SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\mathcal {L} _ {i} (\theta) \geq \mathcal {L} _ {i} \left(\theta^ {\prime}\right) + \left\langle \nabla \mathcal {L} _ {i} \left(\theta^ {\prime}\right), \theta - \theta^ {\prime} \right\rangle + \frac {1}{2 L _ {i}} \left\| \nabla \mathcal {L} _ {i} (\theta) - \nabla \mathcal {L} _ {i} \left(\theta^ {\prime}\right) \right\| ^ {2}.
$$
Substituting  $\theta' = \theta_*$  in the above, we have  $\mathcal{L}_i(\theta) - \mathcal{L}_i(\theta_*) - \langle \nabla \mathcal{L}_i(\theta_%), \theta - \theta_* \rangle \geq 0$  and  $\| \nabla \mathcal{L}_i(\theta) - \nabla \mathcal{L}_i(\theta_*) \|^2 \leq 2L_i(\mathcal{L}_i(\theta) - \mathcal{L}_i(\theta_*) - \langle \nabla \mathcal{L}_i(\theta_%), \theta - \theta_* \rangle)$ . Therefore, as  $L_{\max} := \max_{i \in \mathcal{H}} L_i$ , we have
$$
\frac {1}{| \mathcal {H} |} \sum_ {i \in \mathcal {H}} \| \nabla \mathcal {L} _ {i} (\theta) - \nabla \mathcal {L} _ {i} \left(\theta_ {*}\right) \| ^ {2} \leq \frac {2 L _ {\max }}{| \mathcal {H} |} \sum_ {i \in \mathcal {H}} \left(\mathcal {L} _ {i} (\theta) - \mathcal {L} _ {i} \left(\theta_ {*}\right) - \langle \nabla \mathcal {L} _ {i} \left(\theta_ {*}\right), \theta - \theta_ {*} \rangle\right). \tag {10}
$$
Recall that  $\mathcal{L}_{\mathcal{H}} := \frac{1}{|\mathcal{H}|} \sum_{i \in \mathcal{H}} \mathcal{L}_i$ . Thus,  $\frac{1}{|\mathcal{H}|} \sum_{i \in \mathcal{H}} \mathcal{L}_i(\theta_*) = \mathcal{L}_{\mathcal{H}}(\theta_*)$ , and  $\frac{1}{|\mathcal{H}|} \sum_{i \in \mathcal{H}} \nabla \mathcal{L}_i(\theta_*) = \nabla \mathcal{L}_{\mathcal{H}}(\theta_*) = 0$ . Using this in (10), and then recalling that  $\mathcal{L}_{\mathcal{H}}$  is assumed  $\mu$ -PL, we have
$$
\frac {1}{| \mathcal {H} |} \sum_ {i \in \mathcal {H}} \| \nabla \mathcal {L} _ {i} (\theta) - \nabla \mathcal {L} _ {i} (\theta_ {*}) \| ^ {2} \leq 2 L _ {\max} (\mathcal {L} _ {\mathcal {H}} (\theta) - \mathcal {L} _ {\mathcal {H}} (\theta_ {*})) \leq \frac {L _ {\max}}{\mu} \| \nabla \mathcal {L} _ {\mathcal {H}} (\theta) \| ^ {2}.
$$
Substituting the above in (9), we obtain that
$$
\begin{array}{l} \frac {1}{| \mathcal {H} |} \sum_ {i \in \mathcal {H}} \| \nabla \mathcal {L} _ {i} (\theta) - \nabla \mathcal {L} _ {\mathcal {H}} (\theta) \| ^ {2} \leq 2 \frac {L _ {\max}}{\mu} \| \nabla \mathcal {L} _ {\mathcal {H}} (\theta) \| ^ {2} + 2 \frac {1}{| \mathcal {H} |} \sum_ {i \in \mathcal {H}} \| \nabla \mathcal {L} _ {i} (\theta_ {*}) \| ^ {2} - \| \nabla \mathcal {L} _ {\mathcal {H}} (\theta) \| ^ {2} \\ = \frac {2}{| \mathcal {H} |} \sum_ {i \in \mathcal {H}} \| \nabla \mathcal {L} _ {i} (\theta_ {*}) \| ^ {2} + \left(\frac {2 L _ {\max}}{\mu} - 1\right) \| \nabla \mathcal {L} _ {\mathcal {H}} (\theta) \| ^ {2}. \\ \end{array}
$$
The above proves the proposition.
