SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} \left\langle \mathbf {w} _ {j} ^ {\left(\mathcal {T} _ {1}\right)}, \mathbf {x} _ {i} \right\rangle \leq \left\langle \mathbf {w} _ {j} ^ {(1)}, \mathbf {x} _ {i} \right\rangle - \eta \left(T _ {i j} (1, \mathcal {T} _ {1}) + (\gamma - \rho) G _ {j} ^ {(i)} (1, \mathcal {T} _ {1}) - (\gamma + \rho) B _ {j} ^ {(i)} (1, \mathcal {T} _ {1})\right) \\ \leq 0 - \eta (\mathcal {T} _ {1} - 2) ((n - k) (\gamma - \rho) - 2 k (\gamma + \rho)) \\ = - \frac {(n - k) (\gamma - \rho) - 2 k (\gamma + \rho)}{1 . 0 3 m [ 1 + (\gamma + \rho) (n - k) ]} + O (\eta) \\ \leq - \frac {(n - 3 k) \gamma - (n + k) \rho}{1 . 0 3 m} + O (\eta) \\ \leq - \frac {0 . 9 7 n \gamma - \frac {1 . 0 1}{5} n}{1 . 0 3 m} + O (\eta) \\ \leq - \frac {3 n \gamma}{4 m}. \\ \end{array}
$$
using  $\eta$  is sufficiently small and  $k\leq \frac{n}{100}$  and  $\rho \leq \frac{\gamma}{5}$  (Assumption 3). Likewise, for  $1\leq t <   \mathcal{T}_1$
$$
\begin{array}{l} \ell (t, \mathbf {x} _ {i}) - \ell (t + 1, \mathbf {x} _ {i}) = y _ {i} [ f (t + 1, \mathbf {x} _ {i}) - f (t, \mathbf {x} _ {i}) ] \\ = \sum_ {j = 1} ^ {2 m} (- 1) ^ {i + j} \left[ \phi \left(\left\langle \mathbf {w} _ {j} ^ {(t + 1)}, \mathbf {x} _ {i} \right\rangle\right) - \phi \left(\left\langle \mathbf {w} _ {j} ^ {(t)}, \mathbf {x} _ {i} \right\rangle\right) \right] \\ = \sum_ {j \notin \Gamma_ {(- 1) ^ {i + 1}}} (- 1) ^ {i + j} \left[ \phi \left(\left\langle \mathbf {w} _ {j} ^ {(t + 1)}, \mathbf {x} _ {i} \right\rangle\right) - \phi \left(\left\langle \mathbf {w} _ {j} ^ {(t)}, \mathbf {x} _ {i} \right\rangle\right) \right] \\ = \eta \left(\sum_ {j \in \Gamma_ {(- 1) ^ {i}}} [ \langle \mathbf {w} _ {j} ^ {(t + 1)}, \mathbf {x} _ {i} \rangle - \langle \mathbf {w} _ {j} ^ {(t)}, \mathbf {x} _ {i} \rangle ]\right) \\ + \eta \left(\sum_ {j \notin \Gamma} (- 1) ^ {i + j} \left[ \phi \left(\left\langle \mathbf {w} _ {j} ^ {(t + 1)}, \mathbf {x} _ {i} \right\rangle\right) - \phi \left(\left\langle \mathbf {w} _ {j} ^ {(t)}, \mathbf {x} _ {i} \right\rangle\right) \right]\right) \\ \geq \eta \left(\sum_ {j \in \Gamma_ {(- 1) ^ {i}}} T _ {i j} (t _ {0}, t) + G _ {j} ^ {(i)} (t _ {0}, t) (\gamma - \rho) - B _ {j} ^ {(i)} (t _ {0}, t) (\gamma + \rho)\right) \\ + \eta \left(\sum_ {j \notin \Gamma} T _ {i j} (t, t + 1) - (\gamma + \rho) B _ {j} (t, t + 1)\right) \\ \geq 0. 9 9 \eta m [ 1 + (\gamma - \rho) (n - k - 1) - 2 (\gamma + \rho) k ] - 0. 0 4 \eta m (\gamma + \rho) k \\ \geq 0. 9 9 \eta m [ 1 + (\gamma - \rho) (n - k - 1) ] - 2. 0 2 \eta m (\gamma + \rho) k \\ \end{array}
$$
In the first six lines we use:  $t < \mathcal{T}_0$ , the definition of  $f(t, \mathbf{x})$ , Lemma C.4, Lemma B.2 and Lemma B.3, Lemma C.4 again, and  $|\Gamma_p| \geq 0.99m$  (Assumption 3), respectively.
