To check the continuity of $f$ Let $f:(C^1[a,b],\|.\|_{\infty})\to \mathbb K$  and $c\in (a,b)$ be such that $f(x)=x^{'}(c)$ for all $x\in C^1[a,b]$. Then is $f$ a continuous linear functional? 
I chose $a=0,b=1$ and $c=1$, then it can be easily shown that $f$ is not continuous. But this doesn't serve my purpose. Because $c$ should come from $(a,b)$. Please suggest.
 A: The answer to the question depends on the norm you're equipping $C^1[a,b]$ with.
If you define for a continuously differentiable function $x:[a,b]\rightarrow \mathbb C$ 
$$\|x\|_{C^1}=\|x\|_\infty + \|x'\|_\infty=\sup_{t\in [a,b]}|x(t)|+\sup_{t \in [a,b]}|x'(t)|,$$
then you need to check if there is some constant $M$ such that for all continuous functions $x$ we have
$$|f(x)| \leq M \| x\|_{C^1}.$$
Edit: As we're considering $\|\cdot\|_\infty$ as a norm, it can be shown that $f$ is not continuous.
If $[a,b]=[0,1]$ and $c=1$, consider $x_n(t)= t^n$. Then $x'(t) = n(t)^{n-1}$. 
What does this mean for $|f(x_n)|$? What's $\|x_n\|_\infty$? 
How can you generalize this example to the general case for $c$ and $[a,b]$?
A: The lack of continuity of $f$ can be sensed intuitively by considering that  if real functions  $g, h$ are close, it tells us nothing about the closeness of $g'$ and $h'.$ Consider the case $h=0.$
For example, let $$g_n(t)= n^{-1}\sin [ n^2 (t-c)]$$ for $t\in [0,1]$ and $c\in (0,1)$ and $n\in N.$ We have $\|g_n\|_{\infty}\leq 1/n$ but $g'_n(c)=n.$ So the sequence $(g_n)_{n\in N}$ converges in the $\sup$ norm to the constant function $0,$ but $(g'_n(c))_{n\in N}$ is not even a Cauchy sequence.
