SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\frac {d r (y ; \mu)}{d y} <   0
$$
For  $0 < \mu \leq \frac{a^2}{4}$
$$
\left\{ \begin{array}{l l} \frac {d r (y ; \mu)}{d y} > 0 & y _ {\mathrm {l b}} <   y <   y _ {\mathrm {u b}} \\ \frac {d r (y ; \mu)}{d y} \leq 0 & \text {O t h e r w i s e} \end{array} \right. \tag {32a}
$$
$$
y _ {\mathrm {l b}} = \frac {(4 \mu) ^ {1 / 3}}{2} (a ^ {1 / 3} - \sqrt {a ^ {2 / 3} - (4 \mu) ^ {1 / 3}}) y _ {\mathrm {u b}} = \frac {(4 \mu) ^ {1 / 3}}{2} (a ^ {1 / 3} + \sqrt {a ^ {2 / 3} - (4 \mu) ^ {1 / 3}})
$$
$$
y _ {\mathrm {l b}} \leq \sqrt {\mu} \leq y _ {\mathrm {u b}}
$$
(iv) For  $0 < \mu < \frac{a^2}{4}$ , let  $p(\mu) = r(y_{\mathrm{ub}},\mu)$ , then  $p'(\mu) < 0$  and there exist a unique solution to  $p(\mu) = 0$ , denoted as  $\tau$ . Additionally,  $\tau < \frac{a^2}{4}$ .
(v) There exists a  $\tau > 0$  such that,  $\forall \mu > \tau$ , the equation  $r(y; \mu) = 0$  has only one solution. At  $\mu = \tau$ , the equation  $r(y; \mu) = 0$  has two solutions, and  $\forall \mu < \tau$ , the equation  $r(y; \mu) = 0$  has three solutions. Moreover,  $\mu < \frac{a^2}{4}$ .
(vi)  $\forall \mu < \tau$ , the equation  $r(y; \mu) = 0$  has three solutions, i.e.  $y_{\mu}^{*} < y_{\mu}^{**} < y_{\mu}^{***}$ .
$$
\frac {d y _ {\mu} ^ {*}}{d \mu} > 0 \quad \frac {d y _ {\mu} ^ {* *}}{d \mu} > 0 \quad \frac {d y _ {\mu} ^ {* * *}}{d \mu} <   0 a n d \lim  _ {\mu \rightarrow 0} y _ {\mu} ^ {*} = 0, \lim  _ {\mu \rightarrow 0} y _ {\mu} ^ {* *} = 0, \lim  _ {\mu \rightarrow 0} y _ {\mu} ^ {* * *} = \frac {a}{a ^ {2} + 1}
$$
$$
y _ {\mu} ^ {*} <   y _ {\mathrm {l b}} <   \sqrt {\mu} <   y _ {\mu} ^ {* *} <   y _ {\mathrm {u b}} <   y _ {\mu} ^ {* * *}
$$
Theorem 6 (Detailed Property of  $t(x;\mu)$ ). For  $t(x;\mu)$  in (9), then
(i) For  $\mu > 0$ ,  $\lim_{x \to 0^+} t(x; \mu) = \infty$ ,  $t(a, \mu) < 0$
(ii) If  $\mu < \left(\frac{a(\sqrt{a^2 + 1} - a)}{2(a^2 + 1)}\right)^2$  or  $\mu > \left(\frac{a(\sqrt{a^2 + 1} + a)}{2(a^2 + 1)}\right)^2$ , then  $t(\sqrt{\mu(a^2 + 1)}, \mu) < 0$ .
(iii) For  $\mu >\frac{a^2}{4(a^2 + 1)^3}$
$$
\frac {d t (x ; \mu)}{d x} <   0
$$
For  $0 < \mu \leq \frac{a^2}{4(a^2 + 1)^3}$
$$
\left\{ \begin{array}{l l} \frac {d t (x ; \mu)}{d x} > 0 & x _ {\mathrm {l b}} <   x <   x _ {\mathrm {u b}} \\ \frac {d t (x ; \mu)}{d x} \leq 0 & \text {O t h e r w i s e} \end{array} \right. \tag {33a}
$$
$$
x _ {\mathrm {l b}} = \frac {(4 \mu a) ^ {1 / 3} (1 - \sqrt {1 - \frac {(4 \mu) ^ {1 / 3} (a ^ {2} + 1)}{a ^ {2 / 3}}})}{2} x _ {\mathrm {u b}} = \frac {(4 \mu a) ^ {1 / 3} (1 + \sqrt {1 - \frac {(4 \mu) ^ {1 / 3} (a ^ {2} + 1)}{a ^ {2 / 3}}})}{2}
$$
$$
x _ {\mathrm {l b}} \leq \sqrt {\mu (a ^ {2} + 1)} \leq x _ {\mathrm {u b}}
$$
(iv) For  $0 < \mu < \frac{a^2}{4(a^2 + 1)^3}$  and let  $q(\mu) = t(x_{\mathrm{lb}},\mu)$ , then  $q'(\mu) > 0$  and there exist a unique solution to  $q(\mu) = 0$ , denoted as  $\tau$  and  $\tau < \frac{a^2}{4(a^2 + 1)^3} \leq \frac{1}{27}$ .