We also bound
$$
\begin{array}{l} \ell (1, \mathbf {x} _ {i}) = 1 - y _ {i} f (1, \mathbf {x} _ {i}) \\ = 1 - \sum_ {j = 1} ^ {2 m} (- 1) ^ {i + j} \phi \left(\left\langle \mathbf {w} _ {j} ^ {(1)}, \mathbf {x} _ {i} \right\rangle\right) \\ \leq 1 + \sum_ {i \not \sim j} \phi (\langle \mathbf {w} _ {j} ^ {(1)}, \mathbf {x} _ {i} \rangle) \\ \leq 1 + \sum_ {i \neq j} [ \phi (\langle \mathbf {w} _ {j} ^ {(0)}, \mathbf {x} _ {i} \rangle) - \eta T _ {i j} (0, 1) + \eta (\gamma + \rho) B _ {j} (0, 1) + \eta ] \\ \leq 1 + m \left[ \lambda_ {w} + \eta + 2 \eta (\gamma + \rho) k \right]. \\ \end{array}
$$
Combining these two bounds we see that
$$
\begin{array}{l} \ell \left(\mathcal {T} _ {1}, \mathbf {x} _ {i}\right) \leq \ell \left(1, \mathbf {x} _ {i}\right) - \eta \left(\mathcal {T} _ {1} - 1\right) \left(0. 9 9 m (1 + (\gamma - \rho) (n - k - 1)) - 2. 0 2 m (\gamma + \rho) k\right) \\ \leq 1 - \frac {0 . 9 9 m [ 1 + (\gamma - \rho) (n - k - 1) ] - 2 . 0 2 m (\gamma + \rho) k}{1 . 0 3 m (1 + (\gamma + \rho) (n - k))} + O (\eta) \\ \leq \frac {0 . 9 9 ((\gamma + \rho) + 2 \rho (n - k - 1)) + 0 . 0 8 + \frac {4 . 0 4}{9 9}}{1 . 0 3 m (1 + (\gamma + \rho) (n - k))} + O (\eta) \\ \leq \frac {0 . 9 9 (\frac {1}{1 0 0} + \frac {1}{6} + 0 . 0 8 + \frac {4 . 0 4}{9 9})}{1 . 0 3} + O (\eta) \\ \leq \frac {1}{3}. \\ \end{array}
$$
at this iteration, as desired. Here we use  $(\gamma + \rho) \leq \min \left\{\frac{1}{99k}, \frac{1}{100}, \frac{1}{n - k}\right\}$ ,  $\eta$  is sufficiently small, and  $\rho \leq \frac{1}{6(n - k)}$  (Assumption 3)
C.4 Late training
Lemma C.8. Suppose Assumption 3 holds. Fix  $\varepsilon >0$ . We will say a neuron is aligned (at iteration  $t$ ) if
$$
(- 1) ^ {j} \mathrm {s g n} \left\langle \mathbf {w} _ {j} ^ {(t)}, \mathbf {x} _ {i} \right\rangle = y _ {i}
$$
for all  $i\in \mathcal{S}_T$  . For  $t\geq \mathcal{T}_1$  , if
$$
B (\mathcal {T} _ {1}, t) \leq \frac {5 \varepsilon n}{8 \eta}
$$
than at least  $(1 - \varepsilon)m$  neurons in each  $\Gamma_p$  will be aligned.
Proof. Let  $p \in \{-1, 1\}$  and  $t$  be such that  $\varepsilon m$  different neurons in  $\Gamma_p$  are unaligned at iteration  $t$ . For any neuron index  $j \in \Gamma_p$  and  $i \in S_T$ , we can use Lemma B.2 to bound
$$
\begin{array}{l} (- 1) ^ {i + j} \left\langle \mathbf {w} _ {j} ^ {(t)}, \mathbf {x} _ {i} \right\rangle \geq (- 1) ^ {i + j} \left\langle \mathbf {w} _ {j} ^ {\left(\mathcal {T} _ {1}\right)}, \mathbf {x} _ {i} \right\rangle + \eta T _ {i j} \left(\mathcal {T} _ {1}, t\right) \\ + \eta (\gamma - \rho) G _ {j} ^ {(i)} \left(\mathcal {T} _ {1}, t\right) - \eta (\gamma + \rho) B _ {j} ^ {(i)} \left(\mathcal {T} _ {1}, t\right) \\ = \min  \left\{\frac {3}{4 m}, \frac {3 n \gamma}{4 m} \right\} - \eta (\gamma + \rho) B _ {j} \left(\mathcal {T} _ {1}, t\right) \\ \geq \frac {3 n \gamma}{4 m} - \eta (\gamma + \rho) B _ {j} (\mathcal {T} _ {1}, t). \\ \end{array}
$$
Since  $n\gamma < 1$  (Assumption 3), we see that  $\min \left\{\frac{3}{4m}, \frac{3n\gamma}{4m}\right\} = \frac{3n\gamma}{4m}$ . If the lower bound above is positive then  $(-1)^j \operatorname{sgn} \langle \mathbf{w}_j^{(t)}, \mathbf{x}_i \rangle = y_i$ . Therefore, if a neuron  $j$  is unaligned then  $B_j(\mathcal{T}_1, t) \geq \frac{3n\gamma}{4\eta m(\gamma + \rho)} \geq \frac{5n}{8\eta m}$  (using  $\rho \leq \frac{\gamma}{5}$  from Assumption 3). If there are  $\varepsilon m$  unaligned neurons, then
$$
B \left(\mathcal {T} _ {1}, t\right) = \sum_ {j = 1} ^ {2 m} B _ {j} \left(\mathcal {T} _ {1}, t\right) \geq \frac {5 \varepsilon n}{8 \eta}.
