Linear Algebra Notation $\{f \in C^1[-1,1]; f'(0) + \int_{-1}^1 f(x) \, dx=0\}$ I don't understand how to read the notation for this question (or answer it for that matter) for my university tutorial worksheet a few weeks ago:
It is given that a Vector space must satisfy the ten axioms that define a vector space, and it must be closed under addition and scalar multiplication.
Determine if the following is a vector space: 
$$V = \left\{f \in C^1[-1,1]; f'(0) + \int_{-1}^1 f(x)\,dx=0\right\}$$
 A: The set $V$ consists of all continuously differentiable functions on the interval $[-1,1]$ (this is the $f \in C^1[-1,1]$ part) that satisfy
$$ f'(0) + \int_{-1}^1 f(x) \, dx = 0. $$
Since $V$ consists of (let's say) real functions $f \colon [-1,1] \rightarrow \mathbb{R}$, we can add two such functions as follows:
$$ (f + g)(x) := f(x) + g(x) $$
We can also multiply a function by a scalar $a \in \mathbb{R}$ resulting in a new function defined by
$$ (af)(x) := af(x). $$
Now you need to check whether $(V,+,\cdot)$ satisfies the axioms of being a vector space. Two important things to check are whether the addition and the scalar multiplication result in elements of $V$ (that is, whether $f,g \in V$ and $c \in \mathbb{R}$ implies that $f + g$ and $cf$ belong to $V$). In fact, since $V$ is a subset of the collection of all functions $f \colon [-1,1] \rightarrow \mathbb{R}$ which is a vector space with the operations defined above, those two things together with the fact that $V$ is non-empty guarantee that $V$ is a vector space (as all other axioms follows from the corresponding axioms that hold for the ambient vector space).
A: The set $C^1[-1,1]$ is that set of all real-valued functions whose domain is the interval $[-1,1]$ and that are everywhere differentiable and whose derivatives are everywhere continuous.  Such functions are called continuously differentiable functions.  So you need to check that the sum of two continuously differentiable functions is continuously differentiable.  Two theorems from calculus get you that:


*

*One theorem says the sum of two differentiable functions is differentiable.

*Another says the sum of two continuous functions is continuous. (Apply this one to the derivatives.)


Then you need to check that if $$ f'(0) + \int_{-1}^1 f(x)\,dx=0\quad\text{ and }\quad g'(0) + \int_{-1}^1 g(x)\,dx=0 $$ then $$ (f+g)'(0) + \int_{-1}^1 (f+g)(x)\,dx = 0. $$
That shows that the set in question is closed under addition.
The proof that it is also closed under scalar multiplication is very similar.
Since the set is a subset of something that you already know is a vector space, you don't need to go through the whole list of axioms, but you do need to check closure under addition and scalar multiplication.
