SlowGuess/NeurIPS_2023_OCR huggingface dataset:
-  $x_{r_1}(y,z) \coloneqq \operatorname{argmin} f(x,y) + \frac{r_1}{2}\|x-z\|^2$ .
-  $x^{*}(z) \coloneqq \operatorname{argmin}_{x \in \mathcal{X}} \max_{y \in \mathcal{Y}} f(x, y) + \frac{r_{1}}{2} \| x - z\|^{2}$ .
$y_{+}(z)\coloneqq \mathrm{proj}_{\mathcal{Y}}(y + \alpha \nabla_yf(x_{r_1}(y,z),y)).$
Proposition 3 Suppose that Assumption 2 holds. Then, we have
$$
\left\| x ^ {*} (z) - x (z, v) \right\| ^ {2} \leq \omega_ {2} \| v - y (z, v) \| ^ {\frac {1}{\theta}},
$$
where  $\omega_{2} := \frac{2r_{2}^{\frac{1}{\theta}}}{\tau(r_{1} - L_{x})}$ .
Proof By the strong convexity of  $\phi(\cdot, z) = \max_{y \in \mathcal{Y}} f(\cdot, y) + \frac{r_1}{2} \| \cdot - z \|^2$ , we have
$$
\begin{array}{l} \max  _ {y \in \mathcal {Y}} f (x (z, v), y) + \frac {r _ {1}}{2} \| x (z, v) - z \| ^ {2} - \max  _ {y \in \mathcal {Y}} f (x ^ {*} (z), y) + \frac {r _ {1}}{2} \| x ^ {*} (z) - z \| ^ {2} \\ \geq \frac {r _ {1} - L _ {x}}{2} \| x ^ {*} (z) - x (z, v) \| ^ {2}. \\ \end{array}
$$
$$
\begin{array}{l} \max  _ {y \in \mathcal {Y}} f (x (z, v), y) + \frac {r _ {1}}{2} \| x (z, v) - z \| ^ {2} - \max  _ {y \in \mathcal {Y}} f (x ^ {*} (z), y) + \frac {r _ {1}}{2} \| x ^ {*} (z) - z \| ^ {2} \\ \leq \max  _ {y \in \mathcal {Y}} f (x (z, v), y) + \frac {r _ {1}}{2} \| x (z, v) - z \| ^ {2} - \left(f (x (z, v), y (z, v)) + \frac {r _ {1}}{2} \| x (z, v) - z \| ^ {2}\right) \\ = \max  _ {y \in \mathcal {Y}} f (x (z, v), y) - f (x (z, v), y (z, v)). \\ \end{array}
$$
Recall that  $x(z,v) = \operatorname*{argmin}_{x\in \mathcal{X}}\max_{y\in \mathcal{Y}}F(x,y,z,v)$ . Then, the first inequality is from the fact that
$$
\begin{array}{l} \max  _ {y \in \mathcal {Y}} f (x ^ {*} (z), y) + \frac {r _ {1}}{2} \| x ^ {*} (z) - z \| ^ {2} \geq \max  _ {y \in \mathcal {Y}} \min  _ {x \in \mathcal {X}} f (x, y) + \frac {r _ {1}}{2} \| x - z \| ^ {2} \\ \geq \min  _ {x \in \mathcal {X}} f (x, y (z, v)) + \frac {r _ {1}}{2} \| x - z \| ^ {2} \\ = f (x (z, v), y (z, v)) + \frac {r _ {1}}{2} \| x (z, v) - z \| ^ {2}. \\ \end{array}
$$
Under Assumption 2, we find that
$$
\begin{array}{l} \max  _ {y \in \mathcal {Y}} f (x (z, v), y) - f (x (z, v), y (z, v)) \leq \frac {1}{\tau} \operatorname {d i s t} (\mathbf {0}, - \nabla_ {y} f (x (z, v), y (z, v)) + \partial \mathbf {1} _ {\mathcal {Y}} (y (z, v))) ^ {\frac {1}{\theta}} \\ \leq \frac {1}{\tau} \| r _ {2} (v - y (z, v)) \| ^ {\frac {1}{\theta}}, \\ \end{array}
$$
where the last inequality is due to the first optimality condition of  $y(z, v) = \operatorname{argmax}_{y \in \mathcal{Y}} f(x(y, z, v), y) - \frac{r_2}{2} \| y - v \|^2$ .
Proposition 4 Suppose that Assumption 3 holds. Then, we have
$$
\left\| x ^ {*} (z) - x (z, v) \right\| ^ {2} \leq \omega_ {3} \| v - y (z, v) \|,
$$
where  $\omega_{3} := \frac{4r_{1}\mathrm{diam}(\mathcal{V})}{r_{1} - L_{x}}$ .