B Proof of Theorem 1: Impossibility Result
For convenience, we recall below the theorem statement.
$$
\frac {f}{n} <   \frac {1}{2 + B ^ {2}} \qquad a n d \qquad \varepsilon \geq \frac {1}{8 \mu} \cdot \frac {f}{n - (2 + B ^ {2}) f} \cdot G ^ {2} .
$$
B.1 Proof outline
We prove the theorem by contradiction. We start by assuming that there exists an algorithm  $\mathcal{A}$  that is  $(f, \varepsilon)$ -resilient when the conditions stated in the theorem for the honest workers are satisfied. We consider the following instance of the loss functions where parameters  $\alpha$  and  $K$  are positive real values, and  $z \in \mathbb{R}^d$  is a vector.
$$
\mathcal {L} _ {i} (\theta) = \mathcal {L} _ {\mathrm {I}} (\theta) := \frac {\alpha}{2} \| \theta - z \| ^ {2}, \quad \forall i \in \{1, \dots , f \}, \tag {11}
$$
$$
\mathcal {L} _ {i} (\theta) = \mathcal {L} _ {\mathrm {I I}} (\theta) := \frac {1}{2 K} \| \theta \| ^ {2}, \quad \forall i \in \{f + 1, \dots , n - f \}, \tag {12}
$$
$$
\mathcal {L} _ {i} (\theta) = \mathcal {L} _ {\mathrm {I I I}} (\theta) := \frac {\alpha}{2} \| \theta \| ^ {2}, \quad \forall i \in \{n - f + 1, \dots , n \}. \tag {13}
$$
We then consider two specific scenarios, each corresponding to different identities of honest workers:  $S_{1} = \{1,\dots ,n - f\}$  and  $S_{2} = \{f + 1,\dots ,n\}$ . That is, we let  $S_{1}$  and  $S_{2}$  represent the set of honest workers in the first and second scenarios, respectively. Upon specifying certain conditions on parameters  $\alpha$ ,  $K$  and  $z$  we show that the corresponding honest workers' loss functions in either execution satisfy the assumptions stated in the theorem, i.e., the global loss functions are  $L$ -smooth  $\mu$ -strongly convex and the honest local loss functions satisfy  $(G,B)$ -gradient dissimilarity. Since algorithm  $\mathcal{A}$  is oblivious to the honest identities, it must ensure  $(f,\varepsilon)$ -resilience in both these scenarios. Consequently, we show that  $\varepsilon$  cannot be lower than a value that grows with  $K$  and  $\| z\| ^2$ . Using this approach, we first show that the lower bound on  $\varepsilon$  can be arbitrarily large when  $\frac{f}{n}\geq \frac{1}{2 + B^2}$ . Then, we obtain a non-trivial lower bound on  $\varepsilon$  in the case when  $\frac{f}{n} < \frac{1}{2 + B^2}$ . The two scenarios, and corresponding loss functions, used in our proof are illustrated in Figure 5.
Figure 5: Illustration of the proof of Theorem 1. It is impossible to distinguish between the two scenarios depicted above, corresponding to the local honest losses in scenarios  $S_{1}$  and  $S_{2}$ . We set  $\| z\| ^2$  to grow with  $K$ , and show that  $(G,B)$ -gradient dissimilarity holds in both scenarios. A large  $K$  means that the minimum in the first scenario (left) is close to  $z$ , while it is 0 in the second scenario (right). Thus, any algorithm must make an error  $\varepsilon$  in the order of  $\| z\| ^2$ , which itself grows with  $K$ . When  $f / n\geq \frac{1}{2 + B^2}$ , we show that  $K$  and thus  $\varepsilon$  can be made arbitrarily large.
We make use of the following auxiliary results.
