SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Consequently,  $\mathbb{E}\| Z - \mu \| ^2\leq 13\sigma^2$  when  $D = \hat{\sigma} +L^{3} / \hat{\sigma}^{2}$  and  $\delta \leq \hat{\sigma}^6 /L^6$
Proof. Let  $S$  denote the event that  $\|X_1 - \mu\|_2 \leq \hat{\sigma}$  and let  $T$  denote the event that  $\|X_1 - X_2\|_2 \leq D$ . Then by the law of total expectation
$$
\begin{array}{l} \mathbb {E} \| Z - \mu \| ^ {2} = \Pr [ S ] \mathbb {E} \left[ \| Z - \mu \| _ {2} ^ {2} \mid S \right] + \Pr [ \text {n o t} S ] \mathbb {E} \left[ \| Z - \mu \| _ {2} ^ {2} \mid \text {n o t} S \right] \\ \leq \mathbb {E} \left[ \| Z - \mu \| ^ {2} \mid S \right] + \delta \cdot \mathbb {E} \left[ \| Z - \mu \| ^ {2} \mid \operatorname {n o t} S \right] \\ \end{array}
$$
Where we used that  $\operatorname*{Pr}[S] \leq 1$  and  $\operatorname*{Pr}[\text{not } S] \leq \delta$ . We will prove the lemma by leveraging this inequality and upper bounding both  $\mathbb{E}\left[\|Z-\mu\|_2^2 \mid S\right]$  and  $\mathbb{E}\left[\|Z-\mu\|_2^2 \mid \text{not } S\right]$ .
First, we upper bound  $\mathbb{E}\left[\| Z - \mu \|_2^2 \mid S\right]$ . Again by the law of total expectation and the definition of  $Z$  we have that
$$
\begin{array}{l} \mathbb {E} \left[ \| Z - \mu \| _ {2} ^ {2} \mid S \right] = \Pr [ T \mid S ] \mathbb {E} \left[ \| Z - \mu \| _ {2} ^ {2} \mid T \text {a n d} S \right] (7) \\ + \Pr [ \text {n o t} T \mid S ] \mathbb {E} \left[ \| Z - \mu \| _ {2} ^ {2} \mid T \text {a n d n o t} S \right] \\ \leq \Pr [ T \mid S ] \cdot \hat {\sigma} ^ {2} + \Pr [ \operatorname {n o t} T \mid S ] L ^ {2}. (8) \\ \end{array}
$$
Here we use that if  $T$  and  $S$  both hold then  $Z = X_{1}$  and  $\| X_{1} - \mu \| \leq \hat{\sigma}$  and that if  $S$  holds and not  $T$  then  $Z = X_{3}$  and  $X_{3}$  is independent of  $S$  and  $T$  with variance  $L^2$ . Now, if  $\| X_{2} - \mu \|_{2} \leq D - \hat{\sigma}$  then when  $S$  holds by triangle inequality
$$
\left\| X _ {1} - X _ {2} \right\| \leq \left\| X _ {1} - \mu \right\| + \left\| X _ {2} - \mu \right\| \leq D.
$$
Applying this bound and Chebyshev inequality to  $X_{2}$  (and using that  $X_{2}$  is independent of  $S$ ) we have that
$$
\Pr [ T \mid S ] \geq \Pr [ \| X _ {2} - \mu \| _ {2} \leq D - \hat {\sigma} \mid S ] = \Pr [ \| X _ {2} - \mu \| _ {2} \leq D - \hat {\sigma} ] \geq 1 - \frac {L}{D - \hat {\sigma}}.
$$
Since  $\hat{\sigma} < L$ , combined with (7) this implies that
$$
\mathbb {E} \left[ \| Z - \mu \| _ {2} ^ {2} \mid S \right] \leq \left(1 - \frac {L}{D - \hat {\sigma}}\right) \hat {\sigma} ^ {2} + \frac {L ^ {3}}{D - \hat {\sigma}} = \hat {\sigma} ^ {2} + L \left(\frac {L ^ {2} - \hat {\sigma} ^ {2}}{D - \hat {\sigma}}\right)
$$
Next, we upper bound  $\mathbb{E}\left[\| Z - \mu \|_2^2 \mid \text{not } S\right]$ . Note that if  $T$  holds then
$$
\left\| Z - \mu \right\| ^ {2} \leq 2 \left\| Z - X _ {2} \right\| ^ {2} + 2 \left\| X _ {2} - \mu \right\| ^ {2} \leq 2 D ^ {2} + 2 \left\| X _ {2} - \mu \right\| ^ {2}.
