Let $V = C([a,b])$ be the vector space consisting of all functions $f(t)$ which are defined and continuous on the interval $ 0 \leq t \leq 1.$ Is $f(0) = 2f(1)$ a subspace of V? 
I'm a bit confused on this problem, 
if I can show that $c_1f(0) + c_2g(0) = 2c_1f(1) + 2c_2g(2)$ then I've shown that $f(0) = 2f(1)$ is a vector space? 
So If I write, 
$c_1f(0) + c_2g(0) - 2c_1f(1) - 2c_2g(2) = 0$ 
and let $c_1 = c_2 = 0$ then have I shown that $f(0) = 2f(1)$ is a subspace of V?
 A: No, $f(0)=2f(1)$ is not a subspace, it is an equation.

However, what you probably want to ask is whether $$X=\{f\in V| f(0)=2f(1)\}$$ is a subspace, and to answer that, you need to look at two things:


*

*You have to prove (or disprove) that if $f$ is an element of $X$ and $\alpha$ is a real number, then $\alpha \cdot f$ is also an element of $X$.

*You have to prove (or disprove) that if $f$ and $g$ are elements of $X$, then $f+g$ is also an element of $X$.



Now, how do you go about proving this?
Let's look at the first point. So, you have $f\in X$ and $\alpha \in \mathbb R$. Well, sinc $f\in X$, you know that $f(0)=2f(1)$. Now, what do you want to prove?
Well, you want to prove that $\alpha f\in X$. But you already know that 


*

*$\alpha f$ will be in $X$ if and only if $(\alpha f)(0)=2\cdot ((\alpha f)(0))$

*By definition, for any $x$, $(\alpha f)(x) = \alpha\cdot f(x)$.


Now, it shouldn't be hard to merge the two pieces of information and prove that $\alpha f\in X$, I hope.
The second proof should be just as easy.
A: Hint:
You have to prove that, given two function $f,g \in C$ such that
$$
f(0)=2f(1) \quad \mbox{and} \quad g(0)=2g(1)
$$
than we have:
$$
(f+\alpha g)(0)=2[(f+\alpha g)(1)]
$$
(one multiplication  for a scalar $\alpha$ is sufficient). Can you do this?