B.1.1 Unavoidable error due to anonymity of Byzantine workers
Lemma 1 below establishes a lower bound on the optimization error that any algorithm must incur in at least one of the two scenarios described above. Recall that  $S_{1} = \{1, \dots, n - f\}$  and
$S_{2} = \{f + 1,\dots ,n\}$  . Recall that for any non-empty subset  $S\subseteq [n]$  , we denote
$$
\mathcal {L} _ {S} (\theta) := \frac {1}{| S |} \sum_ {i \in S} \mathcal {L} _ {i} (\theta) \quad \text {a n d} \quad \mathcal {L} _ {*, S} = \min  _ {\theta \in \mathbb {R} ^ {d}} \mathcal {L} _ {S} (\theta).
$$
Lemma 1. Consider the setting where the local loss functions are given by (11), (12), and (13). In this particular case, the following holds for all  $\theta \in \mathbb{R}^d$ :
$$
\max  \left\{\mathcal {L} _ {S _ {1}} (\theta) - \mathcal {L} _ {*, S _ {1}}, \mathcal {L} _ {S _ {2}} (\theta) - \mathcal {L} _ {*, S _ {2}} \right\} \geq \frac {\left(\frac {f}{n - f}\right) ^ {2} \alpha^ {2}}{8 \left(\frac {n - 2 f}{n - f} \frac {1}{K} + \frac {f}{n - f} \alpha\right)} \| z \| ^ {2}.
$$
The proof is deferred to Appendix B.3. We next analyze the  $(G,B)$ -gradient dissimilarity condition for the considered distributed learning setting.
B.1.2 Validity of  $(G,B)$ -gradient dissimilarity
In Lemma 2 below we derive necessary and sufficient conditions on  $\alpha$ ,  $K$  and  $z$  for  $(G, B)$ -gradient dissimilarity when the loss functions are given by (11), (12) and (13).
Lemma 2. Consider the setting where the local loss functions are given by (11), (12), and (13). Denote
$$
A _ {1} := \frac {f (n - 2 f)}{(n - f) ^ {2}} \left(\left(1 - \frac {n - 2 f}{f} B ^ {2}\right) \frac {1}{K ^ {2}} - 2 (1 + B ^ {2}) \frac {\alpha}{K} + \left(1 - \frac {f}{n - 2 f} B ^ {2}\right) \alpha^ {2}\right),
$$
$$
A _ {2} := \frac {f (n - 2 f) \alpha}{(n - f) ((n - 2 f) \frac {1}{K} + f \alpha)} \left(\frac {1}{K} - \alpha\right) \frac {1}{K} z, a n d A _ {3} := \frac {f (n - 2 f) \alpha^ {2}}{\left((n - 2 f) \frac {1}{K} + f \alpha\right) ^ {2}} \frac {\| z \| ^ {2}}{K ^ {2}}. \tag {14}
$$
Suppose that  $S_{1} = \{1, \dots, n - f\}$  denotes the set of honest workers. Then, the honest workers satisfy  $(G, B)$ -gradient dissimilarity if and only if
$$
A _ {1} \leq 0, \quad A _ {3} \leq G ^ {2}, \quad a n d \quad \| A _ {2} \| ^ {2} \leq A _ {1} \left(A _ {3} - G ^ {2}\right). \tag {15}
$$
The proof of Lemma 2 is deferred to Appendix B.4. Since  $\mathcal{L}_{\mathrm{I}}$  corresponds to  $\mathcal{L}_{\mathrm{III}}$  with  $z = 0$ , the result in the lemma above also holds true when the honest workers are represented by set  $S_{2} = \{f + 1, \dots, n\}$ . Specifically, we have the following lemma.
Lemma 3. Consider a specific distributed learning setting where the local loss functions are as defined in (11), (12) and (13). We denote
$$
A _ {1} := \frac {f (n - 2 f)}{(n - f) ^ {2}} \left(\left(1 - \frac {n - 2 f}{f} B ^ {2}\right) \frac {1}{K ^ {2}} - 2 (1 + B ^ {2}) \frac {\alpha}{K} + \left(1 - \frac {f}{n - 2 f} B ^ {2}\right) \alpha^ {2}\right).
$$
Suppose that  $S_{2} = \{f + 1, \dots, n\}$  denotes the set of honest workers. Then, the honest workers satisfy  $(0, B)$ -gradient dissimilarity if and only if  $A_{1} \leq 0$ .
We do not provide a proof of Lemma 3, as it follows the proof of Lemma 2 verbatim upon simply substituting  $z = 0_{\mathbb{R}^d}$  and  $G = 0$
We prove the two assertions of Theorem 1, i.e., the necessity of  $\frac{f}{n} < \frac{1}{2 + B^2}$  and the lower bound on  $\varepsilon$ , separately in sections B.2.1 and B.2.2, respectively.
