SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
O _ {\mathbf {g} ^ {\top} \hat {\mathbf {e}}} | \mathbf {x} \rangle \otimes | 0 \rangle \rightarrow | \mathbf {x} \rangle \otimes \int_ {\mathbf {v} \in \mathbb {R} ^ {d}} \sqrt {p _ {f , \mathbf {x}} (\mathbf {v}) \mathrm {d} \mathbf {v}} \left| \right. \mathbf {v} ^ {\top} \hat {\mathbf {e}} \left. \right\rangle \otimes \left| \right. \operatorname {g a r b a g e} (\mathbf {v}) \left. \right\rangle ,
$$
using one query to  $O_{\mathbf{g}}$ . Hence, Algorithm 4 uses  $\widetilde{\mathcal{O}}(RL / \epsilon')$  queries to the quantum stochastic gradient oracle  $O_{\mathbf{g}}$ , with failure probability at most
$$
\log_ {2} \left(\frac {\| \mathbf {y} _ {r 0} - \mathbf {y} _ {l 0} \|}{\epsilon^ {\prime} / L}\right) \cdot \left(6 \mathcal {T} \log_ {2} \left(\frac {2 R L}{\epsilon^ {\prime}}\right)\right) ^ {- 1} \leq \frac {1}{6 \mathcal {T}} \tag {16}
$$
by union bound. Next, we show that the output  $\hat{\mathbf{y}}$  of Algorithm 4 has a relatively small function value as desired. We denote
$$
\lambda^ {*} := \underset {\lambda \in [ 0, 1 ]} {\arg \min } f (\lambda \mathbf {y} _ {l 0} + (1 - \lambda) \mathbf {y} _ {r 0}) \quad \text {a n d} \quad \mathbf {y} ^ {*} := \lambda^ {*} \mathbf {y} _ {l 0} + (1 - \lambda^ {*}) \mathbf {y} _ {r 0}.
$$
If the algorithm terminates at a point  $\hat{\mathbf{y}}$  satisfying  $|\tilde{g}_{\hat{\mathbf{e}}}(\hat{\mathbf{y}})| \leq \epsilon'/(4R)$ , by convexity and the Cauchy Schwarz inequality we have
$$
f (\hat {\mathbf {y}}) - f (\mathbf {y} ^ {*}) \leq | \nabla f (\hat {\mathbf {y}}) ^ {\top} \hat {\mathbf {e}} | \cdot \| \hat {\mathbf {y}} - \mathbf {y} ^ {*} \| \leq \frac {\epsilon^ {\prime}}{2 R} \cdot 2 R \leq \epsilon^ {\prime}.
$$
Otherwise, by induction we can demonstrate that in each iteration of the algorithm,  $\mathbf{y}^*$  resides within the segment bounded by  $\mathbf{y}_l$  and  $\mathbf{y}_r$ . Assume that this assertion holds true for the  $t$ -th iteration. If  $\tilde{g}_{\mathrm{e}}(\mathbf{y}_m) > \epsilon'/(4R)$ , we have
$$
\begin{array}{l} \langle \nabla f (\mathbf {y} _ {m}), \mathbf {y} _ {l} - \mathbf {y} _ {m} \rangle = - \langle \nabla f (\mathbf {y} _ {m}), \hat {\mathbf {e}} \rangle \cdot \| \mathbf {y} _ {l} - \mathbf {y} _ {m} \| \\ \leq - \left(\tilde {g} _ {\hat {e}} (\mathbf {y} _ {m}) - \frac {\epsilon^ {\prime}}{4 R}\right) \| \mathbf {y} _ {l} - \mathbf {y} _ {m} \| <   0, \\ \end{array}
$$
indicating that the value of  $f$  will decrease along the direction  $\mathbf{y}_l - \mathbf{y}_m$ . Hence,  $\mathbf{y}^*$  lies in the segment between  $\mathbf{y}_l$  and  $\mathbf{y}_m$  of the  $t$ -th iteration, or equivalently, the segment between  $\mathbf{y}_l$  and  $\mathbf{y}_m$  of the  $(t + 1)$ -th iteration, given that  $f$  is convex. A symmetric argument applies in the case of  $\tilde{g}_{\hat{\mathbf{e}}}(\mathbf{y}_m) < \epsilon' / (4R)$ . Then, we have
$$
\left\| \mathbf {y} _ {l} - \mathbf {y} ^ {*} \right\| \leq \left\| \mathbf {y} _ {l} - \mathbf {y} _ {r} \right\|
$$
for every iteration, which leads to
$$
\| \hat {\mathbf {y}} - \mathbf {y} ^ {*} \| \leq \frac {\epsilon^ {\prime}}{L}
$$
when the algorithm terminates. Hence,
$$
f (\hat {\mathbf {y}}) - f (\mathbf {y} ^ {*}) \leq L \cdot \| \hat {\mathbf {y}} - \mathbf {y} ^ {*} \| = \epsilon^ {\prime}
$$
considering that  $f$  is  $L$ -Lipschitz.