$$
Denote the first iteration after  $\mathcal{T}_1$  where more than  $\varepsilon m$  neurons in one of the  $\Gamma_p$  are unaligned as  $\mathcal{T}_{\varepsilon}$ . If no such iteration exists, let  $\mathcal{T}_{\varepsilon} = \infty$ . We will eventually show that indeed  $\mathcal{T}_{\varepsilon} = \infty$ , by showing that the training process reaches zero loss before such an iteration can happen.
Lemma C.9. Assume Assumption 3 holds and also  $\gamma + \rho < \frac{0.99(1 - \varepsilon)}{4k}$ . There is an iteration  $\mathcal{T}_2 \geq \mathcal{T}_1$  so that for all iterations  $t$  satisfying  $\mathcal{T}_2 \leq t < \mathcal{T}_{\varepsilon}$  and all  $i \in S_T$ ,
$$
\ell (t, \mathbf {x} _ {i}) \leq 4 \eta (\gamma + \rho) k m.
$$
Furthermore, we can choose  $\mathcal{T}_2$  so that
$$
\mathcal {T} _ {2} - \mathcal {T} _ {1} \leq \frac {1}{3 \eta m (0 . 9 9 (1 - \varepsilon) - 4 k (\gamma + \rho))} + 1.
$$
Proof. Fix  $i \in S_T$  and  $t < \mathcal{T}_{\varepsilon} - 1$ . Suppose  $\ell(t, \mathbf{x}_i) > 0$ . Using Lemma B.3 and  $t < \mathcal{T}_{\varepsilon}$ ,
$$
\begin{array}{l} y _ {i} f (t + 1, \mathbf {x} _ {i}) - y _ {i} f (t, \mathbf {x} _ {i}) = \sum_ {j = 1} ^ {2 m} (- 1) ^ {i + j} (\phi (\langle \mathbf {w} _ {j} ^ {(t + 1)}, \mathbf {x} _ {i} \rangle) - \phi (\langle \mathbf {w} _ {j} ^ {(t)}, \mathbf {x} _ {i} \rangle)) \\ \geq \sum_ {j = 1} ^ {2 m} \eta \left(T _ {i j} (t, t + 1) - (\gamma + \rho) B _ {j} (t, t + 1)\right) \\ \geq \eta 0. 9 9 (1 - \varepsilon) m - 4 \eta m k (\gamma + \rho). \\ \end{array}
$$
$$
\ell (t + 1, \mathbf {x} _ {i}) \leq \min  \left\{\ell (t, \mathbf {x} _ {i}) - \eta m (0. 9 9 (1 - \varepsilon) - 4 k), 0 \right\}.
$$
By Lemma C.7, the loss of each clean point at  $\mathcal{T}_1$  is at most  $\frac{1}{3}$ , so each clean point reaches zero loss in at most
$$
\left\lceil \frac {1}{3 \eta m (0 . 9 9 (1 - \varepsilon) - 4 k (\gamma + \rho))} \right\rceil
$$
Now suppose  $\ell(t, \mathbf{x}_i) = 0$ . We similarly argue
$$
\begin{array}{l} y _ {i} f (t + 1, \mathbf {x} _ {i}) - y _ {i} f (t, \mathbf {x} _ {i}) = \sum_ {j = 1} ^ {2 m} (- 1) ^ {i + j} \left(\phi \left(\left\langle \mathbf {w} _ {j} ^ {(t + 1)}, \mathbf {x} _ {i} \right\rangle\right) - \phi \left(\left\langle \mathbf {w} _ {j} ^ {(t)}, \mathbf {x} _ {i} \right\rangle\right)\right) \\ \geq \sum_ {j = 1} ^ {2 m} \eta \left(T _ {i j} (t, t + 1) - B _ {j} (t, t + 1)\right) \\ \geq - 4 \eta m k. \\ \end{array}
$$
This implies  $\ell(t + 1, \mathbf{x}_i) \leq 4\eta mk$ . By induction, we see that if  $t < \mathcal{T}_{\varepsilon}$  and  $\ell(t, \mathbf{x}_i) \leq 4\eta mk$ , then  $\ell(t + 1, \mathbf{x}_i) \leq 4\eta mk$ .