B.2.1 Necessity of  $\frac{f}{n} < \frac{1}{2 + B^2}$
In this section, we prove by contradiction the necessity of  $\frac{f}{n} < \frac{1}{2 + B^2}$  by demonstrating that  $\varepsilon$  can be arbitrarily large if  $\frac{f}{n} \geq \frac{1}{2 + B^2}$ . Let  $0 < f < n/2$ ,  $0 < \mu < L$ ,  $G > 0$  and  $B \geq 0$ .
Suppose that  $\frac{f}{n} \geq \frac{1}{2 + B^2}$ , or equivalently  $B^2 \geq \frac{n - 2f}{f}$ . Let  $\mathcal{A}$  be an arbitrary  $(f, \varepsilon)$ -resilient distributed learning algorithm. We consider the setting where the local loss functions are given by (11), (12) and (13), and the corresponding parameters are given by
$$
\alpha = \frac {n - f}{f} \mu , \tag {16}
$$
$K$  is an arbitrary positive real number such that
$$
K \geq \frac {1}{\alpha} \max  \left\{1, \frac {n - 2 f}{f} \cdot \frac {\mu}{L - \mu} \right\}, \tag {17}
$$
and  $z\in \mathbb{R}^d$  is such that
$$
\left\| z \right\| ^ {2} = \frac {f}{n - 2 f} \cdot \frac {G ^ {2}}{2 \alpha} K. \tag {18}
$$
Proof outline. In the proof, we consider two scenarios, each corresponding to two different identities of honest workers:  $S_{1} = \{1,\dots ,n - f\}$  and  $S_{2} = \{f + 1,\dots ,n\}$ . For each of these, we first show that the corresponding local and global honest loss functions satisfy the assumptions made in the theorem, invoking Lemma 2. Then, by invoking Lemma 1, we show that  $\varepsilon$  is proportional to  $K$  which (as per (17)) can be chosen to be arbitrarily large. This yields a contradiction to the assumption that  $\mathcal{A}$  is  $(f,\varepsilon)$ -resilient, proving that  $(f,\varepsilon)$ -resilience is generally impossible when  $\frac{f}{n}\geq \frac{1}{2 + B^2}$ .
First scenario. Suppose that the set of honest workers is represented by  $S_{1} = \{1,\dots ,n - f\}$ . From (11) and (12), we obtain that
$$
\begin{array}{l} \mathcal {L} _ {S _ {1}} (\theta) = \frac {1}{| S _ {1} |} \sum_ {i \in S _ {1}} \mathcal {L} _ {i} (\theta) = \frac {f}{n - f} \mathcal {L} _ {\mathrm {I}} (\theta) + \frac {n - 2 f}{n - f} \mathcal {L} _ {\mathrm {I I}} (\theta) \\ = \frac {f}{n - f} \frac {\alpha}{2} \| \theta - z \| ^ {2} + \frac {n - 2 f}{n - f} \frac {1}{2 K} \| \theta \| ^ {2}. \\ \end{array}
$$
Substituting, from (16),  $\alpha = \frac{n - f}{f}\mu$  in the above we obtain that
$$
\mathcal {L} _ {S _ {1}} (\theta) = \frac {\mu}{2} \| \theta - z \| ^ {2} + \frac {n - 2 f}{n - f} \frac {1}{2 K} \| \theta \| ^ {2}. \tag {19}
$$
$$
\nabla \mathcal {L} _ {S _ {1}} (\theta) = \mu (\theta - z) + \frac {n - 2 f}{n - f} \frac {1}{K} \theta , \text {a n d} \nabla^ {2} \mathcal {L} _ {S _ {1}} (\theta) = \left(\mu + \frac {n - 2 f}{n - f} \frac {1}{K}\right) I _ {d}, \tag {20}
$$
where  $I_{d}$  denotes the identity matrix of size  $d$ . From (20), we deduce that  $\mathcal{L}_{S_1}(\theta)$  is  $\left( \frac{n - 2f}{n - f} \frac{1}{K} + \mu \right)$ -smooth and  $\left( \frac{n - 2f}{n - f} \frac{1}{K} + \mu \right)$ -strongly convex. From (17), we have  $K \geq \frac{1}{\alpha} \cdot \frac{n - 2f}{f} \cdot \frac{\mu}{L - \mu} = \frac{n - 2f}{n - f} \cdot \frac{1}{L - \mu}$ . This implies that  $\frac{n - 2f}{n - f} \frac{1}{K} + \mu \leq L - \mu + \mu = L$ . Clearly,  $\left( \frac{n - 2f}{n - f} \frac{1}{K} + \mu \right) \geq \mu$ . Hence, the honest global loss function  $\mathcal{L}_{S_1}$  is  $L$ -smooth  $\mu$ -strongly convex. Next, invoking Lemma 2, we show that the honest workers satisfy  $(G, B)$ -gradient dissimilarity. We analyze below the terms  $A_1, A_2$  and  $A_3$  introduced in Lemma 2 in this particular scenario.