Using Algorithm 4 as a subroutine, we develop the following Algorithm 5 that solves Problem 5.
Algorithm 5: Stochastic Approximately best point among finite number of points
Input: A set of points  $\{\mathbf{x}_1,\dots ,\mathbf{x}_{\mathcal{T}}\} \subset \mathbb{B}_R(\mathbf{0})$  where  $\mathcal{T}$  is a power of 2, accuracy  $\epsilon$  Output:  $\hat{\mathbf{x}}$  such that  $f(\hat{\mathbf{x}})\leq \min_{i}f(\mathbf{x}_{i}) + \epsilon .$    
1  $\mathbf{y}_{0,i}\gets \mathbf{x}_i$  for all  $i\in [\mathcal{T}]$    
2 for  $\tau = 1,\ldots ,\log_2\mathcal{T}$  do   
3  $\begin{array}{r}\lfloor \textit{for} j = 1,\dots ,\mathcal{T} / 2^{\tau}\textit{do}\textit{y}_{\tau ,j} = \mathrm{StochasticLineSearch}(\mathbf{y}_{\tau -1,2j - 1},\mathbf{y}_{\tau -1,2j},\epsilon /\log_2\mathcal{T}) \end{array}$    
4 return ylog T,1
Proof of Proposition 2. We start by demonstrating that without loss of generality we can assume  $\mathcal{T}$  to be a power of 2. If  $\mathcal{T}$  is not already a power of 2, we can simply augment the set with at most  $\mathcal{T}$  points, each being  $\mathbf{x}_1$ , without changing the algorithm's output.
Next, we show that Algorithm 5 makes  $\mathcal{T} - 1$  calls to the StochasticLineSearch( $\cdot$ ) subroutine (Algorithm 4) with an accuracy of  $\epsilon' = \epsilon / \log_2 \mathcal{T}$  to solve the Problem 5 problem. The algorithm exhibits a hierarchical structure, as illustrated in Figure 1, where each non-leaf node invokes the StochasticLineSearch( $\cdot$ ) subroutine once. Consequently, the total number of calls is  $\mathcal{T} - 1$ .
By Lemma 8, each call to StochasticLineSearch() is executed successfully with probability at least  $1 - 1 / (6\mathcal{T})$ . Then by union bound, we can deduce that the likelihood of all calls to StochasticLineSearch() being successful is at least  $5/6$ . In this case, for any node  $\mathbf{y}_{\tau,j}$  in the tree with  $\tau \in [\log_2\mathcal{T}]$  and  $j \in [\mathcal{T}/2^{\tau}]$ , by convexity and the guarantee of Lemma 8 we have
$$
\begin{array}{l} f (\mathbf {y} _ {\tau , j}) \leq \min  _ {\lambda \in [ 0, 1 ]} f \left(\lambda \mathbf {y} _ {\tau - 1, 2 j - 1} + (1 - \lambda) \mathbf {y} _ {\tau - 1, 2 j}\right) + \epsilon / \log_ {2} \mathcal {T} \\ \leq \min  \left\{f \left(\mathbf {y} _ {\tau - 1, 2 j - 1}\right), f \left(\mathbf {y} _ {\tau - 1, 2 j}\right) \right\} + \epsilon / \log_ {2} \mathcal {T}. \\ \end{array}
$$
Therefore for any leaf node  $\mathbf{x}_j$  with  $j\in [\mathcal{T}]$ , by summing over the path between the root  $\mathbf{y}_{t,1}$  and  $\mathbf{x}_j$  we have
$$
f \left(\mathbf {y} _ {t, 1}\right) \leq f \left(\mathbf {x} _ {j}\right) + t \cdot \frac {\epsilon}{\log_ {2} \mathcal {T}} = f \left(\mathbf {x} _ {j}\right) + \epsilon ,
$$
$$
f \left(\mathbf {y} _ {t, 1}\right) \leq \min  _ {j \in [ T ]} f \left(\mathbf {x} _ {j}\right) + \epsilon ,
$$
Figure 1: The hierarchical structure of Algorithm 5.
