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If I let $V=c([a,b])$ be the vector space consisting of all functions $f(t)$ which are defined and continuous on the interval $0\le t\le1$, what are some conditions that define subspaces of $V$?

For $f(1-t) = -tf(t)$ to be considered a subspace of $V$ I got that $h(1-t)$ such that $-th(t) = f(1-t) + g(1-t) = -tf(t) - tg(t)$, which is a subspace and pretty straightforward, but how will I approach this particular condition, $f(0) =2f(1)$ or even $f(0)f(1)=1$?

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In general, a subset $W$ of a vector space $V$ is a subspace of $V$ if and only if for all scalars $\lambda, \mu$ and all vectors $v,w \in W$ we have $\lambda v + \mu w \in W$.

So, for instance, if $W$ is the subset of $V = \mathcal{C}[a,b]$ consisting of all functions that satisfy $f(1-t)=-tf(t)$, then $W$ is a subspace if and only if for all real numbers $\lambda, \mu \in \mathbb{R}$ we have $$\lambda f(1-t) + \mu g(1-t) = -\lambda tf(t) - \mu tg(t)$$ Is this condition satisfied?

If in addition we introduce the constraint that each $f \in W$ satisfies $f(0)=2f(1)$, or respectively $f(0)f(1)=1$, then we require: $$\lambda f(0) + \mu g(0) = 2\lambda f(1) + 2\mu g(1)$$ or respectively $$(\lambda f(0) + \mu g(0))(\lambda f(1) + \mu g(1)) = 1$$ Do these conditions hold? An affirmative or negative answer to this question will tell you whether or not you have a subspace.

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Thank you very much and no its practice problems I am doing. I am currently in abstract vector space class and wanting to know how to approach these types of problems correctly. Okay, for f(0)=2f(1) we can conclude that it is a subspace of V because it satisfies the two axioms you listed. And for f(0)f(1)=1 we can conclude that it is not a subspace of V because the set of functions will translate to a zero which does not equal 1. Is that correct or am I off? –  diimension Sep 17 '12 at 22:19
    
@user1667553: You're right for the first bit. Your conclusion for the $f(0)f(1)=1$ bit is right, but your reason is not $-$ if all the functions in your subspace were to satisfy $f(0)f(1)=1$ then the zero function wouldn't be in the subspace to start with. However, if you let $f$ and $g$ both be the constant function with value $1$ then you can choose $\lambda, \mu$ appropriately to come up with a counterexample. –  Clive Newstead Sep 17 '12 at 22:22
    
Okay, so if f(1)=1 then it will satisfy the conditions and I am a bit confused on the last sentence you replied with. If we let f and g be constants then lambda,beta will already be constants as well since f and g are one? –  diimension Sep 17 '12 at 22:32
    
sorry for annoying you on this problem but what is the reason why f(0)f(1)=1 not a subspace? I am having difficulty grasping an intuition for these types of problems . –  diimension Sep 24 '12 at 6:27
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You don't need the whole $f-f$ stuff, that was just a mini-proof of the fact that any subspace contains $0$. What you've done is shown that $0$ is not in the subset; and therefore the subset is not a subspace. –  Clive Newstead Sep 24 '12 at 21:24
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