The convolution theorem is much uglier for the hartley transform. I guess this is the main reason. You can see the Fourier transform as a simple variant of the Hartley transform where the linear filtering can be done directly in the frequency domain, without all the fuss of the hartley transform convolutions.
Another advantage of Fourier with respect to Harley is that derivatives become products with polynomials. Thus the Fourier transform transforms linear pde into trivial algebraic equations. For the Hartley transform, you get more complicated functional equations, so that it's not as useful.
> You can blur an image by taking a hamming window and multiplying.
It doesn't seem like you can? (Unless your image has some symmetries).
Another advantage of Fourier with respect to Harley is that derivatives become products with polynomials. Thus the Fourier transform transforms linear pde into trivial algebraic equations. For the Hartley transform, you get more complicated functional equations, so that it's not as useful.
> You can blur an image by taking a hamming window and multiplying.
It doesn't seem like you can? (Unless your image has some symmetries).