Let $A,B\in \mathbb{R}^{d\times d\times d}$. The 3D circular convolution $\text{conv}(A,B)$ can then be computed as $F^{-1}(F(A)\odot F(B))$ where $F(\cdot)$ denotes the 3D discrete Fourier transform and $\odot$ denotes entry-wise multiplication. By using the fast Fourier transform I can then compute $\text{conv}(A,B)$ in $O(d^3\lg d)$ time.
I only need $d^2$ entries of $\text{conv}(A,B)$ and $A,B$ only contain $d^2$ non-zero entries. Is it possible to compute $d^2$ entries of $\text{conv}(A,B)$ faster than $O(d^3)$?
Specifically, $A_{i,j,k}=0$ for all $i\neq 0$ and $B_{i,j,k}=0$ for all $j\neq 0$, and I want the $d^2$ entires $\text{conv}(A,B)_{i,k,d-1}$ for $i,k=0,...,d-1$ .