Lemma C.10. Assume Assumption 3 holds and  $\gamma +\rho < \frac{0.99(1 - \varepsilon)}{4k}$ . For all  $t_1,t_2$  satisfying  $\mathcal{T}_2\leq t_1\leq t_2 < \mathcal{T}_{\varepsilon}$  and  $i\in S_T$
$$
S _ {i} \left(t _ {1}, t _ {2}\right) \leq \frac {4 \left(t _ {2} - t _ {1}\right) (\gamma + \rho) k + 4 k + 3}{0 . 9 9 (1 - \varepsilon)}.
$$
Proof. Recall Lemma B.5 (restated in this setting, using Lemma C.9):
$$
T _ {i} \left(t _ {1}, t _ {2}\right) \leq \frac {4 \eta m k}{\eta} + (\gamma + \rho) B \left(t _ {1}, t _ {2}\right) + 3 m.
$$
Using  $t <   \mathcal{T}_{\varepsilon}$  we bound
$$
\begin{array}{l} 0. 9 9 (1 - \varepsilon) m S _ {i} \left(t _ {1}, t _ {2}\right) \leq T _ {i} \left(t _ {1}, t _ {2}\right) \\ \leq \frac {4 \eta m k}{\eta} + (\gamma + \rho) B (t _ {1}, t _ {2}) + 3 m \\ \leq 4 m k + 4 \left(t _ {2} - t _ {1}\right) (\gamma + \rho) m k + 3 m. \\ \end{array}
$$
From this, the desired inequality follows.
Lemma C.11. Assume Assumption 3 holds and  $\gamma +\rho \leq \min \left\{\frac{0.99(1 - \varepsilon)}{4k},\sqrt{\frac{0.99(1 - \varepsilon)}{8(n - k)k}}\right\}$ . Let  $i\in S_F$ . Suppose there is  $j\not\sim i$  such that
$$
\langle \mathbf {w} _ {j} ^ {\left(\tau_ {2}\right)}, \mathbf {x} _ {i} \rangle > \eta \frac {2 (n - k) (\gamma + \rho) (4 k + 3)}{0 . 9 9 (1 - \varepsilon)}.
$$
Then for all  $t$  satisfying  $\mathcal{T}_2\leq t <   \mathcal{T}_{\varepsilon},\langle \mathbf{w}_{j'}^{(t)},\mathbf{x}_i\rangle >0$  for some neuron  $j^{\prime}$  depending on  $t$
Proof. Let  $\tau_0 \geq \mathcal{T}_2$  be the first iteration after  $\mathcal{T}_2$  where  $\ell(t, \mathbf{x}_i) = 0$ . We will show by induction that for all  $t$  satisfying  $\mathcal{T}_2 \leq t \leq \tau_0$  that  $\langle \mathbf{w}_j^{(t)}, \mathbf{x}_i \rangle > 0$ . The case  $t = \mathcal{T}_2$  follows immediately by the assumption of the lemma. Otherwise, assume  $\langle \mathbf{w}_j^{(t')}, \mathbf{x}_i \rangle > 0$  for all  $\mathcal{T}_2 \leq t' < t$ . By Lemma B.2 and Lemma C.10,
$$
\begin{array}{l} \left\langle \mathbf {w} _ {j} ^ {(t + 1)}, \mathbf {x} _ {i} \right\rangle \geq \left\langle \mathbf {w} _ {j} ^ {\left(\mathcal {T} _ {2}\right)}, \mathbf {x} _ {i} \right\rangle \\ + \eta \left(T _ {i j} \left(\mathcal {T} _ {2}, t + 1\right) - G _ {j} ^ {(i)} \left(\mathcal {T} _ {2}, t + 1\right) (\gamma + \rho) + B _ {j} ^ {(i)} \left(\mathcal {T} _ {2}, t + 1\right) (\gamma - \rho)\right) \\ \geq \left\langle \mathbf {w} _ {j} ^ {\left(\mathcal {T} _ {2}\right)}, \mathbf {x} _ {i} \right\rangle + \eta (t + 1 - \mathcal {T} _ {2}) - \eta (\gamma + \rho) \sum_ {i \in \mathcal {S} _ {T}} S _ {i} \left(\mathcal {T} _ {2}, t + 1\right) \\ \geq \left\langle \mathbf {w} _ {j} ^ {\left(\mathcal {T} _ {2}\right)}, \mathbf {x} _ {i} \right\rangle + \eta (t + 1 - \mathcal {T} _ {2}) \\ - \eta 2 (n - k) (\gamma + \rho) \left(\frac {4 (t + 1 - \mathcal {T} _ {2}) (\gamma + \rho) k + 4 k + 3}{0 . 9 9 (1 - \varepsilon)}\right) \\ = \eta (t + 1 - \mathcal {T} _ {2}) \left(1 - \frac {8 (n - k) k (\gamma + \rho) ^ {2}}{0 . 9 9 (1 - \varepsilon)}\right) + \langle \mathbf {w} _ {j} ^ {\left(\mathcal {T} _ {2}\right)}, \mathbf {x} _ {i} \rangle \\ - \eta \frac {2 (n - k) (\gamma + \rho) (4 k + 3)}{0 . 9 9 (1 - \varepsilon)} \\ > 0. \\ \end{array}
