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  1. Apply the Composite Laplacian Filter first, then apply the gaussian filter.
  2. Apply the gaussian filter first, then apply the composite laplacian filter.

My work as below: Here we assume the original image is function $f(x,y)$ $$ (1): (f-\nabla^2f)\ast G = f \ast G - \nabla^2f \ast G$$ $$ (2): (f \ast G) - \nabla^2(f \ast G) = f \ast G - f \ast \nabla^2G $$

I am wondering the statement $\nabla^2f \ast G = f \ast \nabla^2G$ is true or not.

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Formally, (linear) derivatives and convolutions may commute, as explained on Wikipedia-Convolution/Properties/Differentiation. This is a major "operation-saving" property: if one wants to differentiate many images and convolve then with a fixed kernel (here a Gaussian), one can convolve the images with the differentiate kernel.

You don't have to differentiate all images, just the kernel once.

Caveat: what is written above could be taken with care, notably when derivatives don't exist theoretically, or with subtleties in implementing derivatives.

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    $\begingroup$ Therefore the equality I mentioned is correct? Since I need to prove / disprove this. As you mentioned, seems this is equal $\endgroup$
    – Nic
    Feb 9 at 13:11

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