$$
Since  $Z = X_{3}$  if  $T$  does note hold, we then see that in either case
$$
\left\| Z - \mu \right\| ^ {2} \leq 2 D ^ {2} + 2 \left\| X _ {2} - \mu \right\| ^ {2} + \left\| X _ {3} - \mu \right\| ^ {2}
$$
Since,  $X_{2}$  and  $X_{3}$  are independent of  $S$  with variance  $L$  this therefore implies that
$$
\mathbb {E} \left[ \| Z - \mu \| _ {2} ^ {2} \mid \mathrm {n o t} S \right] \leq \mathbb {E} \left[ 2 D ^ {2} + 2 \| X _ {2} - \mu \| ^ {2} + \| X _ {3} - \mu \| ^ {2} \right] = 2 D ^ {2} + 3 L ^ {2},
$$
$$
\begin{array}{l} \mathbb {E} \| Z - \mu \| ^ {2} \leq \mathbb {E} \left[ \| Z - \mu \| ^ {2} \mid S \right] + \delta \cdot \mathbb {E} \left[ \| Z - \mu \| ^ {2} \mid \operatorname {n o t} S \right] \\ \leq \hat {\sigma} ^ {2} + L \left(\frac {L ^ {2} - \hat {\sigma} ^ {2}}{D - \hat {\sigma}}\right) + \delta (2 D ^ {2} + 3 L ^ {2}). \\ \end{array}
$$
Corollary 2. Given access to the quantum sampling oracle  $O_{X}$  of a  $d$ -dimensional random variable  $X$ , for any  $\hat{\sigma}, \delta \geq 0$ , the procedure QuantumMeanEstimation $^{+}(X, \hat{\sigma})$  (Algorithm 3) uses  $\tilde{\mathcal{O}}(L\sqrt{d}\hat{\sigma}^{-1})$  queries and outputs an estimate  $\hat{\mu}$  of the expectation  $\mu = \mathbb{E}[X]$  satisfying  $\mathbb{E}\|\hat{\mu} - \mu\|^{2} \leq \hat{\sigma}^{2}$ .
At Line 2 and Line 5 of Algorithm 3, we can further refine our approach to obtain  $X_{2}$  and  $X_{4}$  by taking the same number of classical samples as we would have used in the QuantumMeanEstimation procedure, which can reduce the poly-logarithmic factor in the overall query complexity.
Proof. By Lemma 3, the output  $\hat{\mu}$  of QuantumMeanEstimation $^{+}$ $(X,\hat{\sigma})$  (Algorithm 3) satisfies
$$
\mathbb {E} \| \hat {\mu} - \mu \| ^ {2} \leq \hat {\sigma} ^ {2}. \tag {9}
$$
As for the number of queries to  $O_{X}$ , note that it uses
$$
\widetilde {\mathcal {O}} \left(\frac {L \sqrt {d} \log (1 / \delta)}{\hat {\sigma}}\right) = \widetilde {\mathcal {O}} \left(\frac {L \sqrt {d}}{\hat {\sigma}} \log \left(\frac {L ^ {6}}{\hat {\sigma} ^ {6}}\right)\right) = \widetilde {\mathcal {O}} \left(\frac {L \sqrt {d}}{\hat {\sigma}}\right)
$$
queries to prepare  $X_{1}$ . Moreover, we can prepare the classical samples  $X_{2}$  and  $X_{3}$  by measuring the quantum state  $O_{X} |0\rangle$  in the computational basis, which takes 1 query to  $O_{X}$ . Hence, the overall query complexity equals  $\widetilde{\mathcal{O}}(L\sqrt{d}\hat{\sigma}^{-1})$ .
Equipped with Corollary 2, we are ready to prove Theorem 4.
