1
$\begingroup$

I think most students have there first contact with the term support in an early analysis course where one defines $\mathrm{supp} f = \overline{\{x\in D: f(x)\neq 0\}}$. When one later works with almost everywhere defined functions, distributions and such stuff one wants to define a support as well but realizes that the actual question is not where a function does not vanish but where it really (!) vanishes, that is in a neighborhood.

My question is therefore: Why do most people define the support of a classical function from the "wrong" point of view?

  • 0
    At least my first Analysis courses didn't cover measures or what "almost everywhere" stands for. – 2017-02-24
  • 0
    Please read my question again. I said "When one later ..." – 2017-02-24
  • 0
    I did read it. But usually one defines the support in an early analysis course where the usual definition is very intuitive. – 2017-02-24
  • 0
    I think the main reason here is that the "classical approach" is more intuitive and easier to grasp. When, however, you start talking about measure theory, distribution theory, and Lebesgue spaces, you are be sure that the audience is capable of understanding both approaches. – 2017-02-24
  • 0
    1) It doesn't help anybody if something is easier to grasp but the wrong concept 2) I don't see why it is more intuitive that a function is zero then that it is non-zero and that you don't have to require the latter in a neighborhood in the first place but you have to pay for it when taking the closure. – 2017-02-24
  • 0
    Your question does not quite make sense to me, and as worded it seems to me that you are raising a straw man. The question of where a function does not vanish is often very important. – 2017-02-24
  • 0
    But what does "not vanish" mean? Does $x$ vanish in $0$ on $(-1,1)$? No. Does the characteristic function of $\{0\}$ not vanish in $0$ on $(-1,1)$? Classically one would say yes, but weak one would say no! I think a good definition should direct me into the right direction when it comes to generalization. Another example: Linear isomorphisms. When you define them as bijective, linear maps (as it is done in most linear algebra classes), then those students will think of topological isomorphisms as bijective, continuous maps because they were led into the wrong direction by a bad definition. – 2017-02-24

0 Answers 0