Proof We first define  $\psi(x,y,z) = f(x,y) + \frac{r_1}{2}\|x - z\|^2$  and  $Y(z) = \operatorname{argmax}_{y \in \mathcal{V}} \min_{x \in \mathcal{X}} \psi(x,y,z) = \operatorname{argmax}_{y \in \mathcal{V}} \psi(x_{r_1}(y,z),y,z)$ . Let  $y^*(z)$  be an arbitrary solution in  $Y(z)$ . By the strong convexity of  $\psi(\cdot,y,z) = f(\cdot,y) + \frac{r_1}{2}\|\cdot - z\|^2$ , we have
$$
\begin{array}{l} \psi (x (z, v), y ^ {*} (z), z) - \psi (x (z, v), y (z, v), z) \geq \psi (x (z, v), y ^ {*} (z), z) - \psi (x ^ {*} (z), y ^ {*} (z), z) \\ \geq \frac {r _ {1} - L _ {x}}{2} \| x ^ {*} (z) - x (z, v) \| ^ {2}. \\ \end{array}
$$
Here, the first inequality is because  $\psi(x^{*}(z), y^{*}(z), z) = \max_{y \in \mathcal{Y}} \psi(x_{r_{1}}(y, z), y, z) \geq \psi(x_{r_{1}}(y(z, v), z), y(z, v), z) = \psi(x(z, v), y(z, v), z)$ . On the other hand, by the concavity of  $\Psi(x, \cdot, z)$ ,
$$
\begin{array}{l} \psi (x (z, v), y ^ {*} (z), z) - \psi (x (z, v), y (z, v), z) \leq \langle \nabla_ {y} \psi (x (z, v), y (z, v), z), y ^ {*} (z) - y (z, v) \rangle \\ \leq r _ {1} \| y (z, v) - v \| \| y ^ {*} (z) - y (z, v) \| \\ \leq 2 r _ {1} \operatorname {d i a m} (\mathcal {Y}) \| y (z, v) - v \|. \\ \end{array}
$$
The second inequality is due to  $\langle \nabla_y f(x(z,v),y(z,v)) - r_1(y(z,v) - v),y^* (z) - y(z,v)\rangle \leq 0$  and the last inequality is because both  $y^{*}(z)$  and  $y(z,v)$  lie in set  $\mathcal{V}$ .
G Proof of Theorem 1
Before presenting the proof of Theorem 1, we first introduce the following lemma.
Lemma 8 Let  $\epsilon \geq 0$ . Suppose that
$$
\max  \left\{\frac {\| x ^ {t} - x ^ {t + 1} \|}{c}, \frac {\| y ^ {t} - y _ {+} ^ {t} \left(z ^ {t} , v ^ {t}\right) \|}{\alpha}, \frac {\| z ^ {t} - z ^ {t + 1} \|}{\beta}, \frac {\| v ^ {t} - v ^ {t + 1} \|}{\mu} \right\} \leq \epsilon .
$$
Then, there exists a  $\rho > 0$  such that  $(x^{t + 1}, y^{t + 1})$  is a  $\rho \epsilon$ -GS.
Proof In accordance with definition 1, we just need to evaluate the two terms  $\mathrm{dist}(\mathbf{0},\nabla_xf(x^{t + 1},y^{t + 1}) + \partial \mathbf{1}_{\mathcal{X}}(x^{t + 1}))$  and  $\mathrm{dist}(\mathbf{0}, - \nabla_yf(x^{t + 1},y^{t + 1}) + \partial \mathbf{1}_{\mathcal{Y}}(y^{t + 1}))$ . We first note that
$$
x ^ {t + 1} = \operatorname {p r o j} _ {\mathcal {X}} (x ^ {t} - c \nabla_ {x} F (x ^ {t}, y ^ {t}, z ^ {t}, v ^ {t})) = \operatorname * {a r g m i n} _ {x \in \mathcal {X}} \left\{\frac {1}{2} \| x - x ^ {t} + c \nabla_ {x} F (x ^ {t}, y ^ {t}, z ^ {t}, v ^ {t}) \| ^ {2} \right\}.
$$
Its optimality condition yields
$$
\mathbf {0} \in x ^ {t + 1} - x ^ {t} + c \nabla_ {x} f \left(x ^ {t}, y ^ {t}\right) + c r _ {1} \left(x ^ {t} - z ^ {t}\right) + \partial \mathbf {1} _ {\mathcal {X}} \left(x ^ {t + 1}\right),
$$
$$
\frac {1}{c} \left(x ^ {t} - x ^ {t + 1}\right) - r _ {1} \left(x ^ {t} - z ^ {t}\right) + \nabla_ {x} f \left(x ^ {t + 1}, y ^ {t + 1}\right) - \nabla_ {x} f \left(x ^ {t}, y ^ {t}\right) \in \nabla_ {x} f \left(x ^ {t + 1}, y ^ {t + 1}\right) + \partial \mathbf {1} _ {\mathcal {X}} \left(x ^ {t + 1}\right).