Proof of Theorem 4. The structure of our proof is similar to the proof of [7, Proposition 1]. Observe that the output  $\hat{\mu}$  of Algorithm 1 can be expressed as
$$
\hat {\mu} = \tilde {\mu} _ {0} + 2 ^ {J} \left(\tilde {\mu} _ {J} - \tilde {\mu} _ {J - 1}\right), \quad J \sim \operatorname {G e o m} \left(\frac {1}{2}\right) \in \mathbb {N}. \tag {10}
$$
$$
\mathbb {E} [ \hat {\mu} ] = \mathbb {E} [ \tilde {\mu} _ {0} ] + \sum_ {j = 1} ^ {\infty} \operatorname * {P r} \{J = j \} 2 ^ {j} (\mathbb {E} [ \tilde {\mu} _ {j} ] - \mathbb {E} [ \tilde {\mu} _ {j - 1} ]) = \mathbb {E} [ \tilde {\mu} _ {\infty} ] = \mu ,
$$
given that  $\operatorname*{Pr}\{J = j\} = 2^{-j}$ . As for the variance, we have
$$
\mathbb {E} \| \hat {\mu} - \mu \| ^ {2} \leq 2 \mathbb {E} \| \hat {\mu} - \tilde {\mu} _ {0} \| ^ {2} + 2 \mathbb {E} \| \tilde {\mu} _ {0} - \mu \| ^ {2},
$$
$$
\mathbb {E} \| \hat {\mu} - \tilde {\mu} _ {0} \| ^ {2} = \sum_ {j = 1} ^ {\infty} \operatorname * {P r} (J = j) 2 ^ {2 ^ {j}} \mathbb {E} \| \tilde {\mu} _ {j} - \tilde {\mu} _ {j - 1} \| ^ {2} = \sum_ {j = 1} ^ {\infty} 2 ^ {j} \mathbb {E} \| \tilde {\mu} _ {j} - \tilde {\mu} _ {j - 1} \| ^ {2},
$$
and for each  $j$  we have
$$
\mathbb {E} \| \tilde {\mu} _ {j} - \tilde {\mu} _ {j - 1} \| ^ {2} \leq 2 \mathbb {E} \| \tilde {\mu} _ {j} - \mu \| ^ {2} + 2 \mathbb {E} \| \tilde {\mu} _ {j - 1} - \mu \| ^ {2}.
$$
By Corollary 2,
$$
\mathbb {E} \| \tilde {\mu} _ {j} - \mu \| ^ {2} \leq \frac {\hat {\sigma} ^ {2}}{1 0 0 \cdot 2 ^ {3 j / 2}}, \quad \forall j \geq 0,
$$
$$
\mathbb {E} \| \tilde {\mu} _ {j} - \tilde {\mu} _ {j - 1} \| ^ {2} \leq \frac {\sigma^ {2}}{5 0 \cdot 2 ^ {3 (j - 1) / 2}} + \frac {\sigma^ {2}}{5 0 \cdot 2 ^ {3 j / 2}} \leq \frac {\hat {\sigma} ^ {2}}{1 0 \cdot 2 ^ {3 j / 2}},
$$
$$
\mathbb {E} \| \hat {\mu} - \tilde {\mu} _ {0} \| ^ {2} = \frac {\hat {\sigma} ^ {2}}{1 0} \sum_ {j = 1} ^ {\infty} \frac {1}{2 ^ {j / 2}} \leq \frac {1}{3} \hat {\sigma} ^ {2}.
$$
$$
\mathbb {E} \| \hat {\mu} - \mu \| ^ {2} \leq 2 \mathbb {E} \| \hat {\mu} - \tilde {\mu} _ {0} \| ^ {2} + 2 \mathbb {E} \| \tilde {\mu} _ {0} - \mu \| ^ {2} \leq \hat {\sigma} ^ {2},
$$
given that  $\mathbb{E}\| \tilde{\mu}_0 - \mu \| ^2\leq \hat{\sigma}^2 /100$  by Corollary 2. Moreover, the expected number of queries is
$$
\widetilde {\mathcal {O}} \left(\frac {L \sqrt {d} \log d}{\hat {\sigma}}\right) \cdot \left(1 + \sum_ {j = 1} ^ {\infty} \Pr \{J = j \} \cdot \left(2 ^ {3 j / 4} + 2 ^ {3 (j - 1) / 4}\right)\right) = \widetilde {\mathcal {O}} \left(\frac {L \sqrt {d}}{\hat {\sigma}}\right).
$$
We first state a result from [15] which bounds properties of  $F$  defined in (5). The lemma states that, provided the radius  $r$  of the convolution is sufficiently small,  $F$  is pointwise close to  $f$  and hence finding an  $\mathcal{O}(\epsilon)$ -minimum of  $f$  is equivalent to finding an  $\epsilon$ -minimum of  $F$ . Further the lemma shows that  $F$  is Lipschitz and smooth.
