Laplacian of Gaussian operator This might be a silly question. I was reading about the Laplacian of Gaussian (LoG) operator and got confused about the alternative equivelant ways we can make use of it.
Let's assume we have a 2D image $I(x,y)$
Since linear convolution commutes with differential operators the following equality is valid:
$$\nabla^2 (G_{\sigma}\ast I) = (\nabla^2 G_{\sigma})\ast I $$
We also know that convolution is commutative, so we could also write (could we??) :
$$ \nabla^2 (G_{\sigma}\ast I) = \nabla^2 (I \ast G_{\sigma})  \mathop{=}\limits^?G_{\sigma}\ast(\nabla^2 I)$$
which means that we first have to  apply the Laplacian operator on image $I(x,y)$ and then convolve the result with the Gaussian filter.
However, is this approach correct? It seems to me wrong and haven't found it anywhere. I tend to believe it's not valid since it will amplify the noise of initial image $I(x,y)$ (because we first apply the Laplacian operator on the noisy image) and then the smoothing will not be such efficient.
This sounds counter intuitive to many, but as long as the difference operator and the smoothing kernel are linear and space-invariant, they can be applied in any order, and thus are often combined in a single convolution operator (for more computational efficiency), for the same.
For some intuition, consider that, either for the linear smoothing and the derivative, a noisy pixel  is replaced by a linear weighed combination of other pixels. And a linearly-weighed combination of another linearly-weighed combination remains a linear combination, the same due to the commutativity and the associativity properties of the addition and the multiplication.
This is illustrated below for the first derivative only, but you can follow the same argument by differentiating again.


What you describe is commonly known as the LoG, Laplacian of Gaussian.
Note that when one operation becomes nonlinear, this is not true anymore.