$$
Then, we are ready to simplify the stationarity measure as follows:
$$
\begin{array}{l} \operatorname {d i s t} (\mathbf {0}, \nabla_ {x} f (x ^ {t + 1}, y ^ {t + 1}) + \partial \mathbf {1} _ {\mathcal {X}} (x ^ {t + 1})) \\ \leq \left(\frac {1}{c} + L _ {x}\right) \| x ^ {t} - x ^ {t + 1} \| + r _ {1} \| x ^ {t} - z ^ {t} \| + L _ {x} \| y ^ {t} - y ^ {t + 1} \| \\ \leq \left(\frac {1}{c} + L _ {x} + r _ {1} + L _ {x} L _ {y} \alpha \sigma_ {6}\right) \| x ^ {t} - x ^ {t + 1} \| + \frac {r _ {2}}{\beta} \| z ^ {t} - z ^ {t + 1} \| + L _ {x} \| y ^ {t} - y _ {+} ^ {t} (z ^ {t}, v ^ {t}) \| \\ \leq \frac {r _ {1} - L _ {x} + r _ {1} ^ {2} - L _ {x} ^ {2} + (2 r _ {1} + 1) L _ {x} L _ {y}}{c \left(r _ {1} - L _ {x}\right)} \| x ^ {t} - x ^ {t + 1} \| + \frac {r _ {2}}{\beta} \| z ^ {t} - z ^ {t + 1} \| + \frac {L _ {x}}{\alpha} \| y ^ {t} - y _ {+} ^ {t} \left(z ^ {t}, v ^ {t}\right) \|. \\ \end{array}
$$
The last inequality is because  $\max \left\{\frac{\|x^t - x^{t + 1}\|}{c},\frac{\|y^t - y_+^t(z^t,v^t)\|}{\alpha},\frac{\|z^t - z^{t + 1}\|}{\beta}\right\} \leq \epsilon$ . Similarly, we have
$$
\frac {1}{\alpha} (y ^ {t} - y ^ {t + 1}) - r _ {2} (y ^ {t} - v ^ {t}) + \nabla_ {y} f (x ^ {t + 1}, y ^ {t}) - \nabla_ {y} f (x ^ {t + 1}, y ^ {t + 1}) \in - \nabla_ {y} f (x ^ {t + 1}, y ^ {t + 1}) + \partial \mathbf {1} _ {\mathcal {Y}} (y ^ {t + 1}),
$$
which can be directly derived from the update of  $y^{t + 1}$ . Then, we can simplify the expression
$$
\begin{array}{l} \operatorname {d i s t} \left(\mathbf {0}, - \nabla_ {y} f \left(x ^ {t + 1}, y ^ {t + 1}\right) + \partial \mathbf {1} _ {\mathcal {Y}} \left(y ^ {t + 1}\right)\right) \\ \leq \left(\frac {1}{\alpha} + L _ {y}\right) \| y ^ {t} - y ^ {t + 1} \| + r _ {2} \| y ^ {t} - v ^ {t} \| \\ \leq \left(\frac {1}{\alpha} + L _ {y} + r _ {2}\right) \| y ^ {t} - y _ {+} ^ {t} \left(z ^ {t}, v ^ {t}\right) \| + \left(\frac {1}{\alpha} + L _ {y} + r _ {2}\right) L _ {y} \alpha \sigma_ {6} \| x ^ {t} - x ^ {t + 1} \| + \frac {r _ {2}}{\mu} \| v ^ {t} - v ^ {t + 1} \| \\ \leq \frac {1 + L _ {y} + r _ {2}}{\alpha} \| y ^ {t} - y _ {+} ^ {t} (z ^ {t}, v ^ {t}) \| + \frac {L _ {y} (1 + L _ {y} + r _ {2}) (2 r _ {1} + 1)}{c (r _ {1} - L _ {x})} \| x ^ {t} - x ^ {t + 1} \| + \frac {r _ {2}}{\mu} \| v ^ {t} - v ^ {t + 1} \|. \\ \end{array}
$$
The last inequality is due to  $\max \left\{\frac{\|x^t - x^{t + 1}\|}{c},\frac{\|y^t - y_+^t(z^t,v^t)\|}{\alpha},\frac{\|v^t - v^{t + 1}\|}{\mu}\right\} \leq \epsilon$ . The proof is complete.
Proof of Theorem 1 Firstly, it is easy to check that  $\Phi (x,y,z,v)$  is lower bounded by  $\bar{F}$ . Let
$$
\zeta := \max  \left\{\frac {r _ {1}}{3 2} \| x ^ {t + 1} - x ^ {t} \| ^ {2}, \frac {r _ {2}}{1 5} \| y ^ {t} - y _ {+} ^ {t} (z ^ {t}, v ^ {t}) \| ^ {2}, \frac {r _ {1}}{5 \beta} \| z ^ {t} - z ^ {t + 1} \| ^ {2}, \frac {r _ {2}}{4 \mu} \| v _ {+} ^ {t} (z ^ {t + 1}) - v ^ {t} \| ^ {2} \right\}.
$$
Then, we consider the following two cases separately:
- There exists  $t \in  \{ 0,1,\ldots ,T - 1\}$  such that
$$
\frac {1}{2} \zeta \leq 4 r _ {1} \beta \| x (z ^ {t + 1}, v (z ^ {t + 1})) - x (z ^ {t + 1}, v _ {+} ^ {t} (z ^ {t + 1})) \| ^ {2}.
