Are these linear maps bounded? Let $\mathcal{C}^{\infty}_c$ be the complex vector space of $\mathcal{C}^{\infty}$ functions with compact support in $(0,1)$.Define two norms on it , $\|x(t)\|_u=\text{max}_{t\in (0,1)} \ |x(t)|$ and $\|x(t)\|_d=\text{max}_{t\in (0,1)}\ |x'(t)|$ denoted by $X_u$ and $X_d$ respectively and $T(x)=x'(t)$. Are the following maps bounded? $$a)\ T:X_u\to X_u \\ b)\ T:X_u \to X_d$$
 A: The question is essentially asking the following:
Suppose you have a $C_c^\infty(0,1)$ function $x(t)$.
a) $(Tx)(t) = x'(t)$. Can the derivative of $x$ be controlled by bounding $x$?

 Hint: the other direction works: if you can bound $x'$, then you can use the fundamental theorem of calculus to control $x$. Should we expect the converse to hold? If we keep $x(t)$ bounded by $1$, how bad can we make the behavior of $x'(t)$?

b) Can the second derivative of $x$ be controlled by bounding $x$? If you know the answer to a) then this question should be trivial.
A: Edit 2:let $[a,b]\subset(0,1)$.let $g(x)=1$ on$[a,b]$ and $g(x)=0$ on $(0,a-\epsilon)\cup(b+\epsilon,1)$. By "using" proper translation and rescaling of $e^{-1/(x^2-1)}$ defined on $[a-\epsilon,a]$ and $[b,b+\epsilon]$ we can turn $g$ into a function in $C^\infty_c(0,1)$ that has norm less than one. Buy its derivative has rather large norm.by making $\epsilon$ smaller we are not changing the norm of the function however since the two gaps meaning $[b,b+\epsilon]$ and $[a-\epsilon,a]$are getting narrower then derivatives evaluated at the endpoints becomes larger. here
