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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?

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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 $f'(0)=0$. –  Michael Boratko 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
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1 Answer

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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).

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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
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