$$
- For any  $t \in  \{ 0,1,\ldots ,T - 1\}$  ,we have
$$
\frac {1}{2} \zeta \geq 4 r _ {1} \beta \| x (z ^ {t + 1}, v (z ^ {t + 1})) - x (z ^ {t + 1}, v _ {+} ^ {t} (z ^ {t + 1})) \| ^ {2}.
$$
We first consider the first case. If the KL exponent  $\theta \in (\frac{1}{2}, 1)$ , then from Proposition 2,
$$
\begin{array}{l} \left\| v _ {+} ^ {t} \left(z ^ {t + 1}\right) - v ^ {t} \right\| ^ {2} \leq \frac {3 2 r _ {1} \mu \beta}{r _ {2}} \left\| x \left(z ^ {t + 1}, v \left(z ^ {t + 1}\right)\right) - x \left(z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right) \right\| ^ {2} \\ \leq \frac {3 2 r _ {1} \mu \beta \omega_ {0}}{r _ {2}} \| v _ {+} ^ {t} (z ^ {t + 1}) - v ^ {t} \| ^ {\frac {1}{\theta}}. \\ \end{array}
$$
Then, we have  $\| v_{+}^{t}(z^{t + 1}) - v^{t}\| \leq \rho_{1}\beta^{\frac{\theta}{2\theta - 1}}$ , where  $\rho_{1} := \left(\frac{32r_{1}\mu\omega_{0}}{r_{2}}\right)^{\frac{\theta}{2\theta - 1}}$ . Armed with this, we can bound other terms as follows:
$$
\begin{array}{l} \frac {\| x ^ {t + 1} - x ^ {t} \| ^ {2}}{c ^ {2}} \leq \frac {2 5 6 \beta}{c ^ {2}} \| x (z ^ {t + 1}, v (z ^ {t + 1})) - x (z ^ {t + 1}, v _ {+} ^ {t} (z ^ {t + 1})) \| ^ {2} \\ \leq \frac {2 5 6 \beta \omega_ {0}}{c ^ {2}} \| v _ {+} ^ {t} (z ^ {t + 1}) - v ^ {t} \| ^ {\frac {1}{\theta}} = \rho_ {2} \beta^ {\frac {2 \theta}{2 \theta - 1}}, \\ \end{array}
$$
$$
\begin{array}{l} \frac {\| y ^ {t} - y _ {+} ^ {t} (z ^ {t} , v ^ {t}) \| ^ {2}}{\alpha^ {2}} \leq \frac {1 2 0 r _ {1} \beta}{r _ {2} \alpha^ {2}} \| x (z ^ {t + 1}, v (z ^ {t + 1})) - x (z ^ {t + 1}, v _ {+} ^ {t} (z ^ {t + 1})) \| ^ {2} \\ \leq \frac {1 2 0 r _ {1} \beta \omega_ {0}}{r _ {2} \alpha^ {2}} \| v _ {+} ^ {t} (z ^ {t + 1}) - v ^ {t} \| ^ {\frac {1}{\theta}} = \rho_ {3} \beta^ {\frac {2 \theta}{2 \theta - 1}}, \\ \end{array}
$$
$$
\frac {\| z ^ {t + 1} - z ^ {t} \| ^ {2}}{\beta^ {2}} \leq 4 0 \| x (z ^ {t + 1} , v (z ^ {t + 1})) - x (z ^ {t + 1}, v _ {+} ^ {t} (z ^ {t + 1})) \| ^ {2} \leq 4 0 \omega_ {0} \| z _ {+} ^ {t} (v ^ {t + 1}) - z ^ {t} \| ^ {\frac {1}{\theta}} = \rho_ {4} \beta^ {\frac {1}{2 \theta - 1}},
$$
$$
\begin{array}{l} \frac {\| v ^ {t + 1} - v ^ {t} \| ^ {2}}{\mu^ {2}} \leq \frac {2 \| v _ {+} ^ {t} (z ^ {t + 1}) - v ^ {t} \| ^ {2}}{\mu^ {2}} + 2 \left(4 L _ {y} ^ {2} \alpha^ {2} \sigma_ {6} ^ {2} \| x ^ {t} - x ^ {t + 1} \| ^ {2} + 4 (\sigma_ {8} + 1) ^ {2} \| y ^ {t} - y _ {+} ^ {t} (z ^ {t}, v ^ {t}) \| ^ {2} + \right. \\ 2 \sigma_ {3} ^ {2} \| z ^ {t} - z ^ {t + 1} \| ^ {2}) \\ \leq \frac {2 \rho_ {1} ^ {2}}{\mu^ {2}} \beta^ {\frac {2 \theta}{2 \theta - 1}} + 8 L _ {y} ^ {2} \alpha^ {2} \sigma_ {6} ^ {2} c ^ {2} \rho_ {2} \beta^ {\frac {2 \theta}{2 \theta - 1}} + 8 (\sigma_ {8} + 1) ^ {2} \alpha^ {2} \rho_ {3} \beta^ {\frac {2 \theta}{2 \theta - 1}} + 4 \sigma_ {3} ^ {2} \rho_ {3} \beta^ {\frac {4 \theta - 1}{2 \theta - 1}} \\ \leq \rho_ {5} \beta^ {\frac {2 \theta}{2 \theta - 1}}, \\ \end{array}
$$