$$
We now continue the induction past  $\tau_0$ . If  $\ell(t, \mathbf{x}_i) = 0$ , then point  $i$  clearly activates some neuron. Let  $\tau_1$  be the first iteration after  $\tau_0$  where  $\ell(t, \mathbf{x}_i) > 0$ . By Lemma B.3
$$
\begin{array}{l} y _ {i} f (\tau_ {1}, \mathbf {x} _ {i}) - y _ {i} f (\tau_ {1} - 1, \mathbf {x} _ {i}) = \sum_ {j ^ {\prime} = 1} ^ {2 m} - (- 1) ^ {i + j} (\phi (\langle \mathbf {w} _ {j} ^ {(\tau_ {1})}, \mathbf {x} _ {i} \rangle) - \phi (\langle \mathbf {w} _ {j} ^ {(\tau_ {1} - 1)}, \mathbf {x} _ {i} \rangle)) \\ \geq \sum_ {j = 1} ^ {2 m} \eta \left(T _ {i j} (t, \tau_ {1}) - G _ {j} (t, \tau_ {1} - 1)\right) \\ \geq - 4 \eta m (n - k). \\ \end{array}
$$
$$
\sum_ {j ^ {\prime} \not \sim i} (- 1) ^ {j} \phi (\langle \mathbf {w} _ {j ^ {\prime}} ^ {(\tau_ {1})}, \mathbf {x} _ {i} \rangle) \geq y _ {i} f (\tau_ {1}, \mathbf {x} _ {i}) \geq 1 - 4 \eta m (n - k)
$$
and there is some  $j'$  satisfying
$$
\phi \left(\left\langle \mathbf {w} _ {j ^ {\prime}} ^ {\left(\tau_ {1}\right)}, \mathbf {x} _ {i} \right\rangle\right) \geq \frac {1}{m} - 4 \eta (n - k) > \eta \frac {2 (n - k) (\gamma + \rho) (4 k + 3)}{0 . 9 9 (1 - \varepsilon)},
$$
assuming  $\eta$  is sufficiently small (Assumption 3). We can run the original induction argument with  $\tau_{1}$  replacing  $\mathcal{T}_2$  and  $\tau_{2} = \min \{t\geq \tau_{2}:\ell (t,\mathbf{x}_{i}) = 0\}$  replacing  $\tau_0$  to verify the conclusion for  $\tau_{1}\leq t\leq \tau_{2}$ . By switching back and forth between these two arguments, we can show that point  $i$  activates some neuron for all  $\mathcal{T}_2\leq t < \mathcal{T}_{\varepsilon}$ .
Lemma C.12. If Assumption 3 holds, the training process reaches loss.
Proof. In this proof, let  $\varepsilon = \frac{1}{5}$ . The conditions of Lemma C.9, Lemma C.10, and Lemma C.11 hold because  $\gamma + \rho \leq \min \left\{\frac{0.99 / 5}{k}, \sqrt{\frac{0.99 / 5}{2k(n - k)}}\right\}$ .
By Lemma C.2, there is a finite bound on the number of updates, independent of the number of iterations spent training. If we carry out the training procedure for infinitely many iterations, there must be some iteration where we make no updates. Since the training procedure is deterministic, we will not make any updates after this point, and we will have converged. It remains to show that this convergence results in zero training loss. The only way for a point to not update any neurons is for that point's loss to be zero or for that point to activate no neurons.
Lemma C.8 and Lemma C.11 say, under certain conditions, that every clean point and every corrupted point activates some neuron for each iteration  $t \leq \mathcal{T}_{\varepsilon}$ . We need only to verify that these conditions hold and that  $B(\mathcal{T}_1, t)$  remains below the limitation set in Lemma C.8.