Lemma 4 ([15, Lemma 8]). For any  $L$ -Lipschitz convex function  $f \colon \mathbb{R}^d \to \mathbb{R}$ , its Gaussian convolution  $g$  defined in Eq. (5) is convex,  $L$ -Lipschitz, and satisfies
$$
| f (\mathbf {x}) - g (\mathbf {x}) | \leq \sqrt {d} \cdot L r \quad a n d \quad \nabla^ {2} g (\mathbf {x}) \leq (L / r) \cdot \mathbb {I} _ {d}, \quad \forall \mathbf {x} \in \mathbb {R} ^ {d},
$$
where  $\mathbb{I}_d$  is the  $d$ -dimensional identity matrix.
Prior to proving Theorem 5, we show that one query to a QSGO  $O_{\mathbf{g}_F}$  (Definition 3) of  $F_T$  satisfying
$$
O _ {\mathbf {g} _ {F}} | \mathbf {x} \rangle \otimes | 0 \rangle \otimes | 0 \rangle \rightarrow | \mathbf {x} \rangle \otimes \int_ {\mathbf {v} \in \mathbb {R} ^ {d}} \sqrt {p _ {F , \mathbf {x}} (\mathbf {v}) \mathrm {d} \mathbf {v}} | \mathbf {v} \rangle \otimes | \text {g a r b a g e} (\mathbf {v}) \rangle , \tag {11}
$$
can be implemented by one query to  $O_{\mathbf{g}}$  of  $f$  defined in Definition 3.
Lemma 5. The QSGO  $O_{\mathbf{g}_F}$  of  $F_r$  defined in (11) can be implemented with one query to the QSGO  $O_{\mathbf{g}}$  of  $f$ .
Proof. We first prepare the following quantum state
$$
| \psi \rangle = | \mathbf {x} \rangle \otimes \left(\int_ {\mathbf {y} \in \mathbb {R} ^ {d}} \sqrt {\gamma_ {r} (\mathbf {y}) \mathrm {d} \mathbf {y}} | \mathbf {x} - \mathbf {y} \rangle\right) \otimes | 0 \rangle , \quad \forall \mathbf {x} \in \mathbb {R} ^ {d}.
$$
Applying  $O_{\mathbf{g}}$  to the last two registers yields
$$
\begin{array}{l} \left(\mathbb {I} \otimes O _ {\mathbf {g}}\right) | \psi \rangle = | \mathbf {x} \rangle \otimes \left(\int_ {\mathbf {y} \in \mathbb {R} ^ {d}} \sqrt {\gamma_ {r} (\mathbf {y}) \mathrm {d} \mathbf {y}} | \mathbf {x} - \mathbf {y} \rangle \otimes \int_ {\mathbf {v} \in \mathbb {R} ^ {d}} \sqrt {p _ {f , \mathbf {x} - \mathbf {y}} (\mathbf {v}) \mathrm {d} \mathbf {v}} | \mathbf {v} \rangle\right) \\ = | \mathbf {x} \rangle \otimes \int_ {\mathbf {y}, \mathbf {v} \in \mathbb {R} ^ {d}} \sqrt {\gamma_ {r} (\mathbf {y}) p _ {f , \mathbf {x} - \mathbf {y}} (\mathbf {v}) \mathrm {d} \mathbf {y} \mathrm {d} \mathbf {v}} | \mathbf {x} - \mathbf {y} \rangle | \mathbf {v} \rangle . \tag {12} \\ \end{array}
$$
$$
p _ {F, \mathbf {x}} (\mathbf {v}) = \int_ {\mathbf {y} \in \mathbb {R} ^ {d}} \gamma_ {r} (\mathbf {y}) p _ {f, \mathbf {x} - \mathbf {y}} (\mathbf {v}) \mathrm {d} \mathbf {y}, \quad \forall \mathbf {v} \in \mathbb {R} ^ {d},
$$
if we measure the last register, the probability density function of the outcome would be exactly  $p_{F,\mathbf{x}}$ . Hence, the quantum state  $(\mathbb{I} \otimes O_{\mathbf{g}}) |\psi \rangle$  in (12) can also be written as
$$
\left(\mathbb {I} \otimes O _ {\mathbf {g}}\right) | \psi \rangle = | \mathbf {x} \rangle \otimes \int_ {\mathbf {v} \in \mathbb {R} ^ {d}} \sqrt {p _ {F , \mathbf {x}} (\mathbf {v}) \mathrm {d} \mathbf {v}} | \text {g a r b a g e} (\mathbf {v}) \rangle \otimes | \mathbf {v} \rangle .
$$
By swapping the last two quantum registers, we can obtain the desired output state of  $O_{\mathbf{g}_F}$ .