indicating  $\mathbf{y}_{t,1}$  is a valid solution of Problem 5. Note that there are in total  $\mathcal{T} - 1$  non-leaf nodes in Figure 1, and the value of each non-leaf node is computed by one call to the subroutine StochasticLineSearch( $\cdot$ ). Hence, the total number of calls to StochasticLineSearch( $\cdot$ ) equals  $\mathcal{T} - 1$ , and the overall failure probability of Algorithm 5 is at most
$$
(\mathcal {T} - 1) \cdot \frac {1}{6 \mathcal {T}} \leq \frac {1}{6} \tag {17}
$$
by union bound, as the failure probability of Algorithm 4 is at most  $1 / (6\mathcal{T})$ . Since each call to Algorithm 4 takes
$$
\widetilde {\mathcal {O}} \left(\frac {R L}{\epsilon^ {\prime}}\right) = \widetilde {\mathcal {O}} \left(\frac {R L}{\epsilon}\right)
$$
queries to the quantum stochastic gradient oracle  $O_{\mathbf{g}}$  defined in (2), the total number of queries to the quantum stochastic gradient oracle  $O_{\mathbf{g}}$  is then  $\widetilde{\mathcal{O}}(RL\mathcal{T}/\epsilon)$ .
D.3 Proof of Corollary 1
Proof of Corollary 1. We first run cutting plane method with the separation oracle being an  $\frac{\epsilon}{10R}$ -approximate gradient of  $f$  implemented by Algorithm 1, which by Proposition 1 outputs a set  $\Gamma$  of  $\mathcal{T} = \widetilde{\mathcal{O}}(d\log(L/\epsilon))$  points containing at least one  $\mathcal{O}(\epsilon)$ -optimal point of  $f$  using  $\widetilde{\mathcal{O}}\left(d^{3/2}LR/\epsilon\right)$  queries to  $O_{\mathbf{g}}$  with success probability at least  $5/6$ . Then by running Algorithm 5 on  $\Gamma$ , with success probability at least  $5/6$  we can find a point  $\hat{\mathbf{x}} \in \mathbb{R}^d$  satisfying
$$
f (\hat {\mathbf {x}}) \leq \min  _ {\mathbf {x} \in \Gamma} f (\mathbf {x}) + \epsilon \leq f (\mathbf {x} ^ {*}) + \mathcal {O} (\epsilon),
$$
which takes  $\widetilde{\mathcal{O}}(RLT/\epsilon) = \widetilde{\mathcal{O}}(dRL/\epsilon)$  queries to  $O_{\mathbf{g}}$  by Proposition 2. Hence, the overall number of queries to  $O_{\mathbf{g}}$  equals  $\widetilde{\mathcal{O}}(d^{3/2}LR/\epsilon)$ , and the overall success probability is at least  $2/3$ .
The proof of Theorem 6 is based on the following result from [40].
Theorem 9 ([40, Theorem 1]). Consider the bounded variance setting in Problem 2, the output  $\mathbf{x}_{\mathrm{out}}$  of Algorithm 6 is an expected  $\frac{2\Delta\ell}{T} + \hat{\sigma}^2$ -critical point.