Term  $A_{1}$ . Recall from Lemma 2 that
$$
A _ {1} = \frac {f (n - 2 f)}{(n - f) ^ {2}} \left(\left(1 - \frac {n - 2 f}{f} B ^ {2}\right) \frac {1}{K ^ {2}} - 2 (1 + B ^ {2}) \frac {\alpha}{K} + \alpha^ {2} \left(1 - \frac {f}{n - 2 f} B ^ {2}\right)\right).
$$
Since we assume  $B^2 \geq \frac{n - 2f}{f}$ , we have  $1 - \frac{f}{n - 2f} B^2 \leq 0$ . Using this in the above we obtain that
$$
\begin{array}{l} A _ {1} \leq \frac {f (n - 2 f)}{(n - f) ^ {2}} \left(\left(1 - \frac {n - 2 f}{f} B ^ {2}\right) \frac {1}{K ^ {2}} - 2 (1 + B ^ {2}) \frac {\alpha}{K}\right) \\ = \frac {f (n - 2 f)}{(n - f) ^ {2}} \frac {\alpha}{K} \left(\left(1 - \frac {n - 2 f}{f} B ^ {2}\right) \frac {1}{\alpha K} - 2 (1 + B ^ {2})\right). \\ \end{array}
$$
As  $\alpha, K > 0$ , and by (17),  $K \geq \frac{1}{\alpha} \geq \frac{1}{\alpha} \frac{\left(1 - \frac{n - 2f}{f} B^{2}\right)}{1 + B^{2}}$ . Therefore,  $1 + B^{2} \geq \frac{1}{\alpha K} \left(1 - \frac{n - 2f}{f} B^{2}\right)$ . Using this in the above implies that
$$
A _ {1} \leq - \frac {f (n - 2 f)}{(n - f) ^ {2}} \frac {\alpha}{K} (1 + B ^ {2}) \leq - \frac {f (n - 2 f)}{(n - f) ^ {2}} \frac {\alpha}{K}. \tag {21}
$$
Term  $A_{2}$  . Recall from Lemma 2 that
$$
A _ {2} = \frac {f (n - 2 f) \alpha}{(n - f) ((n - 2 f) \frac {1}{K} + f \alpha)} \left(\frac {1}{K} - \alpha\right) \frac {1}{K} z. \tag {22}
$$
Therefore, as  $K > 0$  and we assume  $n > 2f$ , we have
$$
\begin{array}{l} \left\| A _ {2} \right\| ^ {2} = \left(\frac {f (n - 2 f) \alpha}{(n - f) ((n - 2 f) \frac {1}{K} + f \alpha)}\right) ^ {2} \left(\frac {1}{K} - \alpha\right) ^ {2} \frac {1}{K ^ {2}} \| z \| ^ {2} \\ \leq \left(\frac {f (n - 2 f) \alpha}{(n - f) f \alpha}\right) ^ {2} \left(\frac {1}{K} - \alpha\right) ^ {2} \frac {1}{K ^ {2}} \| z \| ^ {2} = \left(\frac {n - 2 f}{n - f}\right) ^ {2} \left(\alpha - \frac {1}{K}\right) ^ {2} \frac {1}{K ^ {2}} \| z \| ^ {2}. \\ \end{array}
$$
As  $K \geq \frac{1}{\alpha} > 0$ , we have  $\left(\alpha - \frac{1}{K}\right)^2 \leq \alpha^2$ . Thus, from above we obtain that
$$
\left\| A _ {2} \right\| ^ {2} \leq \left(\frac {n - 2 f}{n - f}\right) ^ {2} \alpha^ {2} \frac {1}{K ^ {2}} \| z \| ^ {2},
$$
Substituting from (18), i.e.  $\| z\|^2 = \frac{f}{n - 2f} \frac{G^2}{2\alpha} \cdot K$ , in the above implies that
$$
\left\| A _ {2} \right\| ^ {2} \leq \frac {f (n - 2 f)}{(n - f) ^ {2}} \frac {\alpha}{K} \frac {G ^ {2}}{2}. \tag {23}
$$
Term  $A_{3}$ . Recall from Lemma 2 that
$$
A _ {3} := \frac {f (n - 2 f) \alpha^ {2}}{\left((n - 2 f) \frac {1}{K} + f \alpha\right) ^ {2}} \frac {\| z \| ^ {2}}{K ^ {2}}.