The following result from [57] bounds the rate at which Algorithm 2 decreases the function error of  $F_{r}$ . Note that validity of this result relies solely on the fact that, at Line 6, the variance of the unbiased gradient estimate  $\tilde{\mathbf{g}}_t$  does not exceed  $\hat{\sigma}^2$ , irrespective of its implementation.
Lemma 6. The output  $\mathbf{x}_{\mathcal{T} + 1}^{ag}$  of Algorithm 2 satisfies
$$
\mathbb {E} \left[ F _ {r} \left(\mathbf {x} _ {\mathcal {T} + 1} ^ {a g}\right) - F _ {r} ^ {*} \right] \leq \frac {4 L R ^ {2}}{r \mathcal {T} (\mathcal {T} + 2)} + \frac {4 R \hat {\sigma}}{\sqrt {\mathcal {T}}},
$$
where  $F_{r}$  is the convoluted function defined in Line 1 of Algorithm 2 and  $F_{r}^{*}$  is its minimum.
Proof. This lemma follows from [57, Corollary 1], which shows that
$$
\mathbb {E} \left[ F _ {r} \left(\mathbf {x} _ {\mathcal {T} + 1} ^ {a g}\right) - F _ {r} ^ {*} \right] \leq \frac {4 \ell_ {F} R ^ {2}}{\mathcal {T} (\mathcal {T} + 2)} + \frac {4 R \hat {\sigma}}{\sqrt {\mathcal {T}}},
$$
where  $\ell_F$  is the Lipschitz parameter of  $F_{r}$ , which equals  $L / r$  by Lemma 4.
Equipped with above results, we are now ready to present the proof of Theorem 5.
Proof of Theorem 5. By Lemma 6, the output  $\mathbf{x}_{\mathrm{out}}$  of Algorithm 2 satisfies
$$
\mathbb {E} \left[ F \left(\mathbf {x} _ {\text {o u t}}\right) - F ^ {*} \right] \leq \frac {4 \ell_ {F} R ^ {2}}{r \mathcal {T} (\mathcal {T} + 2)} + \frac {4 R \hat {\sigma}}{\sqrt {\mathcal {T}}} \leq \frac {\epsilon}{2}. \tag {13}
$$
Moreover, by Lemma 4 we can derive that
$$
\mathbb {E} [ f (\mathbf {x} _ {\mathrm {o u t}}) - f ^ {*} ] \leq \mathbb {E} [ f (\mathbf {x} _ {\mathrm {o u t}}) - F (\mathbf {x} _ {\mathrm {o u t}}) ] + (f ^ {*} - F ^ {*}) + \mathbb {E} [ F (\mathbf {x} _ {\mathrm {o u t}}) - F ^ {*} ] = \epsilon .
$$
Finally we bound the expected number of queries to  $O_{\mathbf{g}}$  used in the algorithm. Since  $\| \mathbf{g}_F(\mathbf{x})\| \leq L$  for any  $\mathbf{x}\in \mathbb{R}^d$ , by Theorem 4 we know that Algorithm 1 can output an estimate  $\tilde{\mathbf{g}}_F(\mathbf{x}_t^{md})$  of  $\nabla F(\mathbf{x}_t^{md})$  with variance at most  $\hat{\sigma}^2$  using  $\tilde{\mathcal{O}} (L\sqrt{d} /\hat{\sigma})$  queries to the oracle  $O_F$ , and hence the same number of queries to  $O_{\mathbf{g}}$  by Lemma 5. Then, the total number of queries to  $O_{\mathbf{g}}$  equals
$$
\mathcal {T} \cdot \widetilde {\mathcal {O}} \left(\frac {L \sqrt {d}}{\hat {\sigma}}\right) = \frac {d ^ {1 / 4} L R}{\epsilon} \cdot \widetilde {\mathcal {O}} \left(\frac {L \sqrt {d}}{d ^ {1 / 8}} \cdot \sqrt {\frac {R}{L \epsilon}}\right) = \widetilde {\mathcal {O}} \left(d ^ {5 / 8} \cdot \left(\frac {L R}{\epsilon}\right) ^ {3 / 2}\right).
$$
D Proof of quantum stochastic cutting plane method
D.1 Proof of Proposition 1
Definition 5 ( $\delta$ -approximate gradient oracle ( $\delta$ -AGO)). For  $f \colon \mathbb{R}^d \to \mathbb{R}$ , its  $\delta$ -approximate gradient oracle,  $\delta$ -AGO, is defined as a random function that when queried at  $\mathbf{x}$ , returns a vector  $\tilde{\mathbf{g}}(\mathbf{x})$  that satisfies  $\| \tilde{\mathbf{g}}(\mathbf{x}) - \nabla f(\mathbf{x}) \| \leq \delta$ .