(v) There exists a  $\tau > 0$  such that,  $\forall \mu > \tau$ , the equation  $t(x; \mu) = 0$  has only one solution. At  $\mu = \tau$ , the equation  $t(x; \mu) = 0$  has two solutions, and  $\forall \mu < \tau$ , the equation  $t(x; \mu) = 0$  has three solutions. Moreover,  $\tau < \frac{a^2}{4(a^2 + 1)^3} \leq \frac{1}{27}$
(vi)  $\forall \mu <  \tau ,t(x;\mu) = 0$  has three stationary points,i.e.  $x_{\mu}^{**} <   x_{\mu}^{**} <   x_{\mu}^{*}$
$$
\frac {d x _ {\mu} ^ {*}}{d \mu} <   0 \quad \frac {d x _ {\mu} ^ {* * *}}{d \mu} > 0 a n d \lim  _ {\mu \rightarrow 0} x _ {\mu} ^ {*} = a, \lim  _ {\mu \rightarrow 0} x _ {\mu} ^ {* *} = 0, \lim  _ {\mu \rightarrow 0} x _ {\mu} ^ {* * *} = 0
$$
$$
\max  _ {\mu \leq \tau} x _ {\mu} ^ {* *} \leq \frac {a (\sqrt {a ^ {2} + 1} - a)}{2 \sqrt {a ^ {2} + 1}} \quad a n d \quad \frac {a (\sqrt {a ^ {2} + 1} + a)}{2 \sqrt {a ^ {2} + 1}} \leq \min  _ {\mu > 0} x _ {\mu} ^ {*}
$$
It also implies that  $t\left(\frac{a(\sqrt{a^2 + 1} - a)}{2\sqrt{a^2 + 1}};\mu\right) \geq 0$  and  $\max_{\mu \leq \mu_0}x_\mu^{**} < \min_{\mu >0}x_\mu^*$ .
Lemma 7. Algorithm 1 with input  $f = g_{\mu}(x,y), z_0 = (x(0), y(0))$  where  $(x(0), y(0)) \in C_{\mu 3}$  in (41), then  $\forall t \geq 0$ ,  $(x(t), y(t)) \in C_{\mu 3}$ . Moreover,  $\lim_{t \to \infty} (x(t), y(t)) = (x_{\mu}^{*}, y_{\mu}^{*})$
Lemma 8. For any  $(x,y)\in C_{\mu 3}$  in (41), and  $(x,y)\neq (x_{\mu}^{*},y_{\mu}^{*})$
$$
g _ {\mu} (x, y) > g _ {\mu} (x _ {\mu} ^ {*}, y _ {\mu} ^ {*})
$$
Theorem 7 (Detailed Property of  $r_{\epsilon}(y;\mu)$ ). For  $r_{\epsilon}(y;\mu)$  in (15), then
(i) For  $\mu > 0, \epsilon > 0, \lim_{y \to 0^+} r_{\epsilon}(y; \mu) = \infty, y\left(\frac{a}{a^2 + 1}, \mu\right) < 0$
(ii) For  $\mu >\frac{(a - \epsilon / \mu)^4}{4(a + \epsilon / \mu)^2}$ , then  $\frac{dr_{\epsilon}(y;\mu)}{dy} < 0$ . For  $0 < \mu \leq \frac{(a - \epsilon / \mu)^4}{4(a + \epsilon / \mu)^2}$
$$
\left\{ \begin{array}{l l} \frac {d r _ {\epsilon} (y ; \mu)}{d y} > 0 & y _ {\mathrm {l b}, \mu , \epsilon} <   y <   y _ {\mathrm {u b}, \mu , \epsilon} \\ \frac {d r _ {\epsilon} (y ; \mu)}{d y} \leq 0 & \text {O t h e r w i s e} \end{array} \right. \tag {34a}
$$
$$
y _ {\mathrm {l b}, \mu , \epsilon} = \frac {(4 \mu) ^ {1 / 3}}{2} \left(\left(\frac {(a - \epsilon / \mu) ^ {2}}{a - \epsilon / \mu}\right) ^ {1 / 3} - \sqrt {\left(\frac {(a - \epsilon / \mu) ^ {2}}{a - \epsilon / \mu}\right) ^ {2 / 3} - (4 \mu) ^ {1 / 3}}\right)
$$
$$
y _ {\mathrm {u b}, \mu , \epsilon} = \frac {(4 \mu) ^ {1 / 3}}{2} \left(\left(\frac {(a - \epsilon / \mu) ^ {2}}{a - \epsilon / \mu}\right) ^ {1 / 3} + \sqrt {\left(\frac {(a - \epsilon / \mu) ^ {2}}{a - \epsilon / \mu}\right) ^ {2 / 3} - (4 \mu) ^ {1 / 3}}\right)
$$
$$
y _ {\mathrm {l b}, \mu , \epsilon} \leq (2 \mu^ {2}) ^ {1 / 3} \frac {(a + \epsilon / \mu) ^ {1 / 3}}{(a - \epsilon / \mu) ^ {2 / 3}}
$$
$$
y _ {\mathrm {l b}, \mu , \epsilon} \leq \sqrt {\mu} \leq y _ {\mathrm {u b}, \mu , \epsilon}
$$