where  $\rho_{2}:= \frac{256\omega_{0}}{c^{2}}\rho_{1}^{\frac{1}{\theta}},\rho_{3}:= \frac{120r_{1}\omega_{0}}{r_{2}\alpha^{2}}\rho_{1}^{\frac{1}{\theta}},\rho_{4}:= 40\omega_{0}\rho_{1}^{\frac{1}{\theta}}$  and  $\rho_5\coloneqq \frac{2\rho_1^2}{\mu^2} +8L_y^2\alpha^2\sigma_6^2 c^2\rho_2 + 8(\sigma_8 + 1)^2\alpha^2\rho_3 + 4\sigma_3^2\rho_3$ . According to Lemma 8, there exists a  $\rho >0$  such that  $(x^{t + 1},y^{t + 1})$  is a  $\rho \epsilon$  GS, where  $\epsilon = \max \{\sqrt{\rho_2}\beta^{\frac{\theta}{2\theta - 1}},\sqrt{\rho_3}\beta^{\frac{\theta}{2\theta - 1}},\sqrt{\rho_4}\beta^{\frac{\theta}{2\theta - 1}},\sqrt{\rho_5}\beta^{\frac{1}{4\theta - 2}}\}$ . For the general concave case, the above analysis holds with  $\theta = 1$  and  $(x^{t + 1},y^{t + 1})$  is a  $\rho \max \{\sqrt{\rho_2}\beta \sqrt{\rho_4}\beta ,\sqrt{\rho_5}\beta ,\sqrt{\rho_3}\beta^{\frac{1}{2}}\}$  GS by replacing  $\omega_0$  with  $\omega_{1}$ .
One remaining thing is to prove that  $z^{t + 1}$  is an  $\mathcal{O}(\beta^{\frac{1}{4\theta - 2}})$ -OS. By the dual error bounds in Proposition 3 and error bounds in Lemmas 2 and 4, we have
$$
\begin{array}{l} \left\| z ^ {t + 1} - x ^ {*} \left(z ^ {t + 1}\right) \right\| \\ \leq \| z ^ {t + 1} - z ^ {t} \| + \| z ^ {t} - x ^ {t + 1} \| + \| x ^ {t + 1} - x \left(y ^ {t}, z ^ {t}, v ^ {t}\right) \| + \| x \left(y ^ {t}, z ^ {t}, v ^ {t}\right) - x \left(z ^ {t}, v ^ {t}\right) \| + \\ \left\| x \left(z ^ {t}, v ^ {t}\right) - x \left(z ^ {t + 1}, v ^ {t}\right) \right\| + \left\| x \left(z ^ {t + 1}, v ^ {t}\right) - x ^ {*} \left(z ^ {t + 1}\right) \right\| \\ \leq \left(1 + \sigma_ {2}\right) \| z ^ {t + 1} - z ^ {t} \| + \frac {\| z ^ {t} - z ^ {t + 1} \|}{\beta} + \sigma_ {6} \| x ^ {t} - x ^ {t + 1} \| + \sigma_ {1} \sigma_ {8} \| y ^ {t} - y _ {+} ^ {t} (z ^ {t}, v ^ {t}) \| + \\ \omega_ {2} \left\| v ^ {t} - y \left(z ^ {t + 1}, v ^ {t}\right) \right\| ^ {\frac {1}{2 \theta}} \\ \leq \left(1 + \sigma_ {2}\right) \sqrt {\rho_ {4}} \beta^ {\frac {4 \theta - 1}{4 \theta - 2}} + \sqrt {\rho_ {4}} \beta^ {\frac {1}{4 \theta - 2}} + \left(\sigma_ {6} \sqrt {\rho_ {2}} + \sigma_ {1} \sigma_ {8} \sqrt {\rho_ {3}}\right) \beta^ {\frac {\theta}{2 \theta - 1}} + \omega_ {2} \rho_ {1} \beta^ {\frac {1}{4 \theta - 2}} \\ = \mathcal {O} \left(\beta^ {\frac {1}{4 \theta - 2}}\right). \tag {23} \\ \end{array}
$$
Now, we consider the second case. Since
$$
\begin{array}{l} \Phi \left(x ^ {t}, y ^ {t}, z ^ {t}, v ^ {t}\right) - \Phi \left(x ^ {t + 1}, y ^ {t + 1}, z ^ {t + 1}, v ^ {t + 1}\right) \\ \geq \frac {r _ {1}}{6 4} \| x ^ {t + 1} - x ^ {t} \| ^ {2} + \frac {r _ {2}}{3 0} \| y ^ {t} - y _ {+} ^ {t} (z ^ {t}, v ^ {t}) \| ^ {2} + \frac {r _ {1}}{1 0 \beta} \| z ^ {t} - z ^ {t + 1} \| ^ {2} + \frac {r _ {2}}{8 \mu} \| v _ {+} ^ {t} (z ^ {t + 1}) - v ^ {t} \| ^ {2} \\ \end{array}
$$
holds for  $t\in \{0,1,\ldots ,T - 1\}$  and  $\Phi (x,y,z,v)\geq \bar{F}$ , we know that
$$