We show that a  $\delta$ -AGO can be efficiently implemented using our quantum variance reduction algorithm. To obtain this result we can also use [53, Claim 1] by changing some parameters. Nevertheless, compared to their deterministic implementation, our approach has a lower expected time complexity (though is randomized).
Lemma 7. For any  $\delta, \xi \geq 0$ , with success probability at least  $1 - \xi$ , the  $\delta$ -AGO of a Lipschitz function  $f$  can be implemented using  $\mathcal{O}(\log(1/\xi))$  calls to Algorithm 1 with in total  $\mathcal{O}(L\sqrt{d}\log(1/\xi)/\delta)$  queries to the QSGO.
Proof. We implement the  $\delta$ -AGO using the following procedure. First, we obtain  $k = \log(1/\xi) + 10$  independent unbiased estimates  $\mathbf{g}_1, \ldots, \mathbf{g}_k$  of  $\nabla f(\mathbf{x})$  with variance  $\delta^2/16$  by Algorithm 1, using
$$
k \cdot \frac {L \sqrt {d}}{\delta} = \mathcal {O} (L \sqrt {d} \log (1 / \xi) / \delta)
$$
queries in total to the QSGO  $O_{\mathbf{g}}$ , by Theorem 4. Note that any individual sample  $\mathbf{g}_i$  satisfies
$$
\operatorname * {P r} \left[ \| \mathbf {g} _ {i} - \nabla f (\mathbf {x}) \| \geq \frac {\delta}{2} \right] \leq \frac {1}{4}.
$$
Hence, if we let  $S$  denote the subset of  $[k]$  where each  $i \in S$  have  $\ell_2$  distance to  $\nabla f(\mathbf{x})$  at most  $\delta / 2$ , then by Hoeffding's inequality,  $|S| \geq 2k / 3$  with probability at least
$$
1 - \exp \left(- \frac {2 (k / 3 - k / 4) ^ {2}}{k}\right) \geq 1 - \xi .
$$
Observe that for any sample  $\mathbf{g}_i$  satisfying  $\| \mathbf{g}_i - \mathbf{g}_j\| \leq \delta /2$  for at least  $2k / 3$  of  $\mathbf{g}_j$ 's, there must exist at least some  $\mathbf{g}_{j'}\in S$  such that  $\| \mathbf{g}_i - \mathbf{g}_{j'}\| \leq \delta /2$ , and therefore  $\| \mathbf{g}_i - \nabla f(\mathbf{x})\| \leq \delta$  by triangle inequality. Consequently, we just need to find and output any such  $\mathbf{g}_i$  as an  $\delta$ -AGO, which is guaranteed to exist given that any sample in  $S$  satisfies this property. This can be done in expected  $\mathcal{O}(k)$  time by picking a random estimate and checking whether it is close to at least  $2k / 3$  of other estimates.
Next, we establish that we can query an  $\epsilon$ -optimal point by applying the cutting plane method with the separation oracle being an  $\mathcal{O}(\epsilon / R)$ -approximate gradient oracle, as provided by Lemma 7.
