SlowGuess/NeurIPS_2023_OCR huggingface dataset:
which satisfies  $f(\mathbf{x}^{\prime}) \leq f(\mathbf{x})$  and is thus also an  $\epsilon$ -optimum of  $f$ . Then we can derive that
$$
- \frac {1}{3} \left\langle \mathbf {x} ^ {\prime}, \bar {\mathbf {g}} \right\rangle + \frac {1}{3} \left\langle \mathbf {x} ^ {*}, \bar {\mathbf {g}} \right\rangle \leq \epsilon ,
$$
$$
\begin{array}{l} \left\langle \mathbf {x} ^ {\prime}, \mathbf {x} ^ {*} \right\rangle \geq \left\langle \mathbf {x} ^ {*}, \mathbf {x} ^ {*} \right\rangle - \epsilon \cdot \frac {\| \mathbf {x} ^ {*} \|}{\| \bar {\mathbf {g}} \|} \\ \geq 1 - \frac {R \epsilon}{2 \| \bar {\mathbf {g}} \|}, \\ \end{array}
$$
given that  $\frac{R}{2}\frac{\mathbf{x}}{\|\mathbf{x}\|} \geq \|\mathbf{x}'\|$ . As for the last entry, note that
$$
\begin{array}{l} \underset {i, j} {\mathbb {E}} [ \hat {\mathbf {g}} _ {i, j} (\mathbf {x}) ] = - \frac {1}{3} \cdot \underset {i, j} {\mathbb {E}} \mathbf {g} _ {i, j} + \frac {2 L}{3} \cdot \max  \left\{\| \mathbf {x} \| - \frac {R}{2} \right\} \cdot \frac {\mathbf {x}}{\| \mathbf {x} \|} \\ = - \frac {1}{3} \cdot \bar {\mathbf {g}} + \frac {2 L}{3} \cdot \max  \left\{\| \mathbf {x} \| - \frac {R}{2} \right\} \cdot \frac {\mathbf {x}}{\| \mathbf {x} \|}, \\ \end{array}
$$
which is the (sub-)gradient of  $\bar{f}^A$  at  $\mathbf{x}$ . Moreover, for any  $i,j$  we have
$$
\left\| \hat {\mathbf {g}} _ {i, j} (\mathbf {x}) \right\| \leq \frac {1}{3} \left\| \mathbf {g} _ {i, j} \right\| + \frac {2 L}{3} \leq L.
$$
Equipped with Lemma 11, we first prove our quantum lower bound for Problem 1 in the low-dimensional regime by encoding the Search  $\circ$  Parity problem into the task of finding an  $\epsilon$ -optimal point of  $\bar{f}^A$  defined in (20).
Proof of Theorem 8 when  $\epsilon \leq \frac{RL}{100\sqrt{d}}$ . Our proof proceeds by contradiction. Denote  $n = \frac{RL\sqrt{d}}{100\alpha\epsilon}$ . Assume for simplicity that  $d$  is even, and  $\alpha n$  is an even multiple of  $d$  (the other cases can be handled by padding arguments). Then, we show that  $\mathsf{Search}^d \circ \mathsf{Parity}^{\alpha n / d}$  on this instance  $A$  can be solved by finding an  $\epsilon$ -optimum of  $f^A$  defined in (20). In particular, we define the set of vectors  $\{\mathbf{g}_{i,j}\}$  to be
$$
\mathbf {g} _ {i, j} = \frac {\alpha L n}{\sqrt {(2 \alpha n) ^ {2} - d ^ {2}}} (- 1) ^ {1 + A _ {i, j}} \mathbf {e} _ {i}, \tag {22}
$$
where  $i \in [d]$ ,  $j \in [\alpha n / d]$ , and  $\mathbf{e}_i \in \mathbb{R}^d$  is the  $i$ -th indicator vector. Consider the function  $\bar{f}^A$  defined in (20) with the set of vectors  $\{\mathbf{g}_{i,j}\}$  having the values in (22), we have
$$
\begin{array}{l} \bar {\mathbf {g}} = \underset {i, j} {\mathbb {E}} [ \mathbf {g} _ {i, j} ] = \frac {1}{\alpha n} \cdot \frac {\alpha L n}{\sqrt {(2 \alpha n) ^ {2} - d ^ {2}}} \sum_ {i = 1} ^ {d} \mathbf {e} _ {i} \sum_ {j = 1} ^ {\alpha n / d} (- 1) ^ {1 + A _ {i, j}} \\ = \frac {2 L}{\sqrt {(2 \alpha n) ^ {2} - d ^ {2}}} \sum_ {i = 1} ^ {d} b _ {i} ^ {(A)} = \frac {2 L}{\sqrt {(2 \alpha n) ^ {2} - d ^ {2}}} {\bf b} ^ {(A)}, \\ \end{array}