Proof of Theorem 6. By Theorem 9 and the associated parameter setting, the output  $\mathbf{x}_{\mathrm{out}}$  of Algorithm 6 satisfies
$$
\mathbb {E} \| \nabla f (\mathbf {x} _ {\mathrm {o u t}}) \| ^ {2} \leq \frac {2 \Delta \ell}{\mathcal {T}} + \hat {\sigma} ^ {2} \leq \epsilon^ {2},
$$
Algorithm 6: Quantum randomized SGD (Q-SGD)
Input: Function  $f\colon \mathbb{R}^d\to \mathbb{R}$  , precision  $\epsilon$  ,variance  $\sigma$  , smoothness  $\ell$  Parameters:  $\hat{\sigma} = \epsilon /3$  total iteration budget  $\mathcal{T} = 12\Delta \ell \epsilon^{-2}$  Output:  $\epsilon$  -critical point of  $f$    
1 Uniformly randomly select  $N$  from  $1,\dots ,T$    
2 Set  $\mathbf{x}_0\gets \mathbf{0}$    
3 for  $t = 0,1,2,\ldots ,N - 1$  do   
4 Call Algorithm 1 for an unbiased estimate  $\tilde{\mathbf{g}}_t$  of  $\nabla f(\mathbf{x}_t)$  with variance at most  $\hat{\sigma}^2$    
5  $\mathbf{x}_{t + 1}\leftarrow \mathbf{x}_t - \tilde{\mathbf{g}}_t / \ell$    
6 return  $\mathbf{x}_N$
implying that Algorithm 6 solves Problem 2 with probability at least  $2/3$  under the bounded variance setting, and the remaining thing would be to analyze the number of queries it makes to  $O_{\mathbf{g}}$ . By Theorem 4, at each iteration one can run Algorithm 1 to obtain an unbiased estimate  $\tilde{\mathbf{g}}_t$  of  $\nabla f(\mathbf{x}_t)$  with variance at most  $\hat{\sigma}^2$  using
$$
\widetilde {\mathcal {O}} \left(\frac {\sigma \sqrt {d}}{\hat {\sigma}}\right) = \widetilde {\mathcal {O}} \left(\frac {\sigma \sqrt {d}}{\epsilon}\right)
$$
queries to  $O_{\mathbf{g}}$ . Then, the total number of queries to  $O_{\mathbf{g}}$  is at most
$$
\mathcal {T} \cdot \tilde {\mathcal {O}} \left(\frac {\sigma \sqrt {d}}{\epsilon}\right) = \tilde {\mathcal {O}} \left(\frac {\sigma \Delta \ell \sqrt {d}}{\epsilon^ {3}}\right).
$$
F Proof of Theorem 7
Algorithm 7: Q-SPIDER
Input: Function  $f\colon \mathbb{R}^d\to \mathbb{R}$  precision  $\epsilon$  variance  $\sigma$  smoothness  $\ell$  Parameters:  $q = \frac{20\sigma}{\epsilon},\hat{\sigma}_1 = \frac{\epsilon}{40},\hat{\sigma}_2 = \frac{\epsilon}{40}\sqrt{\frac{\epsilon}{10\sigma}}$  total iteration budget  $\mathcal{T} = \frac{1600\ell\Delta}{\sigma^2}$  Output: An  $\epsilon$  -critical point of  $f$    
1 Set  $\mathbf{x}_0\gets \mathbf{0}$    
2 for  $t = 0,1,2,\ldots ,T$  do   
3 if mod  $(t,q) = 0$  then Call Algorithm 1 to obtain an unbiased estimate  $\tilde{\mathbf{g}}_t$  of  $\nabla f(\mathbf{x}_t)$  with variance  $\hat{\sigma}_1^2$    
5  $\mathbf{v}_t\gets \tilde{\mathbf{g}}_t$    
6 else Call Algorithm 1 to obtain an unbiased estimate  $\tilde{\mathbf{g}}_t$  of  $\nabla f(\mathbf{x}_t) - \nabla f(\mathbf{x}_{t - 1})$  with variance  $\hat{\sigma}_2^2$    
8  $\mathbf{v}_t\gets \tilde{\mathbf{g}}_t + \mathbf{v}_{t - 1}$    
9 if  $\| \mathbf{v}_t\| \leq 2\epsilon$  then return  $\mathbf{x}_t$  .   
10 else  $\mathbf{x}_{t + 1}\leftarrow \mathbf{x}_t - \frac{\epsilon}{\ell}\cdot \frac{\mathbf{v}_t}{\|\mathbf{v}_t\|}$    
11 return xT
The proof of Theorem 7 is based on the following result from [29].
