SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
g \left(x _ {0}\right) - g (0) \leq \left(4 \log (c) + 2 \log \left(\eta_ {1} ^ {- 1}\right) + 3. 5 \log 2 + 0. 5\right) L _ {2} ^ {- 1} \leq \Delta_ {0},
$$
it suffices to take
$$
\log \left(\eta_ {1} ^ {- 1}\right) = \frac {L _ {2} \Delta_ {0} - 3 . 5 \log 2 - 0 . 5}{2} - 2 \log (c).
$$
Since we require  $\eta_1 \leq c^2 / 2$ , parameters  $L_2$  and  $\Delta_0$  need to satisfy
$$
\log 2 - 2 \log (c) \leq \frac {L _ {2} \Delta_ {0} - 3 . 5 \log 2 - 0 . 5}{2} - 2 \log (c),
$$
that is,  $L_{2}\Delta_{0} \geq 5.5\log 2 + 0.5$ , which holds because  $L_{2}\Delta_{0} \geq 10$ . Take  $c = \max \{2 / \sqrt{L_0L_2}, 2 / (L_2G_0), \sqrt{8 / L_0}\}$ . Then, as long as  $\eta \leq 2 / L_0$ , the requirement that  $\eta \leq c^2 / 4$  is satisfied. Therefore, on  $g(x)$  with initial point  $x_0$ , gradient descent with a constant stepsize either gets stuck, or takes at least
$$
\begin{array}{l} c \eta_ {1} ^ {- 1 / 2} / 6 = \frac {c}{6} \exp \left(\frac {L _ {2} \Delta_ {0} - 3 . 5 \log 2 - 0 . 5}{4} - \log (c)\right) \\ = \frac {1}{6} \exp \left(\frac {L _ {2} \Delta_ {0} - 3 . 5 \log 2 - 0 . 5}{4}\right) \\ \geq \frac {1}{6} \exp \left(\frac {L _ {2} \Delta_ {0}}{8}\right) \\ \end{array}
$$
steps to reach a 1-stationary point.
On the other hand, if  $\eta > 2 / L_0$ , consider the function  $f(x) = \frac{L_0}{2} x^2$ . For any  $x_t \neq 0$ , we always have  $|x_{t+1}| / |x_t| = |1 - \eta L_0| > 1$ , which means the iterates diverge to infinity.