Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'll go ahead and give you the problem first, and then explain my trouble with it.

Which of the following subsets are subspaces of the vector space C(-$\infty$,$\infty$) defined as follows: Let V be the set of all real-valued continuous functions defnined on $\mathbb{R}^1$. If $f$ and $g$ are in $V$, we define $(f + g)(t) = f(t) + g(t)$. If $f$ is in $V$ and $c$ is a scalar, we define $c$ · $f$ by $(c · f)(t) = cf(t)$. Then $V$ is a vector space, which is donatedby C(-$\infty$,$\infty$).

(a) All nonnegative functions (b) All constant functions

There are more, but I have a feeling that after the (a) I'll be able to get the rest myself. Anyay, my problem is that the problem says it required calculus, and I don't understand how calculus even comes into it. As far as I can tell, (b) would be a subspace (because it's are closed under operations of V), but (a) would not because V could make a negative value positive. I may be completely wrong though; any suggestions?

share|cite|improve this question
You are correct. There isn't really any calculus necessary for the problem you stated. Sums and scalar multiples of constant functions are constant, so (b) is a subspace. But non-negative functions times -1 can be negative, so (a) is not closed under scalar multiplication. – Seth Jun 5 '12 at 20:18
The author may have a broader definition of "calculus" than you do - it need not imply that differentiation or integration is required. Also, if the comment that "it required calculus" referred to a larger set of these problems, it is very possible that later questions may actually require knowledge of differentiation and integration - for instance, the space of all functions $f$ such that $f'(0)=0$. – process91 Jun 5 '12 at 21:16
Your first point is a good one, but the "Calculus Required" bit was referring only to this problem. – Kyle Jun 5 '12 at 22:18
up vote 0 down vote accepted

The set of constant functions makes a subspace since (1) each constant function is continuous, (2) if $f$ and $g$ are constant functions and $c$ is a scalar, then $f+g$ and $c\cdot f$ are also constant functions.

The set of non-negative functions is not a subspace for various reasons: (1) not every non-negative function is continuous, (2) if $f$ is non-negative and non-zero, then $-1\cdot f$ takes a negative value so the set of non-negative functions is not closed under scalar multiplication. (It is closed under addition).

share|cite|improve this answer
The problem implies that (a) means only those functions that are continuous, because (a) must be a subset of V. However, your second argument still applies for (a), so it is still not a subspace. – Kyle Jun 5 '12 at 21:05
It is now fixed :) – Egbert Jun 5 '12 at 21:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.