We apply Lemma C.2 starting at  $t_0 = \mathcal{T}_1$ . By Lemma C.7,  $\ell(\mathcal{T}_1, \mathbf{x}_i) \leq \frac{1}{3}$  for all  $i \in S_T$ . Using Lemma C.6, for  $i \in S_F$ ,
$$
\begin{array}{l} \ell \left(\mathcal {T} _ {1}, \mathbf {x} _ {i}\right) \leq 1 - y _ {i} \sum_ {j = 1} ^ {2 m} (- 1) ^ {j} \phi \left(\left\langle \mathbf {w} _ {j} ^ {\left(\mathcal {T} _ {1}\right)}, \mathbf {x} _ {i} \right\rangle\right) \\ \leq 1 + \sum_ {j \sim i} ^ {2 m} \phi \left(\left\langle \mathbf {w} _ {j} ^ {\left(\mathcal {T} _ {1}\right)}, \mathbf {x} _ {i} \right\rangle\right) \\ \leq 1 + O (\eta). \\ \end{array}
$$
With these bounds, Lemma C.2 shows that for all  $t \geq \mathcal{T}_1$ ,
$$
\begin{array}{l} B \left(\mathcal {T} _ {1}, t\right) \leq \frac {2 k}{1 - 4 k (n - k) (\gamma + \rho) ^ {2}} \left(\frac {1}{\eta} + 2 (n - k) (\gamma + \rho) \frac {1}{3 \eta}\right) + O (1) \\ \leq \frac {2 k (1 + (2 / 3) (n - k) (\gamma + \rho))}{\eta (1 - 4 k (n - k) (\gamma + \rho) ^ {2})} + O (1) \\ \leq \frac {2 k (5 / 3)}{\eta (1 - 4 / 9 9)} + O (1) \\ \leq \frac {n}{1 0 \eta} \\ \end{array}
$$
using  $\gamma + \rho \leq \min \left\{\frac{1}{n - k}, \frac{1}{99k}\right\}$ ,  $\eta$  sufficiently small, and  $k \leq \frac{n}{100}$  (Assumption 3). By Lemma C.8,  $\mathcal{T}_{\varepsilon} = \infty$  if  $\frac{n}{10\eta} < \frac{5\varepsilon n}{8\eta}$ , which is clearly true for  $\varepsilon = \frac{1}{5}$ .
We now show that every training point  $i$  activates at least one neuron each iteration. By Lemma C.9, this is true if  $i \in S_T$ . By Lemma C.11, this is true for  $i \in S_F$  if there is a neuron  $j \not\sim i$  such that  $\langle \mathbf{w}_j^{(\mathcal{T}_2)}, \mathbf{x}_i \rangle > \eta \frac{2(n - k)(\gamma + \rho)(4k + 3)}{0.99(1 - \varepsilon)}$ . Fix  $i \in S_F$ .
First, assume that  $\ell(t, \mathbf{x}_i) > 0$  for all  $t < \mathcal{T}_2$ . By Assumption 3 and Lemma C.3, we know there is  $j \in \Gamma_{y_i}$  such that  $i \in \mathcal{A}_j^{(1)}$ . Using Lemma B.2 and Lemma C.4 we can bound
$$
\begin{array}{l} \langle \mathbf {w} _ {j} ^ {\left(\mathcal {T} _ {1}\right)}, \mathbf {x} _ {i} \rangle \geq \langle \mathbf {w} _ {j} ^ {(1)}, \mathbf {x} _ {i} \rangle + \eta T _ {i j} (1, \mathcal {T} _ {1}) - \eta (\gamma + \rho) G _ {j} ^ {(i)} (1, \mathcal {T} _ {1}) + \eta (\gamma - \rho) B _ {j} ^ {(i)} (1, \mathcal {T} _ {1}) \\ \geq \eta (1 - (\gamma + \rho) (n - k)) \left(\mathcal {T} _ {1} - 1\right) \\ \end{array}
$$
$$
\langle \mathbf {w} _ {j} ^ {(\mathcal {T} _ {2})}, \mathbf {x} _ {i} \rangle \geq \langle \mathbf {w} _ {j} ^ {(\mathcal {T} _ {1})}, \mathbf {x} _ {i} \rangle + \eta T _ {i j} (\mathcal {T} _ {1}, \mathcal {T} _ {2}) - \eta (\gamma + \rho) G _ {j} ^ {(i)} (\mathcal {T} _ {1}, \mathcal {T} _ {2}) + \eta (\gamma - \rho) B _ {j} ^ {(i)} (\mathcal {T} _ {1}, \mathcal {T} _ {2}).
