Prove that the set of all real valued function on [a,b] is a vector space Question - Show that the set of all real valued functions on [a,b] , $\mathrm F $[a,b] under usual addition and scalar multiplication is a vector space. 
What I did - let  $\mathrm L = \{ F:[a,b] \; | \; a,b \in \mathbb R \} \; $ & $\; \mathrm u, v \in L$ s.t  $\mathrm u = f[a,b] \; \& \;  v= g[a,b] $and after that I showed that the 10 axioms do satisfy under these conditionsHowever , my instructor has marked it all wrong and she has highlighted that I have started the problem in the wrong way. She also have added the correct method , which is as follows ;  
  $\mathrm L = \{ F:[a,b]\rightarrow \mathbb R^{2} \; | \; a,b \in \mathbb R \} \; $  let  $\; \mathrm x,y \in [a,b]$ Take any$ \,\mathrm F_1$ & $ \, \mathrm F_2$ in $\mathrm L \, $ s.t  $\mathrm F_1(x,y) \in \mathbb R^{2} \; \& \;  F_2(x,y) \in \mathbb R^{2} $ Now Im really confused  because:  1.what is wrong with my approach ? , what is my mistake ? 2.why my instructor have used functions defined on (x,y) ? I think it should be [a,b][please look into this photo to see if I have missed something else

 A: It seems your instructor has erred and written that  $L$ consists of functions from  $\Bbb R^2$to $\Bbb R^2$, or else  (possibly) this was the original problem.  We can't see the original problem so who knows?
For any vector space $V$ and set $S$, the set of functions from $S$ to $V$, denoted $V^S$, is a vector space in a natural way.
A: The instructor's comments seem to indicate that you were meant to show that the space of $\mathbb{R}^2$-valued (not real valued) functions on $[a,b]^2$ is a vector space.  Without seeing the question as it was originally posed, it is difficult to know what the correct answer is.
That being said, let us suppose that what we actually want to do is prove that the space of real valued functions on the interval $[a,b]$ is a vector space.  That is, suppose that $a,b\in\mathbb{R}$ and show that $L := \{ f : [a,b]\to\mathbb{R} \}$ is a vector space.  We will need to show that the axioms of a vector space hold.  I'm not going to do your homework for you, but let's at least see that addition is commutative.
Suppose that $u,v\in L$.  Remember that $L$ is a collection of functions, hence $u$ and $v$ are both functions that take elements of $[a,b]$ as input, and output real numbers.  Then $u+v$ is also a function that goes from $[a,b]$ into the reals.  Specifically, if $x\in[a,b]$, then
$$(u+v)(x) := u(x) + v(x).$$
To show that addition is commutative in $L$, we would need to show that the two functions $u+v$ and $v+u$ are equal.  To do this, we can show that they are equal for any value $x\in[a,b]$.  But $u(x) + v(x)$ is a sum of real numbers, and we know that addition is commutative over $\mathbb{R}$.  Thus
$$u(x) + v(x) = v(x) + u(x)$$
for any number $x\in[a,b]$.  But the left-hand side is exactly $(u+v)(x)$ (as above), and the right-hand side is exactly $(v+u)(x)$.  Combining these, we have
$$(u+v)(x) = u(x) + v(x) = v(x) + u(x) = (v+u)(x)$$
for any $x\in[a,b]$.  Therefore addition is commutative over $L$.
The other axioms are similar---just remember that elements of $L$ are functions, with addition and multiplication defined pointwise.  If you want to check the axioms, you'll need to evaluate sums and scalar multiples of functions in a pointwise manner.
A: You have a notational error . a and b are fixed numbers like a=2 and b =6 or any other fixed real numbers .a and b are still called variables but they are fixed for the whole problem ;sometimes referred to as parameters .Anyway for fixed a and b you are concerned with functions defined for any  x and y in the interval [a,b] with values u= f(x,y) in R and v=g(x,y) .Now you check as you probably did all the vector space rules .In the proofs you will need to say ___ is valid for every x and y .You probably had the right idea for the proofs for these rules but the notation ,using a and b as the variables varying thru [a,b] is technically an error . Hope this helps .  