Proof of Proposition 1. Define  $K_{\epsilon /2}$  as the set of  $\epsilon /2$ -optimal points of the function  $f$ , and  $K_{\epsilon}$  as the set of  $\epsilon$ -optimal points of  $f$ . We know that  $K_{\epsilon /2}$  contains a ball of radius at least  $r_K = \epsilon /(2L)$  since for any  $\mathbf{x}$  with  $\| \mathbf{x} - \mathbf{x}^* \| \leq r_K$  we have
$$
f (\mathbf {x}) - f \left(\mathbf {x} ^ {*}\right) \leq L \| \mathbf {x} - \mathbf {x} ^ {*} \| \leq \frac {\epsilon}{2}. \tag {14}
$$
We apply the cutting plane method, as described in Lemma 7, to query a point in  $K_{\epsilon /2}$ , which is a subset of the ball  $\mathbb{B}_{2R}(\mathbf{0})$ . To achieve this, we use an  $\epsilon /(10R)$ -approximate gradient oracle  $(\epsilon /(10R)\text{-AGO})$  of  $f$ , implemented using Lemma 7, as the separation oracle for the cutting plane method. Throughout this process, we assume each query to the  $\epsilon /(10R)$ -AGO is successfully executed, and we will later discuss the error probability associated with this assumption. In this case, we show that any query outside of  $K_{\epsilon}$  to the  $\epsilon /(10R)$ -AGO will be a valid separation oracle for  $K_{\epsilon /2}$ . In particular, if we ever queried the  $\epsilon /(10R)$ -AGO at any  $\mathbf{x}\in \mathbb{B}_{2R}(\mathbf{0})\setminus K_{\epsilon}$  with output being  $\tilde{\mathbf{g}}$ , for any  $\mathbf{y}\in K_{\epsilon /2}$  we have
$$
\begin{array}{l} \langle \tilde {\mathbf {g}}, \mathbf {y} - \mathbf {x} \rangle \leq \langle \nabla f (\mathbf {x}), \mathbf {y} - \mathbf {x} \rangle + \| \tilde {\mathbf {g}} - \nabla f (\mathbf {x}) \| \cdot \| \mathbf {y} - \mathbf {x} \| \\ \leq f (\mathbf {y}) - f (\mathbf {x}) + \| \tilde {\mathbf {g}} - \nabla f (\mathbf {x}) \| \cdot \| \mathbf {y} - \mathbf {x} \| \\ \leq - \frac {\epsilon}{2} + \frac {\epsilon}{1 0 R} \cdot 4 R <   0, \\ \end{array}
$$
where the second inequality is due to the convexity of  $f$ , indicating that  $\tilde{\mathbf{g}}$  is a valid separation oracle for the set  $K_{\epsilon /2}$ . Consequently, upon applying Lemma 2, we can deduce that after  $C d\log (dR / r_K)$  iterations, at least one of the queries must lie within  $K_{\epsilon}$ .
To ensure an overall success probability of at least  $5 / 6$ , we employ the union bound, which necessitates that each query to the  $\epsilon /(10R)$ -AGO be implemented with a failure probability of no more than  $(6Cd\log (R / r_K))^{-1}$ , which by Lemma 7 requires
$$
\mathcal {O} \left(\frac {1 0 L R \sqrt {d}}{\epsilon} \log (6 C d \log (d R / r _ {K}))\right) = \widetilde {\mathcal {O}} \left(\frac {L R \sqrt {d}}{\epsilon}\right)
$$
queries to the QSGO  $O_{\mathbf{g}}$ , and the overall query complexity equals
$$
\widetilde {\mathcal {O}} \left(\frac {L R \sqrt {d}}{\epsilon}\right) \cdot C d \log (d R / r _ {K}) = \widetilde {\mathcal {O}} \left(\frac {d ^ {3 / 2} L R}{\epsilon}\right).
$$
D.2 Proof of Proposition 2
We first present the following algorithm that makes pairwise comparison using queries to the QSGO(Definition 3).
Note that in Line 7 of Algorithm 4 the estimation step is carried out in a projected space along the vector  $\hat{\mathbf{e}}$ . This is motivated by the fact that we are specifically interested in the information of the gradient  $\nabla f$  within this projected space, which effectively reduces to a one-dimensional variable. As a result, the mean estimation can be performed without introducing an additional factor of  $\sqrt{d}$  in the query complexity as given in Theorem 4.
Lemma 8. For any  $\mathbf{y}_{l0},\mathbf{y}_{r0}\in \mathbb{B}_R(\mathbf{0})$  and any  $\epsilon^{\prime} > 0$ , with success probability at least  $1 / (6\mathcal{T})$  Algorithm 4 returns a point  $\hat{\mathbf{y}}\in \mathbb{R}^{d}$  satisfying
$$
f (\hat {\mathbf {y}}) \leq \min  _ {\lambda \in [ 0, 1 ]} f \left(\lambda \mathbf {y} _ {l 0} + (1 - \lambda) \mathbf {y} _ {r 0}\right) + \epsilon^ {\prime}
$$
using  $\widetilde{\mathcal{O}}\left(RL / \epsilon^{\prime}\right)$  queries to an  $L$ -bounded QSGO  $O_{\mathbf{g}}$  defined in Definition 3.