$$
By induction on  $t$  we see that
$$
G _ {j} \left(\mathcal {T} _ {1}, t\right) \leq \max  _ {\mathcal {T} _ {1} \leq t ^ {\prime} <   t} \left((n - k) \left(t + 1 - t ^ {\prime}\right) + \frac {(\gamma + \rho) B _ {j} \left(\mathcal {T} _ {1} , t ^ {\prime}\right)}{\gamma - \rho}\right)
$$
for  $t > \mathcal{T}_1$ . The base case  $t = \mathcal{T}_1 + 1$  is clear. Suppose the inequality holds for  $t$ . Either  $G_j(\mathcal{T}_1, t)$  increases by at most  $n - k$  or there is some  $i' \in S_T \cap \mathcal{A}_j^{(t+1)}$  with  $i' \not\sim j$ . By Lemma B.2 and Lemma C.7,
$$
\begin{array}{l} 0 <   \langle \mathbf {w} _ {j} ^ {(t)}, \mathbf {x} _ {i ^ {\prime}} \rangle \leq \langle \mathbf {w} _ {j} ^ {(\mathcal {T} _ {1})}, \mathbf {x} _ {i ^ {\prime}} \rangle - \eta (\gamma - \rho) G _ {j} (\mathcal {T} _ {1}, t + 1) + \eta (\gamma + \rho) B _ {j} (\mathcal {T} _ {1}, t) \\ \leq - \eta (\gamma - \rho) G _ {j} \left(\mathcal {T} _ {1}, t\right) + \eta (\gamma + \rho) B _ {j} \left(\mathcal {T} _ {1}, t\right) \\ \end{array}
$$
from which the inequality follows. Since  $\frac{(\gamma + \rho)B_j(\mathcal{T}_1,t')}{\gamma - \rho} \leq 3k(t' - \mathcal{T}_1) < (n - k)(t' - \mathcal{T}_1)$  (using  $\rho < \frac{\gamma}{5}$  and  $k \leq \frac{n}{100}$  from Assumption 3), this maximum occurs at  $\tau = \mathcal{T}_1$ . This yields
$$
\begin{array}{l} \langle \mathbf {w} _ {j} ^ {\left(\mathcal {T} _ {2}\right)}, \mathbf {x} _ {i} \rangle \geq \eta (1 - (\gamma + \rho) (n - k)) \left(\mathcal {T} _ {2} + 1 - 1\right) \eta (\gamma - \rho) B _ {j} ^ {(i)} \left(\mathcal {T} _ {1}, \mathcal {T} _ {2}\right) \\ \geq \eta (1 - (\gamma + \rho) (n - k)) \mathcal {T} _ {2} \\ \end{array}
$$
We want to show this bound is larger than a quantity that is  $O(\eta)$ . This happens when both of the following hold:
$$
O (1) \leq (1 - (\gamma + \rho) (n - k)) \mathcal {T} _ {2},
$$
which holds when  $\eta$  is sufficiently small.
Now suppose  $\ell(\tau, \mathbf{x}_i) = 0$  for some iteration  $\tau \leq \mathcal{T}_2$ . By Lemma C.7,  $\mathcal{T}_1 < \tau$ . In this case, we see that
$$
\begin{array}{l} y _ {i} f \left(\mathcal {T} _ {2}, \mathbf {x} _ {i}\right) = y _ {i} f \left(\mathcal {T} _ {2}, \mathbf {x} _ {i}\right) + \sum_ {j = 1} ^ {2 m} - (- 1) ^ {i + j} \left[ \phi \left(\left\langle \mathbf {w} _ {j} ^ {\left(\mathcal {T} _ {1}\right)}, \mathbf {x} _ {i} \right\rangle\right) - \phi \left(\left\langle \mathbf {w} _ {j} ^ {(\tau)}, \mathbf {x} _ {i} \right\rangle\right) \right] \\ \geq 1 - \eta (\gamma + \rho) G (\tau , \mathcal {T} _ {2}) \\ \geq 1 - (\gamma + \rho) \frac {2 (n - k)}{1 - 4 k (n - k) (\gamma + \rho) ^ {2}} \left[ \frac {1}{3} + 2 k (\gamma + \rho) \right] + O (\eta) \\ \geq 1 - \frac {2}{1 - 4 / 9 9} \left(\frac {1}{3} + \frac {2}{9 9}\right) + O (\eta) \geq \frac {1}{4} \\ \end{array}
$$
where in the third line we use Lemma C.2 and the fourth line we use  $\gamma + \rho \leq \min \left\{\frac{1}{n - k}, \frac{1}{99k}\right\}$  and  $\eta$  sufficiently small (Assumption 3). Since this is positive, there is some neuron  $j$  with  $i \not\sim j$  such that  $\phi(\langle \mathbf{w}_j^{(\mathcal{T}_2)}, \mathbf{x}_i \rangle)$  is at least  $\frac{1}{m}$  this bound. This is an  $\Omega(1)$  lower bound. Since the required condition is  $\langle \mathbf{w}_j^{(\mathcal{T}_2)}, \mathbf{x}_i \rangle > O(\eta)$ , this can be achieved by taking  $\eta$  sufficiently small.