Theorem 10 ([29, Theorem 1]). Consider the mean-squared smoothness setting in Problem 2, the output  $\mathbf{x}_{\mathrm{out}}$  of Algorithm 7 is an expected  $\epsilon$ -critical point.
Proof of Theorem 7. Theorem 10 shows that Algorithm 7 can solve Problem 2 under the mean-squared smoothness setting and the remaining thing would be to analyze the number of queries it makes to  $O_{\mathbf{g}}^{S}$ .
For each iteration  $t$  with  $\mathrm{mod}(t, q) = 0$  such that Line 5 is executed, note that we can prepare the following quantum oracle
$$
\hat {O} _ {\mathbf {g}} | \mathbf {x} \rangle \otimes | 0 \rangle \rightarrow | \mathbf {x} \rangle \otimes \int_ {\omega} \sqrt {p (\omega) \mathrm {d} \omega} | \mathbf {g} (\mathbf {x}, \omega) \rangle \otimes | \mathrm {g a r b a g e} (\mathbf {x}, \omega) \rangle ,
$$
where  $p(\omega)$  is the probability distribution of the random seed  $\omega$ , by first applying  $O_{\mathbf{g}}^{S}$  to the state
$$
| \mathbf {x} \rangle \otimes \left(\int_ {\omega} \sqrt {p (\omega) \mathrm {d} \omega} | \omega \rangle\right) \otimes | 0 \rangle
$$
and then uncompute the second register. This procedure uses one query to the  $\sigma$ -SQ-QSGO  $O_{\mathbf{g}}^{S}$  (Definition 4). By Theorem 4, one can run Algorithm 1 to obtain an unbiased estimate  $\tilde{\mathbf{g}}_t$  of  $\nabla f(\mathbf{x}_t)$  with variance at most  $\hat{\sigma}_1^2$  using
$$
\widetilde {\mathcal {O}} \left(\frac {\sigma \sqrt {d}}{\hat {\sigma} _ {1}}\right) = \widetilde {\mathcal {O}} \left(\frac {\sigma \sqrt {d}}{\epsilon}\right)
$$
queries to  $\hat{O}_{\mathbf{g}}$  and thus the same number of queries to  $O_{\mathbf{g}}^{S}$ .
Similarly, for each iteration  $t$  with  $\mathrm{mod}(t,q) = 0$  such that Line 7 is executed, note that one can prepare the following quantum oracle
$$
\begin{array}{l} \hat {O} _ {\mathbf {g}} ^ {t} \left| \right. \mathbf {x} _ {t} \left. \right\rangle \otimes \left| \right. \mathbf {x} _ {t - 1} \left. \right\rangle \otimes \left| \right. 0 \left. \right\rangle\rightarrow \left| \right. \mathbf {x} _ {t} \left. \right\rangle \otimes \left| \right. \mathbf {x} _ {t - 1} \left. \right\rangle \\ \otimes \int_ {\omega} \sqrt {p (\omega) \mathrm {d} \omega} | \mathbf {g} (\mathbf {x} _ {t}, \omega) - \mathbf {g} (\mathbf {x} _ {t - 1}, \omega) \rangle \otimes | \operatorname {g a r b a g e} (\mathbf {x} _ {t}, \mathbf {x} _ {t - 1}, \omega) \rangle \\ \end{array}
$$
by first applying  $O_{\mathbf{g}}^{S}$  twice to obtain the state
$$
\begin{array}{l} | \mathbf {x} \rangle \otimes | \mathbf {x} _ {t - 1} \rangle \otimes \int_ {\omega} \sqrt {p (\omega) \mathrm {d} \omega} | \omega \rangle \otimes | \mathbf {g} (\mathbf {x}, \omega) \rangle \otimes | \mathbf {g} (\mathbf {x} _ {t - 1}, \omega) \rangle \\ \left. \otimes | \mathbf {g} (\mathbf {x}, \omega) - \mathbf {g} (\mathbf {x} _ {t - 1}, \omega) \right\rangle \otimes | \text {g a r b a g e} (\mathbf {x} _ {t}, \mathbf {x} _ {t - 1}, \omega) \rangle \\ \end{array}
