Second derivative finite difference of a convolution of two functions.

Question

I have a discrete function $I(V)$ which is defined as $$I(V) = R\big(V\big) * Q \big(V \big) = \int_a^b R\big(V\big) Q \big(V-V' \big) dV'$$ and I want to find $\frac{\partial^2R}{\partial V^2}$ but do not know $Q \big(V \big)$. I am able to take the second derivative of $I(V)$ via finite difference: $$\frac{\partial^2I}{\partial V^2} \approx \frac{I(V+V_{0})+I(V-V_{0})-2I(V)}{V_{0}^2}$$ and I know that derivatives can be applied to either function in a convolution: $$\frac{\partial^2I}{\partial V^2} = \frac{\partial^2R}{\partial V^2} * Q \big(V \big) = R \big(V \big) * \frac{\partial^2Q}{\partial V^2} = \frac{\partial R}{\partial V} * \frac{\partial Q}{\partial V}$$ I also know that the process of taking the second derivative of a discrete function is the equivalent of convolving the function with the 3x3 Toeplitz matrix: $$\begin{bmatrix}-2 & 1 & 0\\1 & -2 & 1\\0 & 1 & -2\end{bmatrix}$$ Can I find either $\frac{\partial^2R}{\partial V^2}$ or $Q \big(V \big)$ if I know $I(V)$ (and therefore $\frac{\partial^2I}{\partial V^2}$)?