Algorithm 4: StochasticLineSearch  $(\mathbf{y}_{l0},\mathbf{y}_{r0},\epsilon^{\prime})$
Input: Endpoints  $\mathbf{y}_l,\mathbf{y}_r\in \mathbb{B}_R(\mathbf{0})$  , accuracy  $\epsilon^\prime$  Output:  $\hat{\mathbf{y}}$  such that  $f(\hat{\mathbf{y}})\leq \min_{\lambda \in [0,1]}f\big(\lambda \mathbf{y}_l + (1 - \lambda)\mathbf{y}_r\big) + \epsilon '$    
1  ${\bf y}_l\gets {\bf y}_{l0},{\bf y}_r\gets {\bf y}_{r0}$    
2 if  $\mathbf{y}_l = \mathbf{y}_r$  then return  $\mathbf{y}_l$    
3 else   
4  $\hat{\mathbf{e}}\gets \frac{\mathbf{y}_r - \mathbf{y}_l}{\|\mathbf{y}_r - \mathbf{y}_l\|}$    
5 repeat   
6  $\mathbf{y}_m\gets (\mathbf{y}_l + \mathbf{y}_r) / 2$    
7 Obtain an estimate  $\tilde{g}_{\hat{\mathbf{e}}}(\mathbf{y}_m)$  of  $\nabla f(\mathbf{y}_m)^{\top}\hat{\mathbf{e}}$  with error at most  $\epsilon^{\prime} / (4R)$    
8 if  $|\tilde{g}_{\hat{\mathbf{e}}}(\mathbf{y}_m)|\leq \epsilon ' / (4R)$  then return  $\hat{\mathbf{y}}\gets \mathbf{y}_m$    
9 else   
10 if  $\tilde{g}_{\hat{\mathbf{e}}}(\mathbf{y}_m) > 0$  then  $\mathbf{y}_r\gets \mathbf{y}_m$  else  $\mathbf{y}_l\gets \mathbf{y}_m$    
11 until  $\| \mathbf{y}_r - \mathbf{y}_l\| \leq \epsilon ' / L$    
12 return  $\hat{\mathbf{y}}\gets \mathbf{y}_l$
Proof. Observe that in each iteration the value  $\| \mathbf{y}_r - \mathbf{y}_l\|$  is decreased by at least  $1 / 2$ . Hence, the total number of iterations is at most
$$
\log_ {2} \left(\frac {\| \mathbf {y} _ {r 0} - \mathbf {y} _ {l 0} \|}{\epsilon^ {\prime} / L}\right) \leq \log_ {2} \left(\frac {2 R L}{\epsilon^ {\prime}}\right).
$$
Within each iteration, we need to estimate the value  $\nabla f(\mathbf{y}_m)^\top \hat{\mathbf{e}}$ , which is the component of  $\nabla f(\mathbf{y}_m)$  along  $\hat{\mathbf{e}}$ . Note that this is essentially a univariate mean estimation problem, given that  $\hat{\mathbf{e}}$  is fixed in this iteration we can define an 1-dimensional random variable  $\nabla f(\mathbf{y}_m)^\top \hat{\mathbf{e}}$  whose variance is at most
$$
\operatorname {V a r} [ \nabla f (\mathbf {y} _ {m}) ^ {\top} \hat {\mathbf {e}} ] \leq \mathbb {E} | \nabla f (\mathbf {y} _ {m}) ^ {\top} \hat {\mathbf {e}} | ^ {2} \leq \mathbb {E} \| \nabla f (\mathbf {y} _ {m}) \| ^ {2} \leq L ^ {2}.
$$
Then by Lemma 7, with success probability at least
$$
1 - \left(6 \mathcal {T} \log_ {2} \left(\frac {2 R L}{\epsilon^ {\prime}}\right)\right) ^ {- 1}, \tag {15}
$$
an estimate of  $\nabla f(\mathbf{y}_m)^\top \hat{\mathbf{e}}$  with error at most  $\epsilon' / (4R)$  can be obtained using
$$
\frac {L}{\epsilon^ {\prime} / (4 R)} \cdot \log \left(6 \mathcal {T} \log_ {2} \left(\frac {2 R L}{\epsilon^ {\prime}}\right)\right) = \widetilde {\mathcal {O}} \left(\frac {R L}{\epsilon^ {\prime}}\right)
$$
queries to the quantum stochastic gradient oracle  $O_{\mathbf{g}}$ , given that we can prepare the following quantum oracle