Theorem 8 (Detailed Property of  $r_{\beta}(y;\mu)$ ). For  $r_{\beta}(y;\mu)$  in (19), then
(i) For  $\mu > 0, \epsilon > 0$ ,  $\lim_{y \to 0^+} r_\beta(y; \mu) = \infty$
(ii) For  $\mu >\frac{a^2(1 - \beta)^4}{4(1 + \beta)^2}$ , then  $\frac{dr_{\beta}(y;\mu)}{dy} < 0$ . For  $0 < \mu \leq \frac{a^2(1 - \beta)^4}{4(1 + \beta)^2}$
$$
\left\{ \begin{array}{l l} \frac {d r _ {\beta} (y ; \mu)}{d y} > 0 & y _ {\mathrm {l b}, \mu , \beta} <   y <   y _ {\mathrm {u b}, \mu , \beta} \\ \frac {d r _ {\beta} (y ; \mu)}{d y} \leq 0 & \text {O t h e r w i s e} \end{array} \right. \tag {35a}
$$
$$
y _ {\mathrm {l b}, \mu , \beta} = \frac {(4 \mu) ^ {1 / 3}}{2} \left(\frac {a (1 - \beta) ^ {2}}{1 + \beta}\right) ^ {1 / 3} \left(1 - \sqrt {1 - \frac {(4 \mu) ^ {1 / 3}}{a ^ {2 / 3}} \left(\frac {1 + \beta}{(1 - \beta) ^ {2}}\right) ^ {2 / 3}}\right)
$$
$$
y _ {\mathrm {u b}, \mu , \beta} = \frac {(4 \mu) ^ {1 / 3}}{2} \left(\frac {a (1 - \beta) ^ {2}}{1 + \beta}\right) ^ {1 / 3} \left(1 + \sqrt {1 - \frac {(4 \mu) ^ {1 / 3}}{a ^ {2 / 3}} \left(\frac {1 + \beta}{(1 - \beta) ^ {2}}\right) ^ {2 / 3}}\right)
$$
$$
y _ {\mathrm {l b}, \mu , \beta} \leq \frac {(4 \mu) ^ {2 / 3}}{2 a ^ {1 / 3}} \frac {(1 + \beta) ^ {1 / 3}}{(1 - \beta) ^ {2 / 3}}
$$
$$
y _ {\mathrm {l b}, \mu , \beta} \leq \sqrt {\mu} \leq y _ {\mathrm {u b}, \mu , \beta}
$$
Theorem 9 (Detailed Property of  $t_{\beta}(x;\mu)$ ). For  $t_{\beta}(x;\mu)$  in (20), then
(i) For  $\mu > 0$ ,  $\lim_{x \to 0^+} t_\beta(x; \mu) = \infty$ ,  $t_\beta(a; \mu) < 0$
(ii) For  $\mu >\frac{a^2}{4(a^2 + 1)^3}\frac{(\beta + 1)^4}{(\beta - 1)^2}$
$$
\frac {d t _ {\beta} (x ; \mu)}{d x} <   0
$$
$$
\begin{array}{l} F o r 0 <   \mu \leq \frac {a ^ {2}}{4 (a ^ {2} + 1) ^ {3}} \frac {(\beta + 1) ^ {4}}{(\beta - 1) ^ {2}} \\ \left\{ \begin{array}{l l} \frac {d t _ {\beta} (x ; \mu)}{d x} > 0 & x _ {\mathrm {l b}, \mu , \beta} <   x <   x _ {\mathrm {u b}, \mu , \beta} \\ \frac {d t _ {\beta} (x ; \mu)}{d x} \leq 0 & \text {O t h e r w i s e} \end{array} \right. \tag {36b} \\ \end{array}
$$
$$
x _ {\mathrm {l b}, \mu , \beta} = \frac {1}{2} \left(\frac {4 a \mu (1 + \beta) ^ {2}}{1 - \beta}\right) ^ {1 / 3} \left(1 - \sqrt {1 - \frac {(4 \mu) ^ {1 / 3} (a ^ {2} + 1)}{a ^ {2 / 3}} \left(\frac {1 - \beta}{(1 + \beta) ^ {2}}\right) ^ {2 / 3}}\right)
$$
$$
x _ {\mathrm {u b}, \mu , \beta} = \frac {1}{2} \left(\frac {4 a \mu (1 + \beta) ^ {2}}{1 - \beta}\right) ^ {1 / 3} \left(1 + \sqrt {1 - \frac {(4 \mu) ^ {1 / 3} (a ^ {2} + 1)}{a ^ {2 / 3}} \left(\frac {1 - \beta}{(1 + \beta) ^ {2}}\right) ^ {2 / 3}}\right)
$$
(iii) If  $0 < \beta < \frac{\sqrt{(a^2 + 1)} - 1}{\sqrt{(a^2 + 1)} + 1}$ , then there exists a  $\tau_{\beta} > 0$  such that,  $\forall \mu > \tau_{\beta}$ , the equation  $r_{\beta}(x; \mu) = 0$  has only one solution. At  $\mu = \tau_{\beta}$ , the equation  $r_{\beta}(x; \mu) = 0$  has two solutions, and  $\forall \mu < \tau_{\beta}$ , the equation  $r_{\beta}(x; \mu) = 0$  has three solutions. Moreover,  $\mu < \frac{a^2}{4(a^2 + 1)^3} \frac{(\beta + 1)^4}{(\beta - 1)^2}$ .