\begin{array}{l} \Phi (x ^ {0}, y ^ {0}, z ^ {0}, v ^ {0}) - \bar {F} \\ \geq \sum_ {T = 0} ^ {T - 1} \frac {r _ {1}}{6 4} \| x ^ {t + 1} - x ^ {t} \| ^ {2} + \frac {r _ {2}}{3 0} \| y ^ {t} - y _ {+} ^ {t} (z ^ {t}, v ^ {t}) \| ^ {2} + \frac {r _ {1}}{1 0 \beta} \| z ^ {t} - z ^ {t + 1} \| ^ {2} + \frac {r _ {2}}{8 \mu} \| v _ {+} ^ {t} (z ^ {t + 1}) - v ^ {t} \| ^ {2} \\ \geq T \min  \left\{\frac {r _ {1} c ^ {2}}{6 4}, \frac {r _ {2} \alpha^ {2}}{3 0}, \frac {r _ {1}}{1 0}, \frac {r _ {2} \mu}{8} \right\} \left(\frac {\| x ^ {t + 1} - x ^ {t} \| ^ {2}}{c ^ {2}} + \frac {\| y ^ {t} - y _ {+} ^ {t} (z ^ {t} , v ^ {t}) \| ^ {2}}{\alpha^ {2}}\right) + \\ T \min \left\{\frac {r _ {1} c ^ {2}}{6 4}, \frac {r _ {2} \alpha^ {2}}{3 0}, \frac {r _ {1}}{1 0}, \frac {r _ {2} \mu}{8} \right\} \left(\frac {\| z ^ {t} - z ^ {t + 1} \| ^ {2}}{\beta} + \frac {\| v _ {+} ^ {t} (z ^ {t + 1}) - v ^ {t} \| ^ {2}}{\mu^ {2}}\right). \\ \end{array}
$$
Since  $\Phi (x,y,z,v)\geq \bar{F}$ , there exists a  $t\in \{0,1,\ldots ,T - 1\}$  such that
$$
\begin{array}{l} \max  \left\{\frac {\| x ^ {t} - x ^ {t + 1} \| ^ {2}}{c ^ {2}}, \frac {\| y ^ {t} - y _ {+} ^ {t} (z ^ {t} , v ^ {t}) \| ^ {2}}{\alpha^ {2}}, \frac {\| z ^ {t} - z ^ {t + 1} \| ^ {2}}{\beta}, \frac {\| v ^ {t} - v _ {+} ^ {t} (z ^ {t + 1}) \| ^ {2}}{\mu^ {2}} \right\} \\ \leq \frac {\Phi (x ^ {0} , y ^ {0} , z ^ {0} , v ^ {0}) - \bar {F}}{T \min \left\{\frac {r _ {1} c ^ {2}}{6 4} , \frac {r _ {2} \alpha^ {2}}{3 0} , \frac {r _ {1}}{1 0} , \frac {r _ {2} \mu}{8} \right\}} =: \frac {\eta}{T}. \\ \end{array}
$$
$$
\begin{array}{l} \frac {\| v ^ {t + 1} - v ^ {t} \| ^ {2}}{\mu^ {2}} \leq \frac {2 \| v _ {+} ^ {t} (z ^ {t + 1}) - v ^ {t} \| ^ {2}}{\mu^ {2}} + 2 \left(4 L _ {y} ^ {2} \alpha^ {2} \sigma_ {6} ^ {2} \| x ^ {t} - x ^ {t + 1} \| ^ {2} + 4 (\sigma_ {8} + 1) ^ {2} \| y ^ {t} - y _ {+} ^ {t} (z ^ {t}, v ^ {t}) \| ^ {2} + \right. \\ 2 \sigma_ {3} ^ {2} \| z ^ {t} - z ^ {t + 1} \| ^ {2}) \leq \frac {\eta (2 + 8 L _ {y} ^ {2} \alpha^ {2} \sigma_ {6} ^ {2} c ^ {2} + 8 (\sigma_ {8} + 1) ^ {2} \alpha^ {2} + 4 \sigma_ {3} ^ {2} \beta)}{T} \leq \frac {\rho_ {6}}{T}, \\ \end{array}
$$
where  $\rho_6 = \eta \left(2 + 8L_y^2\alpha^2\sigma_6^2 c^2 +8(\sigma_8 + 1)^2\alpha^2 +4\sigma_3^2\right)$ . Thus, we know there exists a  $\rho >0$  such that  $(x^{t + 1},y^{t + 1})$  is a  $\rho \epsilon$ -GS, where  $\epsilon = \sqrt{\frac{\rho_6}{T\beta}}$ . Similar argument as (23) shows that  $z^{t + 1}$  is an  $\mathcal{O}(\frac{1}{\sqrt{T\beta}})$ -OS. Moreover, if we choose  $\beta \leq \mathcal{O}(T^{-\frac{2\theta - 1}{2\theta}})$  when the KL exponent  $\theta \in (\frac{1}{2},1)$  and  $\beta = \mathcal{O}(T^{-\frac{1}{2}})$  for general concave case, the two cases coincide. We conclude that under KL
assumption,  $(x^{t + 1},y^{t + 1})$  is an  $\mathcal{O}(T^{-\frac{1}{4\theta}})$  -GS and  $z^{t + 1}$  is an  $\mathcal{O}(T^{-\frac{1}{4\theta}})$  -OS. When the dual function is concave,  $(x^{t + 1},y^{t + 1})$  is an  $\mathcal{O}(T^{-\frac{1}{4}})$  -GS and  $z^{t + 1}$  is an  $\mathcal{O}(T^{-\frac{1}{4}})$  -OS.
