\tilde{\omega }}\right) ^{-\frac{1}{2}}\rho ^{\frac{1}{2}}\bar{\omega }\rho ^{\frac{1}{2}}\left( (1-\delta )\rho ^{\frac{1}{2}}\bar{\omega }\rho ^{\frac{1}{2}}+\delta {\tilde{\omega }}\right) ^{-\frac{1}{2}}\right\| _{\infty }\\&\quad = \inf \left\{ a\in {\mathbb {R}}_{+}:\,\left( (1-\delta )\rho ^{\frac{1}{2}}\bar{\omega }\rho ^{\frac{1}{2}}+\delta {\tilde{\omega }}\right) ^{-\frac{1}{2}}\rho ^{\frac{1}{2}}\bar{\omega }\rho ^{\frac{1}{2}}\left( (1-\delta )\rho ^{\frac{1}{2}}\bar{\omega }\rho ^{\frac{1}{2}}+\delta {\tilde{\omega }}\right) ^{-\frac{1}{2}}\le a\,\text {id}_{\text {A}}\right\} \\&\quad = \inf \left\{ a\in {\mathbb {R}}_{+}:\, \rho ^{\frac{1}{2}}\bar{\omega }\rho ^{\frac{1}{2}}\le a\,\left( (1-\delta )\rho ^{\frac{1}{2}}\bar{\omega }\rho ^{\frac{1}{2}}+\delta {\tilde{\omega }} \right) \right\} \\&\quad \le \frac{1}{1-\delta }\hspace{.1cm}. \end{aligned}$$

Then,

$$\begin{aligned}&\langle X, V_{\rho }^{*}\Delta (\sigma , \tau _{\delta }) V_{\rho } (X)\rangle \nonumber \\&\quad = \langle V_{\rho }(X),\Delta (\sigma ,\tau _{\delta })V_{\rho }(X) \rangle \nonumber \\&\quad = {{\,\textrm{tr}\,}}\Big (\rho ^{\frac{1}{2}}\Phi ^{*}(\Phi (\rho )^{-\frac{1}{2}}X^{*})\sigma \Phi ^{*}(X\Phi (\rho )^{-\frac{1}{2}}) \rho ^{\frac{1}{2}}\tau _{\delta }^{-1}\Big )\nonumber \\&\quad = {{\,\textrm{tr}\,}}\Big (\Phi ^{*}(\Phi (\rho )^{-\frac{1}{2}}X^{*})\sigma \Phi ^{*}(X\Phi (\rho )^{-\frac{1}{2}}) \bar{\omega }^{-\frac{1}{2}}\bar{\omega }^{\frac{1}{2}}\rho ^{\frac{1}{2}}\tau _{\delta }^{-1}\rho ^{\frac{1}{2}}\bar{\omega }^{\frac{1}{2}}\bar{\omega }^{-\frac{1}{2}}\Big )\nonumber \\&\quad \le \left\| \bar{\omega }^{\frac{1}{2}}\rho ^{\frac{1}{2}}\tau _{\delta }^{-1}\rho ^{\frac{1}{2}}\bar{\omega }^{\frac{1}{2}}\right\| _{\infty }{{\,\textrm{tr}\,}}\Big ( \bar{\omega }^{-\frac{1}{2}}\Phi ^{*}(\Phi (\rho )^{-\frac{1}{2}}X^{*})\sigma \Phi ^{*}(X\Phi (\rho )^{-\frac{1}{2}}) \bar{\omega }^{-\frac{1}{2}}\Big )\nonumber \\&\quad = \left\| \tau _{\delta }^{-\frac{1}{2}}\rho ^{\frac{1}{2}}\bar{\omega }\rho ^{\frac{1}{2}}\tau _{\delta }^{-\frac{1}{2}}\right\| _{\infty }{{\,\textrm{tr}\,}}\Big ( \bar{\omega }^{-\frac{1}{2}}\Phi ^{*}(\Phi (\rho )^{-\frac{1}{2}}X^{*})\sigma \Phi ^{*}(X\Phi (\rho )^{-\frac{1}{2}}) \bar{\omega }^{-\frac{1}{2}}\Big ) \end{aligned}$$
(2.17)
$$\begin{aligned}&\quad \le \frac{1}{1-\delta }{{\,\textrm{tr}\,}}\Big ( \bar{\omega }^{-\frac{1}{2}}\Phi ^{*}(\Phi (\rho )^{-\frac{1}{2}}X^{*})\sigma \Phi ^{*}(X\Phi (\rho )^{-\frac{1}{2}}) \bar{\omega }^{-\frac{1}{2}}\Big ) \end{aligned}$$
(2.18)
$$\begin{aligned}&\quad \le \frac{1}{1-\delta }{{\,\textrm{tr}\,}}( \sigma \Phi ^{*}(X^{*}\omega ^{-1} X))\nonumber \\&\quad = \frac{1}{1-\delta }\langle X,\Delta (\Phi (\sigma ), \omega ) (X)\rangle \hspace{.1cm}. \end{aligned}$$
(2.19)

Then, equality (2.17) follows from the fact that \(\Vert XX^{*}\Vert _{\infty }=\Vert X^{*}X\Vert _{\infty }\). The inequality (2.18) follows from (2.16). The inequality (2.19) holds by the Schwarz inequality (2.3). So

$$\begin{aligned} V_{\rho }^{*}\Delta (\sigma ,\tau _{\delta })V_{\rho }\le \frac{1}{1-\delta } \Delta (\Phi (\sigma ),\omega ) \end{aligned}$$

and

$$\begin{aligned} f(V_{\rho }^{*}\Delta (\sigma ,\tau _{\delta })V_{\rho })\ge f(\frac{1}{1-\delta } \Delta (\Phi (\sigma ),\omega ))\hspace{.1cm}. \end{aligned}$$
(2.20)