(iv) If  $0 < \beta < \frac{\sqrt{(a^2 + 1)} - 1}{\sqrt{(a^2 + 1)} + 1}$ , then  $\forall \mu < \tau_{\beta}$ ,  $t_{\beta}(x; \mu) = 0$  has three stationary points, i.e.  $x_{\mu, \beta}^{***} < x_{\mu, \beta}^{**} < x_{\mu, \beta}^{*}$ . Besides,
$$
\begin{array}{l} \max  _ {\mu \leq \tau_ {\beta}} x _ {\mu , \beta} ^ {* *} \leq \frac {a ((1 - \beta) \sqrt {a ^ {2} + 1} - \sqrt {(1 - \beta) ^ {2} (a ^ {2} + 1) - (\beta + 1) ^ {2}})}{2 \sqrt {a ^ {2} + 1}} \\ \frac {a ((1 - \beta) \sqrt {a ^ {2} + 1} + \sqrt {(1 - \beta) ^ {2} (a ^ {2} + 1) - (\beta + 1) ^ {2}})}{2 \sqrt {a ^ {2} + 1}} \leq \min  _ {\mu > 0} x _ {\mu , \beta} ^ {*} \\ \end{array}
$$
It implies that
$$
\max  _ {\mu \leq \tau_ {\beta}} x _ {\mu , \beta} ^ {* *} <   \min  _ {\mu > 0} x _ {\mu , \beta} ^ {*}
$$
Lemma 9. Under the same setting as Corollary 3,
$$
\max_{\mu \leq \tau}x^{**}_{\mu ,\epsilon} <   \min_{\mu >0}x^{*}_{\mu ,\epsilon}
$$
E Technical Proofs
E.1 Proof of Theorem 3
Proof. For the sake of completeness, we have included the proof here. Please note that this proof can also be found in [33].
Proof. We use the fact that  $f$  is  $L$ -smooth function if and only if for any  $W, Y \in \mathrm{dom}(f)$
$$
f (W) \leq f (Y) + \langle \nabla f (Y), Y - W \rangle + \frac {L}{2} \| Y - W \| _ {2} ^ {2}
$$
Let  $W = W^{t + 1}$  and  $Y = W^t$ , then using the updating rule  $W^{t + 1} = W^t -\frac{1}{L}\nabla f(W^t)$
$$
\begin{array}{l} f (W ^ {t + 1}) \leq f (W ^ {t}) + \langle \nabla f (W ^ {t}), W ^ {t + 1} - W ^ {t} \rangle + \frac {L}{2} \| W ^ {t + 1} - W ^ {t} \| _ {2} ^ {2} \\ = f (W ^ {t}) - \frac {1}{L} \| \nabla f (W ^ {t}) \| _ {2} ^ {2} + \frac {1}{2 L} \| \nabla f (W ^ {t}) \| _ {2} ^ {2} \\ = f (W ^ {t}) - \frac {1}{2 L} \| \nabla f (W ^ {t}) \| _ {2} ^ {2} \\ \end{array}
$$
$$
\begin{array}{l} \min  _ {0 \leq t \leq n - 1} \| \nabla f (W ^ {t}) \| _ {2} ^ {2} \leq \frac {1}{n} \sum_ {t = 0} ^ {n - 1} \| \nabla f (W ^ {t}) \| _ {2} ^ {2} \leq \frac {2 L (f (W ^ {0}) - f (W ^ {n}))}{n} \leq \frac {2 L (f (W ^ {0}) - f (W ^ {*}))}{n} \\ \min  _ {0 \leq t \leq n - 1} \| \nabla f (W ^ {t}) \| _ {2} ^ {2} \leq \frac {2 L (f (W ^ {0}) - f (W ^ {*}))}{n} \leq \epsilon^ {2} \Rightarrow n \geq \frac {2 L (f (W ^ {0}) - f (W ^ {*}))}{\epsilon^ {2}} \\ \end{array}
$$
E.2 Proof of Theorem 5
(i) For any  $\mu > 0$
$$
\lim  _ {y \to 0 ^ {+}} r (y; \mu) = \lim  _ {y \to 0 ^ {+}} \frac {a}{y} - \frac {a ^ {2}}{\mu} - (a ^ {2} + 1) = \infty
$$
$$
r (\frac {a}{a ^ {2} + 1}) = - \frac {\mu a ^ {2}}{(\frac {a}{a ^ {2} + 1}) ^ {2} + \mu} <   0.
$$
$$
\begin{array}{l} r (\sqrt {\mu}, \mu) = \frac {a}{\sqrt {\mu}} - \frac {a ^ {2}}{4 \mu} - (a ^ {2} + 1) \\ = - \frac {a ^ {2}}{4} (\frac {1}{\sqrt {\mu}} - \frac {2}{a}) ^ {2} - a ^ {2} <   0 \\ \end{array}
$$
$$
\begin{array}{l} \frac {d r (y ; \mu)}{d y} = - \frac {a}{y ^ {2}} + \frac {4 a ^ {2} \mu y}{(y ^ {2} + \mu) ^ {3}} \\ = \frac {4 a ^ {2} \mu y ^ {3} - a (y ^ {2} + \mu) ^ {3}}{y ^ {2} (y ^ {2} + \mu) ^ {3}} \\ = \frac {a ((4 a \mu) ^ {2 / 3} y ^ {2} + (4 a \mu) ^ {1 / 3} y (y ^ {2} + \mu) + (y ^ {2} + \mu) ^ {2}) ((4 a \mu) ^ {1 / 3} y - y ^ {2} - \mu)}{y ^ {2} (y ^ {2} + \mu) ^ {3}} \\ \end{array}
$$
For  $\mu \geq \frac{a^2}{4}$ ,  $(4a\mu)^{1/3}y - y^2 - \mu) < 0 \Leftrightarrow \frac{dr(y; \mu)}{dy} < 0$ .