$$
where the vector  $\mathbf{b}^{(A)}$  is defined in (18). By Lemma 11, every  $\epsilon$ -optimal point  $\mathbf{x}$  of  $\bar{f}^A$  satisfies
$$
\left\langle \frac {R}{2} \frac {\mathbf {x}}{\| \mathbf {x} \|}, \mathbf {x} ^ {*} \right\rangle \geq 1 - \frac {R \epsilon}{2 \| \bar {\mathbf {g}} \|},
$$
by which we can derive that
$$
\begin{array}{l} \left\langle \frac {R}{2} \frac {\mathbf {x}}{\| \mathbf {x} \|}, \mathbf {b} ^ {(A)} \right\rangle \geq \frac {\| \mathbf {b} ^ {(A)} \|}{\| \mathbf {x} ^ {*} \|} \cdot \left(1 - \frac {R \epsilon}{2 \| \bar {\mathbf {g}} \|}\right) \\ \geq \frac {R}{2} \sqrt {\frac {d}{2}} - \frac {3 \alpha n \epsilon}{L} = \frac {R \sqrt {d}}{4}, \\ \end{array}
$$
$$
\left\langle \sqrt {\frac {d}{2}} \cdot \frac {\mathbf {x}}{\| \mathbf {x} \|}, \mathbf {b} ^ {(A)} \right\rangle \geq \frac {d}{2 \sqrt {2}}.
$$
$$
\left\| \sqrt {\frac {d}{2}} \cdot \frac {\mathbf {x}}{\| \mathbf {x} \|} \right\| = \left\| \mathbf {b} ^ {(A)} \right\| = \sqrt {\frac {d}{2}},
$$
$$
\left\| \sqrt {\frac {d}{2}} \cdot \frac {\mathbf {x}}{\| \mathbf {x} \|} - \mathbf {b} ^ {(A)} \right\| \leq \frac {\sqrt {d}}{2},
$$
indicating that we can obtain an estimate  $\widetilde{\mathbf{b}} = \sqrt{\frac{d}{2}}\frac{\mathbf{x}}{\|\mathbf{x}\|}$  of  $\mathbf{b}^{(A)}$  with  $\ell_2$ -error at most  $\sqrt{d}/2$  by obtaining an  $\epsilon$ -optimal point of  $\bar{f}^A$ .
Next, we show that we can implement a quantum stochastic gradient oracle of  $\bar{f}^A$  using only one query to the oracle  $O_A$ , given that one can use one query to  $O_A$  to implement the following oracle
$$
O _ {\mathbf {g}} ^ {A} \left| \right. i \left. \right\rangle\left| \right. j \left. \right\rangle\left| \right. 0 \left. \right\rangle\rightarrow \left| \right. i \left. \right\rangle\left| \right. j \left. \right\rangle\left| \right. \mathbf {g} _ {i, j} \left. \right\rangle
$$
and use one query to  $O_{\mathbf{g}}^{A}$  to implement
$$
O _ {\hat {\mathbf {g}}} ^ {A} | \mathbf {x} \rangle | i \rangle | j \rangle | 0 \rangle \rightarrow | \mathbf {x} \rangle | i \rangle | j \rangle | \hat {\mathbf {g}} _ {i, j} (\mathbf {x}) \rangle ,
$$
where  $\hat{\mathbf{g}}_{i,j}(\mathbf{x})$  is defined in (21). Then by applying  $O_{\hat{\mathbf{g}}}^{A}$  to the state
$$
\frac {1}{\sqrt {\alpha n}} | \mathbf {x} \rangle \sum_ {i \in [ d ]} \sum_ {j \in [ \alpha n / d ]} | i \rangle | j \rangle | 0 \rangle
$$
and uncompute the second and the third register, we construct the following stochastic gradient oracle  $O_{\hat{\mathbf{g}}}$  of  $\bar{f}^A$
$$
O _ {\hat {\mathbf {g}}} | \mathbf {x} \rangle \otimes | 0 \rangle \rightarrow \frac {1}{\sqrt {\alpha n}} | \mathbf {x} \rangle \otimes \sum_ {i \in [ d ]} \sum_ {j \in [ \alpha n / d ]} | \hat {\mathbf {g}} _ {i, j} (\mathbf {x}) \rangle .
