# Conditions that define a subspace

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.
@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
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