# CRF message passing as convolution operation

I was reading the paper _Efficient Inference in Fully Connected CRFs with Gaussian Edge Potentials (Philipp Krähenbül and Vladen Koltun, in Proceedings of 25th Annual Conference on Neural Information Processing Systems (NIPS), 2011 pdf), and I didn't understand this equation (eq 5) in the paper: I understand the first equation, but not the second one. If $$k(f_i,f_j) = e^{-1*(f_i-f_j)^{2}}$$, then $$k(f_i,f_j)$$ would return a scalar value between 0 and 1 ; 0 if $$f_i$$ and $$f_j$$ are far apart in the feature space, 1 otherwise. So, $$k(f_i,f_j)Q_j$$ adds a fraction of $$Q_j$$ to $$Q_i$$ . The first equation is clear. I don't understand how this leads to the 2nd equation. In the 2nd equation, the Gaussian kernel is now over $$Q$$ instead of being over $$f_i,f_j$$ and then multiplied by $$f_i$$. Where is the $$f_j$$? Can someone explain how the 2nd equation is derived from the first one?

$$G \otimes Q$$ would again return a value between 0 and 1, which is multiplied by $$f_i$$ (which is an n-dimensional feature vector). So the result of $$(G \otimes Q)(f_i)$$ would be an n-dimensional vector, and we subtract $$Q_i$$ (a scalar) from that. Am I misunderstanding something here?

I read the article. The meaning of the equation: $$[G\otimes Q](f_i)$$ stand for $G$ convolution with $Q$ in the feature point $f_i$ and not multiplied by $f_i$.
The $f_j$ is hidden inside the convolution operation.