$$
Hence, any quantum algorithm that can find an  $\epsilon$ -optimum of  $\tilde{f}^A$  using  $T$  queries to  $O_{\hat{\mathbf{g}}}$  can be transformed into a quantum algorithm for the Search $^d$ $\circ$  Parity $^{\alpha n / d}$  problem that uses  $T$  queries to the oracle  $O_A$  defined in (19) and returns an estimate  $\tilde{\mathbf{b}}$  with  $\ell_2$ -error at most  $\sqrt{d} / 2$ . Then by Lemma 10, for any quantum algorithm making less than
$$
d \cdot {\frac {\alpha n}{d}} \cdot {\frac {1}{\alpha}} = {\frac {R L}{2 \epsilon \alpha}} {\sqrt {\frac {d}{2}}}
$$
queries to the quantum stochastic gradient oracle  $O_{\hat{\mathbf{g}}}$ , there exists an  $A \in \mathcal{A}_{d,\alpha n / d}$  and corresponding  $\bar{f}^A$  such that the output of the algorithm is not an  $\epsilon$ -optimum of  $\bar{f}^A$  with probability at least  $2 / 3$ .
Similarly, we can obtain our quantum lower bound in the high-dimensional regime by encoding the Search  $\circ$  Parity problem into the task of finding an  $\epsilon$ -optimal point of  $f^A$  defined in (20).
Proof of Theorem 8 when  $\frac{RL}{100\sqrt{d}} \leq \epsilon \leq 1$ . Our proof proceeds by contradiction. Denote  $n = \frac{R^2L^2}{10000\alpha\epsilon^2}$ . Consider an instance  $A \in \mathcal{A}_{\alpha n,1}$  of the Search $^{\alpha n} \circ \mathrm{Parity}^1$  problem where  $d$  is a power of 2 and  $\alpha$  is even. Then, we show that  $\mathrm{Search}^d \circ \mathrm{Parity}^{\alpha n / d}$  on this instance  $A$  can be solved by finding an  $\epsilon$ -optimum of  $\tilde{f}^A$  defined in (20). In particular, we define the set of vectors  $\{\mathbf{g}_{i,j}\}$  where  $j \equiv 1$  to be
$$
\mathbf {g} _ {i, 1} = \alpha L A _ {i, 1} \sqrt {\frac {n}{2 \alpha^ {2} n - 2 \alpha}} \mathbf {e} _ {i}, \tag {23}
$$
where  $i \in [\alpha n]$  and  $\mathbf{e}_i \in \mathbb{R}^d$  is the  $i$ -th indicator vector. Consider the function  $\bar{f}^A$  defined in (20) with the set of vectors  $\{\mathbf{g}_{i,1}\}$  having the values in (23), we have
$$
\mathbf {\bar {g}} = \underset {i} {\mathbb {E}} [ \mathbf {g} _ {i, 1} ] = \frac {\alpha L}{\alpha n} \sqrt {\frac {n}{2 \alpha^ {2} n - 2 \alpha}} \sum_ {i = 1} ^ {d} A _ {i, 1} \mathbf {e} _ {i} = \frac {L}{n} \sqrt {\frac {n}{2 \alpha^ {2} n - 2 \alpha}} \mathbf {b} ^ {(A)},
$$
$$
\begin{array}{l} \left\langle \frac {R}{2} \frac {\mathbf {x}}{\| \mathbf {x} \|}, \mathbf {b} ^ {(A)} \right\rangle \geq \frac {\| \mathbf {b} ^ {(A)} \|}{\| \mathbf {x} ^ {*} \|} \cdot \left(1 - \frac {R \epsilon}{2 \| \bar {\mathbf {g}} \|}\right) \\ \geq \frac {R}{2} \sqrt {\frac {\alpha n}{2}} - \frac {6 \alpha n \epsilon}{L} = \frac {R \sqrt {\alpha n}}{4}, \\ \end{array}
$$
$$
\left\langle \sqrt {\frac {\alpha n}{2}} \cdot \frac {\mathbf {x}}{\| \mathbf {x} \|}, \mathbf {b} ^ {(A)} \right\rangle \geq \frac {\alpha n}{2 \sqrt {2}}.