For  $\mu < \frac{a^2}{4}$ ,  $y_{\mathrm{lb}} < y < y_{\mathrm{ub}}, ((4a\mu)^{1/3}y - y^2 - \mu) > 0 \Leftrightarrow \frac{dr(y; \mu)}{dy} > 0$ . For  $\mu < \frac{a^2}{4}$ ,
$y <   y_{\mathrm{lb}}$  or  $y_{\mathrm{ub}} <   y,((4a\mu)^{1 / 3}y - y^2 -\mu)\leq 0\Leftrightarrow \frac{dr(y;\mu)}{dy}\leq 0.$
$$
\frac {d r (y ; \mu)}{d \mu} = 0 \Leftrightarrow ((4 a \mu) ^ {1 / 3} y - y ^ {2} - \mu) = 0 \Leftrightarrow (4 a \mu) ^ {1 / 3} = y + \frac {\mu}{y}
$$
The intersection between line  $(4a\mu)^{1/3}$  and function  $y + \frac{\mu}{y}$  are exactly  $y_{\mathrm{lb}}$  and  $y_{\mathrm{ub}}$ , and  $y_{\mathrm{lb}} < \sqrt{\mu} < y_{\mathrm{ub}}$ .
(iv) Note that for  $0 < \mu < \frac{a^2}{4}$ ,
$$
\frac {\partial r}{\partial \mu} = - a ^ {2} \frac {y ^ {2} - \mu}{(\mu + y ^ {2}) ^ {3}} \quad \text {a n d} \quad y _ {\mathrm {l b}} <   \sqrt {\mu} <   y _ {\mathrm {u b}}
$$
then  $\left.\frac{\partial r}{\partial\mu}\right|_{y = y_{\mathrm{ub}}} < 0$ . Let  $p(\mu) = r(y_{\mathrm{ub}},\mu)$ , because  $\left.\frac{\partial r}{\partial y}\right|_{y = y_{\mathrm{ub}}} = 0$ , then
$$
\frac {d p (\mu)}{d \mu} = \frac {d r (y _ {\mathrm {u b}} , \mu)}{d \mu} = \left. \frac {\partial r}{\partial y} \right| _ {y = y _ {\mathrm {u b}}} \frac {d y _ {\mathrm {u b}}}{d \mu} + \left. \frac {\partial r}{\partial \mu} \right| _ {y = y _ {\mathrm {u b}}} = \left. \frac {\partial r}{\partial \mu} \right| _ {y = y _ {\mathrm {u b}}} <   0
$$
Also note that when  $\mu = \frac{a^2}{4}$ ,  $y_{\mathrm{ub}} = \sqrt{\mu}$ ,  $p(\mu) = r(y_{\mathrm{ub}},\mu) = r(\sqrt{\mu},\mu) < 0$ , and also if  $\mu < \frac{a^2}{4}$ , then
$$
y _ {\mathrm {u b}} <   \frac {(4 \mu) ^ {1 / 3}}{2} 2 a ^ {1 / 3} = (4 \mu a) ^ {1 / 3}
$$
$$
\begin{array}{l} r ((4 \mu a) ^ {1 / 3}, \mu) = \frac {a}{(4 \mu a) ^ {1 / 3}} - \frac {\mu a ^ {2}}{((4 \mu a) ^ {2 / 3} + \mu) ^ {2}} - (a ^ {2} + 1) \\ = \frac {a}{(4 \mu a) ^ {1 / 3}} - \frac {a ^ {2}}{(\mu) ^ {1 / 3} ((4 a) ^ {2 / 3} + \mu^ {1 / 3}) ^ {2}} - (a ^ {2} + 1) \\ > \frac {1}{\mu^ {1 / 3}} \left(\frac {a}{(4 a) ^ {1 / 3}} - \frac {a ^ {2}}{(4 a) ^ {4 / 3}}\right) - (a ^ {2} + 1) \\ \end{array}
$$
Because  $\frac{a}{(4a)^{1/3}} > \frac{a^2}{(4a)^{4/3}}$ , it is easy to see when  $\mu \to 0$ ,  $r((4\mu a)^{1/3}, \mu) \to \infty$ . We know  $r(y_{\mathrm{ub}}, \mu) > r((4\mu a)^{1/3}, \mu) \to \infty$  as  $\mu \to 0$  because of the monotonicity of  $r(y; \mu)$  in Theorem 5(iii). Combining all of these, i.e.
$$
\frac {d p (\mu)}{d \mu} <   0, \quad \lim _ {\mu \to 0 ^ {+}} p (\mu) = \infty , \quad p \left(\frac {a ^ {2}}{4}\right) <   0
$$
There exists a  $\tau <  \frac{a^2}{4}$  such that  $p(\tau) = 0$
(v) From Theorem 5(iv), for  $\mu >\tau$  , then  $p(\mu) = r(y_{\mathrm{ub}},\mu) > 0$  , and for  $\mu = \tau$  , then  $p(\mu) = r(y_{\mathrm{ub}},\mu) = 0$  . For  $\mu <  \tau$  , then  $p(\mu) = r(y_{\mathrm{ub}},\mu) <   0$  , combining Theorem 5(i),5(iii), we get the conclusions.
