small generalization of a linear algebra exercise I came aross the following exercise and was wondering if the conditions are actually necessary or if we could generalize this. Here Hom$(\mathbb R, \mathbb R)$ is the Vectorspace of linear functions from $\mathbb R$ to $\mathbb R$
let $F$ be a function  Hom$(\mathbb R,\mathbb R) \rightarrow \mathbb R$ 
with $F$: $u \mapsto u(1)$ show that $F$ is linear.
The point is that I think this is valid also for $F$ defined on any function space. For instance
we know that $(f+g)(x)=f(x)+g(y)$ and $(\alpha f)(x)=\alpha \cdot f(x)$ from how functions are generally defined. 
so $F(\alpha u)=(\alpha  u)(1)=\alpha \cdot u(1)=\alpha F(u)$
and $F(\alpha u+\beta v)=(\alpha u+ \beta v)(1)=(\alpha u)(1)+(\beta v)(1)=\alpha u(1)+\beta u(1)=\alpha F(u)+ \beta F(v)$ and hence we have linearity of $F$.
It appears a bit strange to me that the condition Hom$(\mathbb R, \mathbb R)$ is added so maybe I misunderstood the exercise or the definition of a Functionspace in general. Any clarification would be really great. 
For me a funtion space is a Vectorspace wich has functions as vectors. 

Yes, it works much more generally. If $V$ is a $K$-vector space ($K$ any field), and $X$ any set, for every $x\in X$, the evaluation map
$$\operatorname{ev}_x \colon \mathscr{F}(X,V) \to V;\quad \operatorname{ev}_x (f) = f(x)$$
is $K$-linear. That indeed follows directly from the definition of addition and scalar multiplication of functions as the pointwise operations.
($\mathscr{F}(A,B)$ denotes the set of all maps $A\to B$)
