Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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 $f(0)f(1)=1$?

share|cite|improve this question
up vote 3 down vote accepted

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.

share|cite|improve this answer
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
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

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.