(vi) By Theorem 5(v),  $\forall \mu < \tau$ , there exists three stationary points such that  $0 < y_{\mu}^{*} < y_{\mathrm{lb}} < \sqrt{\mu} < y_{\mu}^{**} < y_{\mathrm{ub}} < y_{\mu}^{***}$ . Because  $\left.\frac{dr(y;\mu)}{dy}\right|_{y = y_{\mathrm{lb}}} = \left.\frac{dr(y;\mu)}{dy}\right|_{y = y_{\mathrm{ub}}} = 0$ , then
$$
\left. \frac {d r (y ; \mu)}{d y} \right| _ {y = y _ {\mu} ^ {*}} \neq 0, \quad \left. \frac {d r (y ; \mu)}{d y} \right| _ {y = y _ {\mu} ^ {* *}} \neq 0, \quad \left. \frac {d r (y ; \mu)}{d y} \right| _ {y = y _ {\mu} ^ {* * *}} \neq 0
$$
By implicit function theorem [14], for solution to equation  $r(y;\mu) = 0$ , there exists a unique continuously differentiable function such that  $y = y(\mu)$  and satisfies  $r(y(\mu),\mu) = 0$ . Therefore,
$$
\frac {\partial r}{\partial \mu} = - a ^ {2} \frac {y ^ {2} - \mu}{(\mu + y ^ {2}) ^ {3}}, \quad \frac {\partial r}{\partial y} = - \frac {a}{y ^ {2}} + \frac {4 a ^ {2} \mu y}{(y ^ {2} + \mu) ^ {3}}, \quad \frac {d y (\mu)}{d \mu} = - \frac {\partial r / \partial \mu}{\partial r / \partial y}
$$
Therefore by Theorem 5(iii),
$$
\left. \frac {d y}{d \mu} \right| _ {y = y _ {\mu} ^ {*}} > 0 \quad \left. \frac {d y}{d \mu} \right| _ {y = y _ {\mu} ^ {* *}} > 0 \quad \left. \frac {d y}{d \mu} \right| _ {y = y _ {\mu} ^ {* * *}} <   0
$$
Because  $\lim_{\mu \to 0^{+}}y_{\mathrm{lb}} = \lim_{\mu \to 0^{+}}y_{\mathrm{ub}} = 0$  , then  $\lim_{\mu \to 0^{+}}y_{\mu}^{*} = \lim_{\mu \to 0^{+}}y_{\mu}^{**} = 0$  . Let us consider  $r(\frac{a}{a^2 + 1} (1 - c\mu),\mu)$  where  $c = 32\frac{(a^2 + 1)^3}{a^2}$  and  $\mu <  \frac{1}{2c}$
$$
\begin{array}{l} r \left(\frac {a}{a ^ {2} + 1} (1 - c \mu), \mu\right) \\ = \frac {a}{\frac {a}{a ^ {2} + 1} (1 - c \mu)} - \frac {\mu a ^ {2}}{(\frac {a ^ {2}}{(a ^ {2} + 1) ^ {2}} (1 - c \mu) ^ {2} + \mu) ^ {2}} - (a ^ {2} + 1) \\ = (a ^ {2} + 1) \left(\frac {c \mu}{1 - c \mu}\right) - \frac {\mu a ^ {2}}{\left(\frac {a ^ {2}}{(a ^ {2} + 1) ^ {2}} (1 - c \mu) ^ {2} + \mu\right) ^ {2}} \\ \geq c (a ^ {2} + 1) \mu - \frac {\mu a ^ {2}}{(\frac {a ^ {2}}{(a ^ {2} + 1) ^ {2}} (1 - c \mu) ^ {2}) ^ {2}} \\ = c (a ^ {2} + 1) \mu - \frac {1 6 (a ^ {2} + 1) ^ {4}}{a ^ {2}} \mu \\ = \frac {1 6 (a ^ {2} + 1) ^ {4}}{a ^ {2}} \mu > 0 \\ \end{array}
$$
By Theorem 5(iii), then  $\frac{a}{a^2 + 1} (1 - c\mu) < y_\mu^{***}$ , then
$$
\frac {a}{a ^ {2} + 1} = \lim  _ {\mu \rightarrow 0 ^ {+}} \frac {a}{a ^ {2} + 1} (1 - c \mu), \mu) \leq \lim  _ {\mu \rightarrow 0 ^ {+}} y _ {\mu} ^ {* * *} \leq \frac {a}{a ^ {2} + 1}
$$
$$
\lim  _ {\mu \rightarrow 0 ^ {+}} y _ {\mu} ^ {* * *} = \frac {a}{a ^ {2} + 1}
$$
E.3 Proof of Theorem 6
Proof. (i) For  $\mu > 0$ ,
$$
\lim  _ {x \rightarrow 0 ^ {+}} t (x; \mu) = \lim  _ {x \rightarrow 0 ^ {+}} \frac {a}{x} - \frac {a ^ {2}}{\mu \left(a ^ {2} + 1\right) ^ {2}} - 1 = \infty
$$
$$
t (a, \mu) = - \frac {\mu a ^ {2}}{(\mu (a ^ {2} + 1) + a ^ {2}) ^ {2}} <   0
$$
$$
