Renormalized interaction potential ($V_{\mathrm{ren}}(x)=\int [W(x-z)-W(-z)]d\mu (z)$)

For a twice continuously differentiable interaction potential with $\Vert D^{2}W\Vert \le L$ and a probability measure with finite first absolute moment, this subtraction produces a finite potential even if $W\ast \mu$ is infinite. Taylor's formula and linear growth of the gradient giveDominated differentiation under the integral sign gives $\nabla V_{\mathrm{ren}}=(\nabla W)\ast \mu$ and $D^2{V}_{\mathrm{ren}}=(D^{2}W)\ast \mu$. If the unrenormalized convolution is finite, the two potentials differ only by a constant, so they generate the same velocity field when that field depends on their gradient.