$$
As  $K > 0$  and  $n > 2f$ , we have
$$
A _ {3} \leq \frac {f (n - 2 f) \alpha^ {2}}{f ^ {2} \alpha^ {2}} \frac {\| z \| ^ {2}}{K ^ {2}} = \frac {n - 2 f}{f} \frac {\| z \| ^ {2}}{K ^ {2}}.
$$
Substituting  $\| z\|^2 = \frac{f}{n - 2f} \frac{G^2}{2\alpha} \cdot K$  (see (18)), and then recalling that  $K \geq \frac{1}{\alpha}$  (see (17)), yields
$$
A _ {3} \leq \frac {G ^ {2}}{2 \alpha K} \leq \frac {G ^ {2}}{2}.
$$
$$
G ^ {2} - A _ {3} \geq \frac {G ^ {2}}{2} \geq 0. \tag {24}
$$
**Invoking Lemma 2.** Using the results obtained above in (21), (23) and (24), we show below that the conditions stated in (15) of Lemma (2) are satisfied, i.e.,  $A_{1} \leq 0$ ,  $A_{3} \leq G^{2}$  and  $\| A_{2}\|^{2} \leq A_{1}(A_{3} - G^{2})$ . Hence, we prove that the local loss functions for the honest workers in this particular scenario satisfy  $(G,B)$ -gradient dissimilarity.
First, from (21), we note that
$$
A _ {1} \leq - \frac {f (n - 2 f)}{(n - f) ^ {2}} \frac {\alpha}{K} \leq 0. \tag {25}
$$
Second, as a consequence of (24),  $A_3 \leq G^2$ . Lastly, recall from (23) that
$$
\left\| A _ {2} \right\| ^ {2} \leq \frac {f (n - 2 f)}{(n - f) ^ {2}} \frac {\alpha}{K} \frac {G ^ {2}}{2}.
$$
As  $\frac{f(n - 2f)}{(n - f)^2}\frac{\alpha}{K} \leq -A_1$  (from (21)), from above we obtain that  $\| A_2\|^2 \leq -A_1\frac{G^2}{2}$ . Therefore, as  $G^2 \geq 0$  and  $G^2 \leq 2(G^2 - A_3)$  (from (24)),  $\| A_2\|^2 \leq A_1(A_3 - G^2)$ .
This concludes the analysis for the first scenario. We have shown that the conditions on smoothness, strong convexity and  $(G,B)$ -gradient dissimilarity hold in this scenario.
Second scenario. Consider the set of honest workers to be  $S_{2} = \{f + 1,\dots ,n\}$ . Identical to (19) with  $z = 0$ , from (12) and (13), we obtain that
$$
\mathcal {L} _ {S _ {2}} (\theta) = \left(\frac {\mu}{2} + \frac {n - 2 f}{n - f} \frac {1}{2 K}\right) \| \theta \| ^ {2}.
$$
Therefore, similar to the analysis of the first scenario, the global loss  $\mathcal{L}_{S_2}$  satisfies  $L$ -smoothness and  $\mu$ -strong convexity. Moreover, Lemma 3, in conjunction with the deduction in (25) that  $A_{1} \geq 0$ , implies that the loss functions for the honest workers in this particular scenario satisfy  $(0, B)$ -gradient dissimilarity, thereby also satisfying  $(G, B)$ -gradient dissimilarity.