t (\sqrt {\mu (a ^ {2} + 1)}, \mu) = \frac {a}{\sqrt {a ^ {2} + 1}} \frac {1}{\sqrt {\mu}} - \frac {a ^ {2}}{4 \mu (a ^ {2} + 1) ^ {2}} - 1
$$
If  $t(\sqrt{\mu(a^2 + 1)},\mu) = 0$  , then
$$
\frac {1}{\sqrt {\mu}} = 2 \frac {(a ^ {2} + 1) ^ {3 / 2}}{a} \pm 2 (a ^ {2} + 1) \Rightarrow \mu = \left(\frac {a (\sqrt {a ^ {2} + 1} \mp a)}{2 (a ^ {2} + 1)}\right) ^ {2}
$$
so when  $\mu < \left(\frac{a(\sqrt{a^2 + 1} - a)}{2(a^2 + 1)}\right)^2$  or  $\mu > \left(\frac{a(\sqrt{a^2 + 1} + a)}{2(a^2 + 1)}\right)^2$ , then  $t(\sqrt{\mu(a^2 + 1)}, \mu) < 0$
$$
\begin{array}{l} \frac {d t (x , \mu)}{d x} \\ = - \frac {a}{x ^ {2}} + \frac {4 \mu a ^ {2} x}{(\mu (a ^ {2} + 1) + x ^ {2}) ^ {3}} \\ = \frac {4 \mu a ^ {2} x ^ {3} - a (\mu (a ^ {2} + 1) + x ^ {2}) ^ {3}}{x ^ {2} (\mu (a ^ {2} + 1) + x ^ {2}) ^ {3}} \\ = \frac {a ((\mu (a ^ {2} + 1) + x ^ {2}) ^ {2} + (\mu (a ^ {2} + 1) + x ^ {2}) (4 \mu a) ^ {1 / 3} x + (4 \mu a) ^ {2 / 3} x ^ {2}) ((4 \mu a) ^ {1 / 3} x - \mu (a ^ {2} + 1) - x ^ {2})}{x ^ {2} (\mu (a ^ {2} + 1) + x ^ {2}) ^ {3}} \\ \end{array}
$$
For  $\mu >\frac{a^2}{4(a^2 + 1)^3}$  , then  $(4\mu a)^{1 / 3}x - \mu (a^{2} + 1) - x^{2} <   0\Leftrightarrow \frac{dt(x,\mu)}{dx} <  0.$  For  $\mu <  \frac{a^2}{4(a^2 + 1)^3}$
and  $x_{\mathrm{lb}} < x < x_{\mathrm{ub}}$ , then  $(4\mu a)^{1/3}x - \mu(a^2 + 1) - x^2 > 0 \Leftrightarrow \frac{dt(x, \mu)}{dx} > 0$ . For  $\mu < \frac{a^2}{4(a^2 + 1)^3}$ ,
$x <   x_{\mathrm{lb}}$  or  $x > x_{\mathrm{ub}},(4\mu a)^{1 / 3}x - \mu (a^2 +1) - x^2 <  0\Leftrightarrow \frac{dt(x,\mu)}{dx} <  0.$
We use the same argument as before to show that
$$
x _ {\mathrm {l b}} <   \sqrt {\mu (a ^ {2} + 1)} <   x _ {\mathrm {u b}}
$$
(iv) Note that for  $0 < \mu < \frac{a^2}{4(a^2 + 1)^3}$
$$
\frac {\partial t}{\partial \mu} = - a ^ {2} \frac {x ^ {2} - \mu (a ^ {2} + 1)}{(\mu (a ^ {2} + 1) + x ^ {2}) ^ {3}} \quad \text {a n d} \quad x _ {\mathrm {l b}} <   \sqrt {\mu (a ^ {2} + 1)} <   x _ {\mathrm {u b}}
$$
then  $\left.\frac{\partial t}{\partial\mu}\right|_{x = x_{1b}} > 0$ . Let  $q(\mu) = t(x_{1b},\mu)$ , because  $\left.\frac{\partial t}{\partial x}\right|_{x = x_{1b}} = 0$ , then
$$
\frac {d q (\mu)}{d \mu} = \frac {d t (x _ {\mathrm {l b}} , \mu)}{d \mu} = \left. \frac {\partial t}{\partial x} \right| _ {x = x _ {\mathrm {l b}}} \frac {d x _ {\mathrm {l b}}}{d \mu} + \left. \frac {\partial t}{\partial \mu} \right| _ {x = x _ {\mathrm {l b}}} = \left. \frac {\partial t}{\partial \mu} \right| _ {x = x _ {\mathrm {l b}}} > 0
$$
Note that  $\mu = \frac{a^2}{4(a^2 + 1)^3}$ ,  $x_{\mathrm{ub}} = x_{\mathrm{lb}} = \frac{(4\mu a)^{1 / 3}}{2}$ ,  $t\left(\frac{(4\mu a)^{1 / 3}}{2},\frac{a^2}{4(a^2 + 1)^3}\right) = \frac{a}{(4\mu a)^{1 / 3}} -1 > 0$
When  $\mu < \left(\frac{a(\sqrt{a^2 + 1} - a)}{2(a^2 + 1)}\right)^2$ , then  $t(\sqrt{\mu(a^2 + 1)}, \mu) < 0$  by Theorem 6(ii). It implies that  $q(\mu) < 0$  when  $\mu \to 0^+$ . By Theorem 6(iii),  $q(\mu) = t(x_{\mathrm{lb}}, \mu) < t(\sqrt{\mu(a^2 + 1)}, \mu) < 0$ . Combining all of the theses, i.e.
