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I don't think you can really call it a generalized dot product, because it doesn't map to a scalar. The inner product is the well accepted definition of a gener
by Tomminn 6y ago
I don't think you can really call it a generalized dot product, because it doesn't map to a scalar. The inner product is the well accepted definition of a generalized dot product, and convolution does not follow the axioms that an inner product must follow.
- rualca 6y ago> I don't think you can really call it a generalized dot product, because it doesn't map to a scalar. This. Convolution is a function which results from applying to convolution operator to two functions. A dot product is nothing of the sort.
- enriquto 6y ago> A dot product is nothing of the sort. Its nearly the same thing, isn't it? If you denote by Tx the left-shift operator defined by (Tx f)(y) = f(y+x), then the correlation of f and g evaluated at x is precisely the dot product of f and Tx g. If you evaluate your function at a certain point, you obtain a scalar product.
- rualca 6y ago> (...) then the correlation of f and g evaluated at x It really isn't the same, and oddly enough you unknowingly show that off, by mentioning that convolution is the function that maps input functions to the output function, but the dot product is at best a single point evaluated with the output function.
- tprice7 6y agoFurthermore, it doesn't involve complex conjugation, and (closely related) it doesn't have any property akin to positive-definiteness.
- kazinator 6y agoYes, thank you; you are right. The integral which is involved at the kernel of it is like a dot product, not the entire convolution itself.