$$
and then uncompute the fourth and the fifth register. Observe that
$$
\underset {\boldsymbol {\omega}} {\mathbb {E}} \| \mathbf {g} (\mathbf {x} _ {t}, \boldsymbol {\omega}) - \mathbf {g} (\mathbf {x} _ {t - 1}, \boldsymbol {\omega}) \| ^ {2} \leq \ell^ {2} \| \mathbf {x} _ {t} - \mathbf {x} _ {t - 1} \| \leq \epsilon^ {2},
$$
then by Theorem 4, one can run Algorithm 1 to obtain an unbiased estimate  $\tilde{\mathbf{g}}_t$  of  $\nabla f(\mathbf{x}_t) - \nabla f(\mathbf{x}_{t-1})$  with variance at most  $\hat{\sigma}_1^2$  using
$$
\widetilde {\mathcal {O}} \left(\frac {\epsilon \sqrt {d}}{\hat {\sigma} _ {2}}\right) = \widetilde {\mathcal {O}} \left(\sqrt {\frac {d \sigma}{\epsilon}}\right)
$$
queries to  $\hat{O}_{\mathbf{g}}^t$  and thus twice the number of queries to  $O_{\mathbf{g}}^{S}$ . Then, the total number of queries to  $O_{\mathbf{g}}^{S}$  can be expressed as
$$
\frac {\mathcal {T}}{q} \cdot \widetilde {\mathcal {O}} \left(\frac {\sigma \sqrt {d}}{\epsilon}\right) + \mathcal {T} \cdot \widetilde {\mathcal {O}} \left(\sqrt {\frac {d \sigma}{\epsilon}}\right) = \widetilde {\mathcal {O}} \left(\frac {\ell \Delta \sqrt {d}}{\epsilon^ {2}} \left(1 + \sqrt {\frac {\sigma}{\epsilon}}\right)\right).
$$
G Proof of lower bounds
G.1 Proof of Proposition 3
Note that Problem 3 is strictly harder than the multivariate mean estimation problem considered in [25]. In the latter, the objective is to produce an estimate of the expected value within a bounded  $\ell_2$  error, without the requirement of unbiasedness. A query lower bound for solving quantum mean estimation was established in [25].
Lemma 9 ([25, Theorem 3.8]). Let  $\sigma \geq 0$  and denote  $P_{\sigma}$  to be the set of all  $d$ -dimensional quantum random variables with covariance matrix  $\Sigma$  such that  $\mathrm{Tr}(\Sigma) = \sigma^2$ . Suppose every  $X \in P_{\sigma}$  is indexed by some random seeds  $\omega$ , i.e.,
$$
X = X (\omega), \quad \omega \sim p _ {\Omega},
$$
where  $p_{\Omega}$  denotes the probability distribution of the random seed  $\omega$ . Then there exists a constant  $\alpha$  such that for any  $n > \alpha d$  and any quantum algorithm that uses at most  $n$  queries to the following two oracles
$$
O _ {\Omega} \left| \right. 0 \left. \right\rangle\rightarrow \int_ {\omega} \sqrt {p _ {\Omega} (\omega) \mathrm {d} \omega} \left| \right. \omega \left. \right\rangle \quad a n d \quad \mathcal {B} _ {X} \left| \right. \omega \left. \right\rangle\left| \right. 0 \left. \right\rangle\rightarrow \left| \right. \omega \left. \right\rangle\left| \right. X (\omega) \left. \right\rangle ,
$$
there exists an instance  $X \in P_{\sigma}$  such that the quantum algorithm returns a mean estimate  $\tilde{\mu}$  of the mean  $\mu$  of  $X$  that satisfies
$$
\| \tilde {\mu} - \mu \| \geq \Omega \left(\frac {\sigma \sqrt {d}}{n}\right)
$$
with probability at least  $2/3$ .
By employing an oracle reduction argument, we obtain our lower bound result based on Lemma 9.