This concludes the analysis for the second scenario. We have shown that the conditions on smoothness, strong convexity and  $(G,B)$ -gradient dissimilarity indeed hold true in this scenario.
Final step: lower bound on  $\varepsilon$  in terms of  $K$ . We have established in the above that the conditions of smoothness, strong convexity and  $(G,B)$ -gradient dissimilarity are satisfied in both the scenarios. Therefore, by assumptions, algorithm  $\mathcal{A}$  must guarantee  $(f,\varepsilon)$ -resilience in either scenario. Specifically, the output of  $\mathcal{A}$ , denoted by  $\hat{\theta}$  must satisfy the following.
$$
\max  \left\{\mathcal {L} _ {S _ {1}} (\hat {\theta}) - \mathcal {L} _ {*, S _ {1}}, \mathcal {L} _ {S _ {2}} (\hat {\theta}) - \mathcal {L} _ {*, S _ {2}} \right\} \leq \varepsilon .
$$
Due to Lemma 1, the above holds true only if
$$
\varepsilon \geq \frac {\left(\frac {f}{n - f}\right) ^ {2} \alpha^ {2}}{8 \left(\frac {n - 2 f}{n - f} \frac {1}{K} + \frac {f}{n - f} \alpha\right)} \| z \| ^ {2}. \tag {26}
$$
Recall from (17) that  $K \geq \frac{1}{\alpha} \cdot \frac{n - 2f}{f} \cdot \frac{\mu}{L - \mu}$ . Therefore, we have  $(n - 2f)\frac{1}{K} \leq \frac{L - \mu}{\mu} f\alpha$ , which implies that
$$
\frac {\left(\frac {f}{n - f}\right) ^ {2} \alpha^ {2}}{\frac {n - 2 f}{n - f} \frac {1}{K} + \frac {f}{n - f} \alpha} \geq \frac {\left(\frac {f}{n - f}\right) ^ {2} \alpha^ {2}}{\left(\frac {L - \mu}{\mu} + 1\right) \frac {f}{n - f} \alpha} = \frac {\mu}{L} \cdot \frac {f}{n - f} \alpha .
$$
Using the above in (26), and then substituting  $\alpha = \frac{n - f}{f}\mu$  (from (16)), implies that
$$
\varepsilon \geq \frac {\mu^ {2}}{8 L} \| z \| ^ {2}.
$$
Recall from (18) that  $\| z \|^2 = \frac{f}{n - 2f} \frac{G^2}{2\alpha} \cdot K$  where  $\alpha = \frac{n - f}{f} \mu$ , from above we obtain that
$$
\varepsilon \geq \frac {\mu^ {2}}{1 6 L} \frac {f}{n - 2 f} \frac {G ^ {2}}{\alpha} K = \frac {\mu}{1 6 L} \cdot \frac {f ^ {2}}{(n - 2 f) (n - f)} G ^ {2} K.
$$
That is,  $\varepsilon$  grows with  $K$ . Note that the above holds for any arbitrarily large value of  $K$  satisfying (17). Therefore,  $\varepsilon$  can be made arbitrarily large. This contracts the assumption that  $\mathcal{A}$  is  $(f,\varepsilon)$ -resilient with a finite  $\varepsilon$ . Hence, we have shown that  $(f,\varepsilon)$ -resilience is impossible in general under  $(G,B)$ -gradient dissimilarity when  $\frac{f}{n} \geq \frac{1}{2 + B^2}$ .
Concluding remark. A critical element to the above inference on the unboundedness of  $\varepsilon$  is the condition that  $A_{1} \leq 0$ , shown in (21). Recall that
$$
A _ {1} := \frac {f (n - 2 f)}{(n - f) ^ {2}} \left(\left(1 - \frac {n - 2 f}{f} B ^ {2}\right) \frac {1}{K ^ {2}} - 2 (1 + B ^ {2}) \frac {\alpha}{K} + \alpha^ {2} \left(1 - \frac {f}{n - 2 f} B ^ {2}\right)\right).
$$
The right-hand side in the above is negative for any large enough value of  $K$  as soon as  $1 - \frac{f}{n - 2f} B^2 \leq 0$  or, equivalently,  $\frac{f}{n} \geq \frac{1}{2 + B^2}$ . However, in the case when  $\frac{f}{n} < \frac{1}{2 + B^2}$ ,  $K$  cannot be arbitrarily large if we were to ensure  $A_1 \leq 0$ , i.e.,  $K$  must be bounded from above. This constraint on  $K$  yields a non-trivial lower bound on  $\varepsilon$ . We formalize this intuition in the following.