$$
\frac {d q (\mu)}{d \mu} > 0, \quad \lim _ {\mu \to 0 ^ {+}} q (\mu) <   0, \quad q (\frac {a ^ {2}}{4 (a ^ {2} + 1) ^ {3}}) > 0
$$
There exists a  $\tau < \frac{a^2}{4(a^2 + 1)^3}$ ,  $q(\tau) = 0$ . Such  $\tau$  is the same as in Theorem 5(iv).
(v) We follow the same proof from the proof of Theorem 5(v).
(vi) By Theorem 6(v),  $\forall \mu < \mu_0$ , there exists three stationary points such that  $0 < x_{\mu}^{**} < x_{\mathrm{lb}} < x_{\mu}^{**} < x_{\mathrm{ub}} < x_{\mu}^{*} < a$ . Because  $\left.\frac{dt(x;\mu)}{dx}\right|_{x = x_{\mathrm{lb}}} = \left.\frac{dt(x;\mu)}{dx}\right|_{x = x_{\mathrm{ub}}} = 0$ , then
$$
\left. \frac {d t (x ; \mu)}{d x} \right| _ {x = x _ {\mu} ^ {*}} \neq 0, \quad \left. \frac {d t (x ; \mu)}{d x} \right| _ {x = x _ {\mu} ^ {* *}} \neq 0, \quad \left. \frac {d t (x ; \mu)}{d x} \right| _ {x = x _ {\mu} ^ {* * *}} \neq 0
$$
By implicit function theorem [14], for solutions to equation  $t(x; \mu) = 0$ , there exists a unique continuously differentiable function such that  $x = x(\mu)$  and satisfies  $t(x(\mu), \mu) = 0$ . Therefore,
$$
\frac {d x}{d \mu} = - \frac {\partial t / \partial \mu}{\partial t / \partial x} = a ^ {2} \frac {\frac {x ^ {2} - \mu (a ^ {2} + 1)}{(\mu (a ^ {2} + 1) + x ^ {2}) ^ {3}}}{- \frac {a}{x ^ {2}} + \frac {4 \mu a ^ {2} x}{(\mu (a ^ {2} + 1) + x ^ {2}) ^ {3}}}
$$
Therefore, by Theorem 6(iii)
$$
\left. \frac {d x}{d \mu} \right| _ {x = x _ {\mu} ^ {*}} <   0 \quad \left. \frac {d x}{d \mu} \right| _ {x = x _ {\mu} ^ {* * *}} > 0
$$
Because  $0 < x_{\mu}^{***} < x_{\mathrm{lb}} < x_{\mu}^{**} < x_{\mathrm{ub}}$  and  $\lim_{\mu \to 0^+}x_{\mathrm{lb}} = \lim_{\mu \to 0^+}x_{\mathrm{ub}} = 0$
$$
\lim  _ {\mu \to 0} x _ {\mu} ^ {* *} = \lim  _ {\mu \to 0} x _ {\mu} ^ {* * *} = 0
$$
Let us consider  $t(a(1 - c\mu),\mu)$  where  $c = \frac{32}{a^2}$  and  $\mu < \frac{1}{2c}$ .
$$
\begin{array}{l} t (a (1 - c \mu); \mu) \\ = \frac {a}{a (1 - c \mu)} - \frac {\mu a ^ {2}}{(\mu (a ^ {2} + 1) + a ^ {2} (1 - c \mu) ^ {2}) ^ {2}} - 1 \\ = \frac {c \mu}{1 - c \mu} - \frac {\mu a ^ {2}}{(\mu (a ^ {2} + 1) + a ^ {2} (1 - c \mu) ^ {2}) ^ {2}} \\ \geq c \mu - \frac {\mu a ^ {2}}{(a ^ {2} (1 - c \mu) ^ {2}) ^ {2}} \\ \geq c \mu - \frac {1 6}{a ^ {2}} \mu > 0 \\ \end{array}
$$
By Theorem 6(iii). It implies
$$
a (1 - c \mu) \leq x _ {\mu} ^ {*}
$$
taking  $\mu \rightarrow 0^{+}$  on both side,
$$
a = \lim  _ {\mu \rightarrow 0 ^ {+}} a (1 - c \mu) \leq \lim  _ {\mu \rightarrow 0 ^ {+}} x _ {\mu} ^ {*} \leq a
$$
Hence,  $\lim_{\mu \to 0}x_{\mu}^{*} = a$