Proof of Proposition 3. We first show that one query to the quantum sampling oracle  $O_X$  defined in Definition 1 can be implemented using one query to  $O_{\Omega}$  and one query to  $\mathcal{B}_X$ . In particular, observe that
$$
\mathcal {B} _ {X} O _ {\Omega} | 0 \rangle \otimes | 0 \rangle = \int_ {\omega} \sqrt {p _ {\Omega} (\omega) \mathrm {d} \omega} | \omega \rangle \otimes | X (\omega) \rangle .
$$
By measuring the first register in the computational basis spanned by all the random seeds  $\{|\omega\rangle\}$ , we can obtain the desired output of  $O_X$ . Hence, if there exists a quantum algorithm making at most  $\mathcal{O}(L\sqrt{d}/\hat{\sigma})$  queries to the oracle  $O_X$  that solves Problem 3 with success probability at least  $2/3$ , there also exists a quantum mean estimation algorithm that uses at most  $\mathcal{O}(L\sqrt{d}/\hat{\sigma})$  queries to  $O_\Omega$  and  $\mathcal{B}_X$  and for any  $X \in P_\sigma$  it outputs an estimate  $\tilde{\mu}$  of the mean  $\mu$  of  $X$  that satisfies  $\|\tilde{\mu} - \mu\| \leq \mathcal{O}(\hat{\sigma})$  with probability at least  $2/3$ , which contradicts to Lemma 9.
G.2 Proof of Theorem 8
Our proof follows a similar approach to the previous proof of the quantum mean estimation lower bound [25]. Specifically, our results are derived through a reduction to a specific multivariate mean estimation problem. In this problem, all vectors are sampled from a set of orthonormal bases and there might be duplicated vectors. We choose to reduce to this particular problem instead of directly reducing to mean estimation problem because the construction of our hard instances require a tighter control over the norm of the vectors as well as the norm of their expected value. This problem can be equivalently represented as the following composition problem, as defined in [25].
Problem 6 (Search $^N$ $\circ$  Parity $^M$ ). Let  $N, M \geq 1$  be two integers. Let  $\mathcal{A}_{N,M}$  denote the set of matrices  $A \in \{0,1\}^{N \times M}$  such that  $\lfloor N / 2 \rfloor$  rows have Hamming weights  $\lfloor M / 2 \rfloor$ , and the other rows have Hamming weights  $\lfloor M / 2 \rfloor + 1$ . Define the vector  $\mathbf{b}^{(A)}$  such that,
$$
b _ {i} ^ {(A)} = \left\{ \begin{array}{l l} 0, & \text {i f t h e i - t h r o w o f A h a s H a m m i n g w e i g h t} \lfloor M / 2 \rfloor , \\ 1, & \text {i f t h e i - t h r o w o f A h a s H a m m i n g w e i g h t} \lfloor M / 2 \rfloor + 1, \end{array} \right. \tag {18}
$$
for each  $i \in [N]$ . Then, the Search $^{N}$ $\circ$  Parity $^{M}$  problem consists of finding a vector  $\widetilde{\mathbf{b}} \in \mathbb{R}^{N}$  that minimizes  $\|\widetilde{\mathbf{b}} - \mathbf{b}^{(A)}\|_{1}$  given a quantum oracle
$$
O _ {A} | i, j \rangle \rightarrow (- 1) ^ {A _ {i j}} | i, j \rangle . \tag {19}
$$
[25] showed that  $\Omega(NM)$  queries are needed to approximate the vector  $\mathbf{b}^{(A)}$  with small error.
Lemma 10 ([25, Lemma 3.6]). There exists a constant  $\alpha > 1$  such that, for any quantum algorithm for the Search $^N$  or Parity $^M$  problem that uses at most  $NM / \alpha$  queries there exists an  $A \in \mathcal{A}_{N,M}$  such that this algorithm returns a vector  $\widetilde{\mathbf{b}}$  satisfying  $\|\widetilde{\mathbf{b}} - \mathbf{b}^{(A)}\| \geq \sqrt{N}/2$  with probability at least  $2/3$ .