C.5 Proof of Lemma 3.5
Lemma C.13 (Lemma 3.5). Assume Assumption 3 holds. Let  $y \in \{-1,1\}$  chosen uniformly and  $\mathbf{x} := y\sqrt{\gamma}\mathbf{v} + \sqrt{1 - \gamma}\mathbf{n}$ , where  $\mathbf{n} \sim \mathrm{Uniform}(S^{d-1} \cap \mathrm{span}\{\mathbf{v}\}^{\perp})$ . Suppose that  $|\langle \mathbf{n}, \mathbf{n}_{\ell} \rangle| < \frac{\rho}{1 - \gamma}$  for all  $l \in [2n]$ , then  $yf(T_{end}, \mathbf{x}) > 0$ .
Proof. Following the same steps as in (5) for any  $j \in [2m]$
$$
\begin{array}{l} \langle \mathbf {w} _ {j} ^ {\left(\mathcal {T} _ {\text {e n d}}\right)}, \mathbf {x} \rangle = \langle \mathbf {w} _ {j} ^ {(1)}, \mathbf {x} _ {i} \rangle + (- 1) ^ {j} \eta \sum_ {\ell = 1} ^ {2 n} T _ {\ell j} (1, \mathcal {T} _ {\text {e n d}}) y _ {\ell} \langle \mathbf {x} _ {\ell}, \mathbf {x} _ {i} \rangle \\ = \left\langle \mathbf {w} _ {j} ^ {(t _ {0})}, \mathbf {x} _ {i} \right\rangle + (- 1) ^ {j} y \eta \sum_ {\ell = 1} ^ {2 n} T _ {\ell j} (t _ {0}, t) (- 1) ^ {\ell} y \beta (\ell) \left\langle \mathbf {x} _ {\ell}, \mathbf {x} \right\rangle \\ = \left\langle \mathbf {w} _ {j} ^ {(t _ {0})}, \mathbf {x} _ {i} \right\rangle + (- 1) ^ {j} y \eta \sum_ {\ell = 1} ^ {2 n} T _ {\ell j} (t _ {0}, t) \lambda_ {i \ell} ^ {\prime}, \\ \end{array}
$$
where  $\lambda_{\ell}^{\prime}\coloneqq (-1)^{l}y\beta (\ell)\langle \mathbf{x}_{\ell},\mathbf{x}\rangle = \beta (\ell)\gamma +(1 - \gamma)\langle \mathbf{n}_{\ell},\mathbf{n}\rangle$  . Then as in Lemma B.1
$$
\begin{array}{l} \gamma - \rho \leq \lambda_ {\ell} ^ {\prime} \leq \gamma + \rho , \text {i f} i \in S _ {T}, \\ - (\gamma + \rho) \leq \lambda_ {\ell} ^ {\prime} \leq - (\gamma - \rho), \text {i f} i \in \mathcal {S} _ {F}. \\ \end{array}
$$
Recall, from Lemma C.4 for any  $j \in \Gamma_p$  then  $G_j(1, \mathcal{T}_{\mathrm{end}}) \geq G_j(1, \mathcal{T}_1) = \mathcal{T}_1(n - k)$ . As a consequence, for  $j \in \Gamma_p$  we have
$$
\begin{array}{l} \langle \mathbf {w} _ {j} ^ {\left(\mathcal {T} _ {\text {e n d}}\right)}, \mathbf {x} \rangle \geq \langle \mathbf {w} _ {j} ^ {(1)}, \mathbf {x} _ {i} \rangle + \eta G _ {j} (1, \mathcal {T} _ {\text {e n d}}) (\gamma - \rho) - \eta B _ {j} (1, \mathcal {T} _ {\text {e n d}}) (\gamma + \rho) \\ \geq O (\eta) + \mathcal {T} _ {1} (n - k) (\gamma - \rho) - \eta B _ {j} (1, \mathcal {T} _ {\text {e n d}}) (\gamma + \rho). \\ \end{array}
$$
For  $j$  such that  $(-1)^{j} = y$  then
$$
\begin{array}{l} \phi \left(\left\langle \mathbf {w} _ {j} ^ {\left(\mathcal {T} _ {\text {e n d}}\right)}, \mathbf {x} \right\rangle\right) \leq \phi \left(\left\langle \mathbf {w} _ {j} ^ {(1)}, \mathbf {x} _ {i} \right\rangle - \eta G _ {j} (1, \mathcal {T} _ {\text {e n d}}) (\gamma - \rho) + \eta B _ {j} (1, \mathcal {T} _ {\text {e n d}}) (\gamma + \rho)\right) \\ \leq O (\eta) + \eta B _ {j} (1, \mathcal {T} _ {\mathrm {e n d}}) (\gamma + \rho). \\ \end{array}
$$