Closed map from $C^1[0,1]$ to $C[0,1]$ and non closed functional on $C^1[0,1]$ Let $X=C^1[0,1]$ and $Y=C[0,1]$, both with the sup norm. Define $F: X \rightarrow Y$ by $F(x)=x+x'$ and $f: X \rightarrow K$ $f(x)=x(1)+x'(1)$ for $x \in X$. (Where $K$ is scalar)
I want to prove the following ''$F$ is closed map but $f$ is not a closed map.''
My attempt:
In order to prove that $F$ is closed map first i will show that $F^{-1} : \text{range } F \rightarrow X$ exist and then $F^{-1} $ is bounded. Let $F(x)=F
(y) \implies x+x' = y+y' \implies x-y = -x' + y' \implies (x-y)=-(x-y)'$. why it imply $F$ is one one? I stuck here how to proceed further?
Is my approach right?
In order to show $f$ is not closed. I have to find a closed set in $X$ whose image under $f$ is not closed in $K$. I don't have any intuition how to think such set.
Note : Here $x \in X$ so consider $x$ as a function. Really sorry for bad notations
 A: Answer For the first part: In the context of Functional Analysis a closed linear map is one whose graph is closed. [This is not equivalent to  the topological definition  of a map which sends closed sets to closed sets].
Suppose $(x_n, F(x_n))$ is a sequence in the graph of $F$ converging to $(x,y)$.  Then $x_n \to x$ uniformly and  $x_n+x_n' \to y$ uniformly.  But then $x_n' \to y-x$ uniformly. This implies that $y-x=x'$ so $y+x' =F(x)$. This proves the first part.
Second part: For the second part: Let $x_n(t)=\frac {t^{n}} n$. Then $(x_n, f(x_n)) \to (0,1)$ and $(0,1)$ does not belong to the graph of $f$.
A: To complement Kavi's answer, here is the topological version:
Let $C = \{ x_n \}_{n \in \mathbb{N}}$, where $x_n(t)= n e^{-t} + {1 \over n} (1-e^{-t})$. Each pair of distinct elements of $C$ are separated by at least one, hence $C$ is closed. Note that $F(x_n) = f(x_n) = {1 \over n}$, but $0 \notin F(C)$ and so neither $F$ nor $f$ is a closed map in the topological sense.
The key here is that the kernel and range are non trivial.