Combining the inequalities above, we have:

$$\begin{aligned} {\tilde{Q}}_{f}(\rho \Vert \sigma )&\ge \sup _{\tau \text { defined in} (2.15)} {\tilde{Q}}_{f}(\rho \Vert \sigma {;}\,\tau _{\delta })\nonumber \\&=\left\langle V_{\rho }\big (\Phi (\rho )^{\frac{1}{2}}\big ),f(\Delta (\sigma ,\tau _{\delta }))V_{\rho }\big (\Phi (\rho )^{\frac{1}{2}}\big )\right\rangle \end{aligned}$$
(2.21)
$$\begin{aligned}&=\left\langle \Phi (\rho )^{\frac{1}{2}}, V_{\rho }^{*}f(\Delta (\sigma ,\tau _{\delta }))V_{\rho }\big (\Phi (\rho )^{\frac{1}{2}}\big )\right\rangle \nonumber \\&\ge \left\langle \Phi (\rho )^{\frac{1}{2}}, f\big (V_{\rho }^{*}\Delta (\sigma ,\tau _{\delta }\big )V_{\rho })\Phi (\rho )^{\frac{1}{2}}\right\rangle \end{aligned}$$
(2.22)
$$\begin{aligned}&\ge \left\langle \Phi (\rho )^{\frac{1}{2}}, f\bigg (\frac{1}{1-\delta }\Delta (\Phi (\sigma ),\omega )\bigg )\Phi (\rho )^{\frac{1}{2}}\right\rangle \hspace{.1cm}, \end{aligned}$$
(2.23)

where (2.21) follows from the observation that \(V_{\rho }(\Phi (\rho )^{\frac{1}{2}})=\rho ^{\frac{1}{2}}\) and (2.23) is from (2.20). Again (2.22) is an application of Jensen operator inequality. Taking the limit \(\delta \rightarrow 0^{+}\), the inequality

$$\begin{aligned} {\tilde{Q}}_{f}(\rho \Vert \sigma )\ge \langle \Phi (\rho )^{\frac{1}{2}}, f(\Delta (\Phi (\sigma ),\omega ))\Phi (\rho )^{\frac{1}{2}}\rangle \end{aligned}$$

holds for any invertible \(\omega >0\) with \({{\,\textrm{tr}\,}}(\omega )\le 1\) by the continuity of f. Similar to Case 1, we obtain the monotonicity by the definition (2.2). \(\square \)

The map \(\Phi \) is not necessarily a quantum channel. For example, let \(\Phi ^{*}(\rho )=\rho ^{T}\) be the transpose map, then \(\Phi \) satisfy the properties in Theorem 2.2. It is well known that \(\Phi \) is not completely positive. It is worth mentioning that any 2-positive map \(\Phi ^{*}\) satisfies the Schwarz type inequality, which was mentioned in the email from Mark M. Wilde. We need the following fact [16]:

Fact

Let A and C be positive semi-definite matrices and A be invertible. Then, \(\left( \begin{matrix} A &{} B\\ B^{*}&{} C\end{matrix}\right) \ge 0\) if and only if \(C\ge B^{*}A^{-1}B\).

By the fact \(\left( \begin{matrix} A &{} B\\ B^{*} &{} B^{*}A^{-1}B \end{matrix} \right) \ge 0\). For a 2-positive \(\Phi ^{*}\), we have \(\left( \begin{matrix} \Phi ^{*}(A) &{} \Phi ^{*}(B)\\ \Phi ^{*}(B^{*}) &{} \Phi ^{*}(B^{*}A^{-1}B) \end{matrix} \right) \ge 0\). Using the fact again yields \(\Phi ^{*}(B^{*}A^{-1}B)\ge \Phi ^{*}(B^{*})\Phi ^{*}(A)^{-1}\Phi ^{*}(B)\).

Remark 2.3

In Theorem 2.2, three quantum states \(\sigma ,\Phi (\rho ),\Phi (\sigma )\) are all required to be invertible since we are using the equivalent definition (2.2). The invertibility is not necessary in the proof of [4, Proposition 6], where the proof relies on the original definition of optimized quantum f-divergence. It will be interesting to generalize our results to non-invertible quantum states and operator convex functions (for example, see [17]). It also remains open whether we can prove the monotonicity of the optimized quantum f-divergence without the Schwarz inequality (2.3).

Remark 2.4

To remove the condition \(\Phi (\rho )>0\), it is natural to consider

$$\begin{aligned} V_{\rho }(X):=\Phi ^{*}\left( X\Phi (\rho )^{-\frac{1}{2}}+\Pi _{\Phi }^{\perp }\right) \rho ^{\frac{1}{2}}\hspace{.1cm}. \end{aligned}$$

Then, \(V_{\rho }\) is still a partial isometry and \(V_{\rho }(\Phi (\rho )^{\frac{1}{2}})=\rho ^{\frac{1}{2}}\). However, this only works for \(\rho >0\). If \(\rho \) is not invertible, we use \(\tau _{\delta }\) defined as in Case 2. Then, we have

$$\begin{aligned} \frac{1}{1-\delta }{{\,\textrm{tr}\,}}\left( \sigma \Phi ^{*}\left( X\omega ^{-1}X^{*}+\Pi _{\Phi (\rho )}^{\perp }\omega ^{-1}\Pi _{\Phi (\rho )}^{\perp }\right) \right) \end{aligned}$$

instead of the term (2.19). A different choice of \(V_{\rho }\) (and \(\tau _{\delta }\)) might resolve this problem, and we also leave it an open problem.