Next, suppose that the KL exponent  $\theta \in (0,\frac{1}{2} ]$ . From Proposition 2, we obtain
$$
\begin{array}{l} \left\| x \left(z ^ {t + 1}, v \left(z ^ {t + 1}\right)\right) - x \left(z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right) \right\| ^ {2} \leq \omega_ {0} \left\| v _ {+} ^ {t} \left(z ^ {t + 1}\right) - v ^ {t} \right\| ^ {\frac {1}{\theta}} \tag {24} \\ \leq \omega_ {0} (2 \operatorname {d i a m} (\mathcal {Y})) ^ {\frac {1}{\theta} - 2} \| v _ {+} ^ {t} (z ^ {t + 1}) - v ^ {t} \| ^ {2}. \\ \end{array}
$$
The last inequality is from  $\| v_{+}^{t}(z^{t + 1}) - v^{t}\| = \beta \| y(z^{t + 1},v^{t}) - v^{t}\| \leq 2\mathrm{diam}(\mathcal{V})$  since  $\| v^t\| \leq$  diam(y). Armed with the homogeneous proximal error bound (24), we have
$$
\begin{array}{l} \Phi \left(x ^ {t}, y ^ {t}, z ^ {t}, v ^ {t}\right) - \Phi \left(x ^ {t + 1}, y ^ {t + 1}, z ^ {t + 1}, v ^ {t + 1}\right) \\ \geq \frac {r _ {1}}{3 2} \| x ^ {t + 1} - x ^ {t} \| ^ {2} + \frac {r _ {2}}{1 5} \| y ^ {t} - y _ {+} ^ {t} (z ^ {t}, v ^ {t}) \| ^ {2} + \frac {r _ {1}}{5 \beta} \| z ^ {t} - z ^ {t + 1} \| ^ {2} + \frac {r _ {2}}{4 \mu} \| v _ {+} ^ {t} (z ^ {t + 1}) - v ^ {t} \| ^ {2} - \\ 4 r _ {1} \beta \omega_ {0} (2 \operatorname {d i a m} (\mathcal {Y})) ^ {\frac {1}{\theta} - 2} \| v _ {+} ^ {t} (z ^ {t + 1}) - v ^ {t} \| ^ {2} \\ \geq \frac {r _ {1}}{3 2} \| x ^ {t + 1} - x ^ {t} \| ^ {2} + \frac {r _ {2}}{1 5} \| y ^ {t} - y _ {+} ^ {t} (z ^ {t}, v ^ {t}) \| ^ {2} + \frac {r _ {1}}{5 \beta} \| z ^ {t} - z ^ {t + 1} \| ^ {2} + \\ \left(\frac {r _ {2}}{4 \mu} - 4 r _ {1} \beta \omega_ {0} (2 \operatorname {d i a m} (\mathcal {Y})) ^ {\frac {1}{\theta} - 2}\right) \| v _ {+} ^ {t} (z ^ {t + 1}) - v ^ {t} \| ^ {2} \\ \geq \frac {r _ {1}}{3 2} \| x ^ {t + 1} - x ^ {t} \| ^ {2} + \frac {r _ {2}}{1 5} \| y ^ {t} - y _ {+} ^ {t} (z ^ {t}, v ^ {t}) \| ^ {2} + \frac {r _ {1}}{5 \beta} \| z ^ {t} - z ^ {t + 1} \| ^ {2} + + \frac {r _ {2}}{8 \mu} \| v _ {+} ^ {t} (z ^ {t + 1}) - v ^ {t} \| ^ {2}, \\ \end{array}
$$
where the last inequality is because  $\beta \leq \frac{r_2}{32r_1\mu\omega_0(2\mathrm{diam}(\mathcal{Y}))^{\frac{1}{\theta} - 2}}$ . The rest of the proof is similar to that in the second case when  $\theta \in (\frac{1}{2}, 1)$ , and we conclude that  $(x^{t + 1}, y^{t + 1})$  is a  $\mathcal{O}(T^{-\frac{1}{2}})$ -GS and  $z^{t + 1}$  is an  $\mathcal{O}(T^{-\frac{1}{2}})$ -OS.