$$
$$
\left\| \sqrt {\frac {\alpha n}{2}} \cdot \frac {\mathbf {x} ^ {\prime}}{\| \mathbf {x} ^ {\prime} \|} \right\| = \left\| \mathbf {b} ^ {(A)} \right\| = \sqrt {\frac {\alpha n}{2}},
$$
$$
\left\| \sqrt {\frac {\alpha n}{2}} \cdot \frac {\mathbf {x}}{\| \mathbf {x} \|} - \mathbf {b} ^ {(A)} \right\| \leq \frac {\sqrt {\alpha n}}{2}.
$$
indicating that we can obtain an estimate  $\widetilde{\mathbf{b}} = \sqrt{\frac{\alpha n}{2}}\frac{\mathbf{x}}{\|\mathbf{x}\|}$  of  $\mathbf{b}^{(A)}$  with  $\ell_2$ -error at most  $\sqrt{d}/2$  by obtaining an  $\epsilon$ -optimal point of  $\bar{f}^A$ .
$$
O _ {\mathbf {g}} ^ {A} | i \rangle | 0 \rangle \rightarrow | i \rangle | \mathbf {g} _ {i, 1} \rangle
$$
$$
O _ {\hat {\mathbf {g}}} ^ {A} | \mathbf {x} \rangle | i \rangle | 0 \rangle \rightarrow | \mathbf {x} \rangle | i \rangle | \hat {\mathbf {g}} _ {i, 1} (\mathbf {x}) \rangle ,
$$
where  $\hat{\mathbf{g}}_{i,1}(\mathbf{x})$  is defined in (21). Then by applying  $O_{\hat{\mathbf{g}}}^{A}$  to the state
$$
\frac {1}{\sqrt {\alpha n}} | \mathbf {x} \rangle \sum_ {i \in [ \alpha n ]} | i \rangle | 0 \rangle
$$
and uncompute the second and the third register, we construct the following stochastic gradient oracle  $O_{\mathbf{g}}$  of  $\bar{f}^A$ .
$$
O _ {\hat {\mathbf {g}}} | \mathbf {x} \rangle \otimes | 0 \rangle \rightarrow \frac {1}{\sqrt {\alpha n}} | \mathbf {x} \rangle \otimes \sum_ {i \in [ \alpha n ]} | \hat {\mathbf {g}} _ {i, 1} (\mathbf {x}) \rangle .
$$
Hence, any quantum algorithm that can find an  $\epsilon$ -optimum of  $\bar{f}^A$  using  $T$  queries to  $O_{\hat{\mathbf{g}}}$  can be transformed into a quantum algorithm for the Search $^{\alpha n}$ $\circ$  Parity $^1$  problem that uses  $T$  queries to the oracle  $O_A$  defined in (19) and returns an estimate  $\widetilde{\mathbf{b}}$  with  $\ell_2$ -error at most  $\sqrt{d}/2$ . Then by Lemma 10, for any quantum algorithm making less than
$$
\alpha n = \frac {R ^ {2} L ^ {2}}{1 0 0 \epsilon^ {2}}
$$
queries to the quantum stochastic gradient oracle  $O_{\hat{\mathbf{g}}}$ , there exists an  $A \in \mathcal{A}_{\alpha n,1}$  and corresponding  $\bar{f}^A$  such that the output of the algorithm is not an  $\epsilon$ -optimum of  $\bar{f}^A$  with probability at least  $2/3$ .