4 ms·
> But these excuses are not rigorously justified, even at the graduate level, in my experience. Imo, the informal use is already pretty close to the formal def
by creata 2y ago
> But these excuses are not rigorously justified, even at the graduate level, in my experience.
Imo, the informal use is already pretty close to the formal definition. Formally, a distribution is defined purely by its inner products against certain smooth functions (usually the ones with compact support) which is what the OP alluded to when he said:
> The formal definition of a generalized function is: an element of the continuous dual space of a space of smooth functions.
That "element of the continuous dual space" is just a function that takes in a smooth function with compact support f, and returns what we take to be the inner product of f with our generalized function.
So (again, imo) "we don’t need to represent it directly, only the result of an inner product against a smooth function" isn't that distant to the formal definition.
- mturmon 2y agoI hear you, and I admit I'm drawing a fuzzy line (is the conventional approach “rigorous”). Here are two “test functions”- - we learned much about impulse responses, and sometimes considered responses to dipoles, etc. However, if I read the Wikipedia article correctly (it’s not great…), the theory implies that a distribution (in the technical sense) has derivatives of any order. I’m not sure I really knew that I could count on that. A rigorous treatment would have given me that assurance. - if I understand correctly, the concept of introducing an impulse to a system that has an identity impulse response, which implies an inner product of delta with itself, is not well-defined. Again, I’m not sure if we covered that concept. (Admittedly, it’s been a long time.)
- mturmon 2y agooops, I realize I completely mis-stated the second point. What it should say is: - If delta(x) is OK, why is delta^2(x) not OK?