H Convergence Analysis for (Non)convex-Nonconcave Problems
Similar convergence results for (non)convex-nonconcave problems can be established with the symmetric property of our Lyapunov function. Specifically, we can derive the following Proposition 5 by doing an easy transformation of  $(x,z,r_1,c,\beta,L_x)$  to  $(y,v,r_2,\alpha,\mu,L_y)$ :
Proposition 5 (Basic descent estimate) Suppose that Assumption 1 holds and  $r_1 \geq 2L$ ,  $r_2 \geq 2\lambda L$  with the parameters
$$
0 <   \alpha \leq \min  \left\{\frac {4}{3 (\lambda L + r _ {2})}, \frac {1}{6 L} \right\}, 0 <   c \leq \min  \left\{\frac {2}{3 L \sigma^ {\prime 2}}, \frac {1}{6 L _ {h}}, \frac {1}{5 \sqrt {\lambda + 5} L} \right\}
$$
$$
0 <   \mu \leq \min  \left\{\frac {2 4 r _ {2}}{3 6 0 r _ {2} + 5 r _ {2} ^ {2} \lambda + (2 L + 5 r _ {2}) ^ {2}}, \frac {c L ^ {2}}{3 8 4 r _ {2} (\lambda + 5) (\lambda + 1) ^ {2}} \right\},
$$
$$
0 <   \beta \leq \min \left\{\frac {2 (\lambda + 5)}{2 (\lambda + 5) + L ^ {2}}, \frac {c L ^ {2}}{6 4 r _ {1} (\lambda + 5)} \right\}.
$$
Then, for any  $t \geq 0$ ,
$$
\begin{array}{l} \Psi^ {t} - \Psi^ {t + 1} \geq \frac {r _ {2}}{3 2} \| x ^ {t} - x _ {+} ^ {t} (z ^ {t}, v ^ {t}) \| ^ {2} + \frac {r _ {1}}{1 5} \| y ^ {t} - y ^ {t + 1} \| ^ {2} + \frac {r _ {2}}{5 \mu} \| v ^ {t} - v ^ {t + 1} \| ^ {2} + \frac {r _ {1}}{4 \beta} \| z _ {+} ^ {t} (v ^ {t + 1}) - z ^ {t} \| ^ {2} \\ - 4 r _ {2} \mu \| y (z (v ^ {t + 1}), v ^ {t + 1}) - y (z _ {+} ^ {t} (v ^ {t + 1}), v ^ {t + 1}) \| ^ {2}, \\ \end{array}
$$
where  $\sigma' := \frac{2\alpha r_2 + 1}{\alpha(r_2 - \lambda L)}$  and  $L_h := \left(\frac{L}{r_2 - \lambda L} + 2\right)L + r_1$ . Moreover, we set  $x_+(z, v) := \operatorname{proj}_{\mathcal{X}}(z + c\nabla_x F(x, y(x, z, v), z, v))$  and  $z_+(v) := z + \mu(x(y(z, v), z, v) - z)$  with the following definitions: (i)  $y(z, v) := \operatorname{argmin}_{y \in \mathcal{Y}} \max_{x \in \mathcal{X}} F(x, y, z, v)$ , (ii)  $z(v) := \operatorname{argmin}_{z \in \mathbb{R}^n} P(z, v)$ .
To handle the negative term in the basic decrease estimate (Proposition 5), we would impose the following assumptions on primal function:
Assumption 4 (KL property for primal function) For any fixed point  $y \in \mathcal{V}, z \in \mathbb{R}^n$ , and  $v \in \mathbb{R}^d$ , the problem  $\min_{x \in \mathcal{X}} F(x, y, z, v)$  has a nonempty solution set and a finite optimal value. Moreover, there exist  $\tau_1 > 0, \theta_1 \in (0, 1)$  such that
$$
\left(f (x, y) - \min  _ {x ^ {\prime} \in \mathcal {X}} f \left(x ^ {\prime}, y\right)\right) ^ {\theta_ {1}} \leq \frac {1}{\tau_ {1}} \operatorname {d i s t} (\mathbf {0}, \nabla_ {x} f (x, y) + \partial \mathbf {1} _ {\mathcal {X}} (x))
$$
holds for any  $y\in \mathcal{V}$  and  $x\in \mathcal{X}$
Assumption 5 (Convexity of the primal function) For any fixed point  $y \in \mathcal{V}$ ,  $f(\cdot, y)$  is convex.
Proposition 6 (Proximal error bound) Under the setting of Proposition 5 with Assumption 4 or 5, for any  $z \in \mathbb{R}^n$ ,  $v \in \mathbb{R}^d$ , we have
(i) KL exponent  $\theta_{1}\in (0,1)$
$$
\left\| y \left(z _ {+} ^ {t} \left(v ^ {t + 1}\right), v ^ {t + 1}\right) - y \left(z \left(v ^ {t + 1}\right), v ^ {t + 1}\right) \right\| ^ {2} \leq \omega_ {2} \left\| z _ {+} ^ {t} \left(v ^ {t + 1}\right) - z ^ {t} \right\| ^ {\frac {1}{\theta_ {1}}};
$$
(ii) Convex:
$$
\left\| y \left(z _ {+} ^ {t} \left(v ^ {t + 1}\right), v ^ {t + 1}\right) - y \left(z \left(v ^ {t + 1}\right), v ^ {t + 1}\right) \right\| ^ {2} \leq \omega_ {3} \left\| z _ {+} ^ {t} \left(v ^ {t + 1}\right) - z ^ {t} \right\|,
$$