B.2.2 Lower Bound on  $\varepsilon$
In this section, we prove that  $\varepsilon \geq \frac{1}{8\mu} \cdot \frac{f}{n - (2 + B^2)f} G^2$ . Let  $0 < f < n/2$ , and  $G, B \geq 0$ . Owing to the arguments presented in Section B.2.1, the assertion holds true when  $\frac{f}{n} \geq \frac{1}{2 + B^2}$ . In the following, we assume that  $\frac{f}{n} < \frac{1}{2 + B^2}$ , or equivalently  $B^2 < \frac{n - 2f}{f}$ .
Let  $\mathcal{A}$  be an  $(f,\varepsilon)$ -resilient algorithm. We consider a distributed learning setting where the workers' loss functions are given by (11), (12) and (13) with parameters  $\alpha$ ,  $K$  and  $z$  set as follows.
$$
\alpha = \mu (1 + B ^ {2}), \quad \text {a n d} \quad K = \frac {1}{\mu (1 - \frac {f}{n - 2 f} B ^ {2})}. \tag {27}
$$
We let  $z$  be an arbitrary point in  $\mathbb{R}^d$  such that
$$
\| z \| ^ {2} = \frac {(n - f) ^ {2}}{f (n - 2 f)} \cdot \frac {G ^ {2}}{\alpha (1 + B ^ {2})} \cdot K.
$$
Note that, as  $\alpha = \mu (1 + B^2)$ , we have
$$
K = \frac {1}{\mu \left(1 - \frac {f}{n - 2 f} B ^ {2}\right)} = \frac {1 + B ^ {2}}{1 - \frac {f}{n - 2 f} B ^ {2}} \cdot \frac {1}{\alpha}. \tag {28}
$$
$$
\left\| z \right\| ^ {2} = \frac {(n - f) ^ {2}}{f (n - 2 f)} \cdot \frac {1 - \frac {f}{n - 2 f} B ^ {2}}{(1 + B ^ {2}) ^ {2}} \cdot G ^ {2} K ^ {2}. \tag {29}
$$
From (28), we also obtain that
$$
\begin{array}{l} \frac {n - 2 f}{n - f} \frac {1}{K} + \frac {f}{n - f} \alpha = \frac {n - 2 f}{n - f} \frac {\left(1 - \frac {f}{n - 2 f} B ^ {2}\right)}{1 + B ^ {2}} \alpha + \frac {f}{n - f} \alpha = \frac {n - 2 f - f B ^ {2} + f + f B ^ {2}}{(n - f) (1 + B ^ {2})} \alpha \\ = \frac {\alpha}{1 + B ^ {2}}. \tag {30} \\ \end{array}
$$
Proof outline. We consider two scenarios, each corresponding to two different identities of honest workers:  $S_{1} = \{1,\dots ,n - f\}$  and  $S_{2} = \{f + 1,\dots ,n\}$ . For each of these, we prove that the corresponding local and global loss functions satisfy the assumptions of the theorem, mainly by invoking Lemma 2. Finally, by invoking Lemma 1, we show that  $(f,\varepsilon)$ -resilience implies the stated lower bound on  $\varepsilon$ .
First scenario. Consider the set of honest workers to be  $S_{1} = \{1, \dots, n - f\}$ . From (11) and (12) we obtain that
$$
\mathcal {L} _ {S _ {1}} (\theta) = \frac {f}{n - f} \frac {\alpha}{2} \| \theta - z \| ^ {2} + \frac {n - 2 f}{n - f} \frac {1}{2 K} \| \theta \| ^ {2}. \tag {31}
$$
$$
\nabla \mathcal {L} _ {S _ {1}} (\theta) = \frac {f}{n - f} \alpha (\theta - z) + \frac {n - 2 f}{n - f} \frac {1}{K} \theta , \text {a n d} \nabla^ {2} \mathcal {L} _ {S _ {1}} (\theta) = \left(\frac {f}{n - f} \alpha + \frac {n - 2 f}{n - f} \frac {1}{K}\right) I _ {d},
$$