One thing I cannot wrap my head around is that there are so many many many conditions for different function spaces, how can you quickly determine which function space a vector/function belongs to?

I think there are 11 conditions for vectors space alone, Hilbert space adds one more, Banach space adds two more, Sobolev space adds....how many more I've lost count.

To know if a function belongs to a certain space, you will have to prove it satisfies each and every single condition for that function space. Even for a function simple as $f(x) = x$, you'd have to prove it for 11 + conditions and how can you ever remember all these conditions?

How do you remember all these conditions and is there a good way to know exactly which function space that a vector/function belongs to upon a glance?


1 Answer 1


It looks vaguely like you're confusing the conditions for something to be a vector space with the conditions for something to be an element of a particular already known space.

For example, take the vector space of all real polynomials. The only thing we need to check whether $f(x)=x$ is a member of that space or not is to figure out whether $x$ is a polynomial or not. Since it is, the identity function is in the space.

Similarly for the space of all continuous functions $\mathbb R\to\mathbb R$, or the space of all functions $\mathbb R\to\mathbb R$. In each case it's quick and easy to see that $f(x)=x$ is such a function and therefore in the space.

On the other hand, there's a space of all bounded functions $\mathbb R\to\mathbb R$ or all functions that have a finite limit for $x\to\infty$, and it is similarly easy to see that $f(x)=x$ doesn't satisfy those conditions and therefore is not in those particular spaces.

In neither of these cases do you have to check "11+ conditions" in order to find out whether the function is in a given space.

You do have to check a number of conditions in order to verify whether a particular set with such-and-such addition and multiplication operations is a vector space at all (or is a particularly nice kind of vector space), but you don't need to redo all that every time you want to determine whether a given function (or other object) is in the set or not.

  • $\begingroup$ Thank you, but can you clarify one thing for me. For example, I claim that $f(x) = x$ is an element of Hilbert space. To prove this claim, wouldn't I have to first check additivity, multiplicity, existence of inverse, identity, 0 element, linearity with respect to field,.... ... until I reach a conclusion that all is satisfied therefore $x$ is an element of hilbert space. Sort of tedious don't you think? $\endgroup$
    – Olórin
    Feb 22, 2015 at 6:18
  • $\begingroup$ @MathNewb: There's no such thing as "is an element of Hilbert space". A Hilbert space is a particular kind of vector space, but there are many different Hilbert spaces. Only once you have a particular Hilbert space in mind can you ask whether $f(x)=x$ is one of its elements. $\endgroup$ Feb 22, 2015 at 6:21
  • $\begingroup$ @MathNewb: All of those conditions you list are conditions for something to be a vector space. They are not conditions for something to be an element of a vector space. You can quickly see that $f(x)=x$ is not a vector space (a vector space is a set together with an addition and multiplication operator, and $f(x)=x$ doesn't even look like that), but it can be an element of a vector space. Each particular vector space will tell you how to recognize its elements. $\endgroup$ Feb 22, 2015 at 6:23
  • $\begingroup$ For example, you can ask whether $f(x)=x$ is a member of the particular Hilbert space of square integrable functions, for example. To test that, all you need is to answer one question: does $\int_{-\infty}^\infty |x|^2\,dx$ exist or not? If yes, then the function is in that particular Hilbert space. If no, then it isn't. (It isn't). $\endgroup$ Feb 22, 2015 at 6:25
  • $\begingroup$ I think I see it better now. Just one more thing, suppose $f(x)$ is a square integrable function, why do we need to know whether it is an element of Hilbert space or not? $\endgroup$
    – Olórin
    Feb 22, 2015 at 6:32

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