We establish our quantum lower bounds for stochastic convex optimization by establishing a correspondence between the  $\mathsf{Search}^N\circ \mathsf{Parity}^M$  problem and Problem 1. Specifically, for any input  $A\in \mathcal{A}_{N\times M}$  to the  $\mathsf{Search}^N\circ \mathsf{Parity}^M$  problem, we design the following convex function whose optimal point corresponds to the solution of the  $\mathsf{Search}^N\circ \mathsf{Parity}^M$  problem under certain sets of parameters,
$$
\bar {f} ^ {A} (\mathbf {x}) := \underset {i, j} {\mathbb {E}} [ f _ {i, j} (\mathbf {x}) ], \tag {20}
$$
$$
f _ {i, j} ^ {A} (\mathbf {x}) := - \frac {1}{3} \left\langle \mathbf {x}, \mathbf {g} _ {i, j} \right\rangle + \frac {2 L}{3} \cdot \max \left\{0, \| \mathbf {x} \| - \frac {R}{2} \right\}
$$
for some  $\mathbf{g}_{i,j} \in \mathbb{R}^d$  satisfying  $\| \mathbf{g}_{i,j} \| \leq L$  for all  $i \in [N]$  and  $j \in [M]$ .
Lemma 11. Denote  $\bar{\mathbf{g}}\coloneqq \mathbb{E}_{i,j}[\mathbf{g}_{i,j}]$ . Then, the function  $\bar{f}^A$  defined in (20) has the following properties if  $\bar{\mathbf{g}}$  is a non-zero vector.
1.  $\bar{f}^A$  is convex.
2.  $\bar{f}^A$  is minimized at  $\mathbf{x}^{*} = \frac{R}{2}\frac{\bar{\mathbf{g}}}{\|\bar{\mathbf{g}}\|}$ .
3. Every  $\epsilon$ -optimum  $\mathbf{x}$  of  $\bar{f}^A$  satisfies
$$
\left\langle \frac {R}{2} \cdot \frac {\mathbf {x}}{\| \mathbf {x} \|}, \mathbf {x} ^ {*} \right\rangle \geq 1 - \frac {R \epsilon}{2 \| \bar {\mathbf {g}} \|}
$$
4. For
$$
\hat {\mathbf {g}} _ {i, j} (\mathbf {x}) := - \frac {1}{3} \cdot \mathbf {g} _ {i, j} + \frac {2 L}{3} \cdot \max  \left\{\| \mathbf {x} \| - \frac {R}{2}, 0 \right\} \cdot \frac {\mathbf {x}}{\| \mathbf {x} \|} \tag {21}
$$
it is the case that
$$
\underset {i, j} {\mathbb {E}} \left[ \hat {\mathbf {g}} _ {i, j} (\mathbf {x}) \right] \in \partial \bar {f} ^ {A} (\mathbf {x}) \quad a n d \quad \| \hat {\mathbf {g}} _ {i, j} (\mathbf {x}) \| \leq L, \quad \forall \mathbf {x} \in \mathbb {R} ^ {d}.
$$
Proof. Since  $\bar{f}^A$  is the sum of a linear function and a maximum function, both of which are convex,  $\bar{f}^A$  itself is convex.
Observe that the minimum of  $\bar{f}^A$  cannot be achieved when  $\| \mathbf{x}\| >R / 2$  for otherwise the function value can be further decreased when moving towards  $\mathbf{0}$  given that
$$
\| \bar {\mathbf {g}} \| = \left\| \frac {\sum_ {i , j} \mathbf {g} _ {i , j}}{N M} \right\| \leq \frac {\sum_ {i , j} \| \mathbf {g} _ {i , j} \|}{N M} \leq L.
$$
When  $\| \mathbf{x} \| \leq R / 2$ ,  $\bar{f}^A$  can be expressed as
$$
\bar {f} ^ {A} (\mathbf {x}) = - \frac {1}{3} \left<   \mathbf {x}, \bar {\mathbf {g}} \right > ,
$$
which is minimized at
$$
\frac {R}{2} \cdot \frac {\bar {\mathbf {g}}}{\| \bar {\mathbf {g}} \|} = \frac {R \mathbf {b} ^ {(A)}}{\sqrt {2 d}}.
$$
For any  $\mathbf{x}$  that is an  $\epsilon$ -optimum of  $\bar{f}^A$ , we define
$$
\mathbf {x} ^ {\prime} = \left\{ \begin{array}{l l} \mathbf {x}, & \| \mathbf {x} \| \leq \frac {R}{2}, \\ \frac {R}{2} \cdot \frac {\mathbf {x}}{\| \mathbf {x} \|}, & \text {o t h e r w i s e}, \end{array} \right.
$$