Krdytkyu

Consider the vector spaces $C([0, 1])$ and $C^1([0, 1])$ with norm,
begin{align*}
displaystyle |f|_infty=sup_{xin [0, 1]}|f(x)|,
end{align*}
and let $T:C^1([0, 1])to C([0, 1])$ the operator given by,
begin{align*}
displaystyle Tf=f'.
end{align*}
Is $T$ defined this way an open mapping? If it's, how to prove it?

Yes it is. Take arbitrary $gin C([0,1])$ and consuder $f(t)=int_{0}^tg(tau)dtau$. Then $T(f)=g$ and
$$
Vert fVert_infty
=max_{tin[0,1]}left|int_{0}^tg(tau)dtauright|
leqmax_{tin[0,1]}int_{0}^t|g(tau)|dtau
leqint_{0}^1|g(tau)|dtau
leqint_{0}^1max_{tauin[0,1]}|g(tau)|dtau=Vert gVert_infty
$$
Thus there exist $C=1$ such that for each $gin C([0,1])$ there exist $fin C^1([0,1])$ with $T(f)=g$ and $Vert fVertleq CVert gVert$. Hence $T$ is an open mapping.

Though $T$ is surjective we can not apply open mapping theorem to conclude that $T$ is open mapping, because $(C^1([0,1]),VertcdotVert_infty)$ is not a Banach space.

Here is a characterization of open maps I used in my answer.

Theorem. Let $E$, $F$ be normed spaces and $Tinmathcal{B}(E,F)$, then the following conditions are equivalent.

1) $T$ is an open mapping

2) there exist $r>0$ such that $mathrm{Ball}_F(0,r)subset T(mathrm{Ball}_E(0,1))$

3) there exist $C>0$ such that for all $yin F$ there exist $xin E$ with $T(x)=y$ and $Vert xVertleq CVert yVert$.

Proof. $1)implies 2)$ Let condition $(1)$ holds, then $T(mathrm{Ball}_E(0,1))$ is open. Obviously $0in T(mathrm{Ball}_E(0,1))$, so from previous we have $r>0$ such that
$mathrm{Ball}_F(0,r)subset T(mathrm{Ball}_E(0,1))$

$2)implies 3)$ Now set $C=r^{-1}$. Take arbitrary $yin F$ and consider $hat{y}=rVert yVert^{-1}y$. Then $hat{y}inmathrm{Ball}_E(0,r)$. From $(1)$ we get some $hat{x}inmathrm{Ball}_E(0,1)$ such that $T(hat{x})=hat{y}$. Define $x=r^{-1}Vert yVert hat{x}$. It is easy to check that $T(x)=y$ and $Vert xVertleq CVert yVert$.

$3)implies 1)$ Let $Usubset E$ be an open set. Take arbitrary $y_0in T(U)$, then there exist $x_0in U$ such that $T(x_0)=y_0$. Since $U$ is open there exist $R>0$ such that $mathrm{Ball}_E(x_0,R)subset U$. Define $r=(2C)^{-1}R$. Consider arbitrary $yin mathrm{Ball}(y_0,r)$, then there exist $hat{x}in E$ such that $T(hat{x})=y-y_0$ and $Vert hat{x}Vertleq CVert y-y_0Vertleq R/2$. Consider $x=hat{x}+x_0$. It is easy to check that $T(x)=y$ and $xinmathrm{Ball}_E(x_0,R)$. Thus for each $yinmathrm{Ball}_F(y_0,r)$ there exist $xinmathrm{Ball}_E(x_0,R)$ such that $T(x)=y$. This means that $mathrm{Ball}_F(y_0,r)subset T(mathrm{Ball}_E(x_0,R))subset T(U)$. Thus for each $y_0in T(U)$ we found $r>0$ such that $mathrm{Ball}_F(y_0,r)subset T(U)$, hence $T(U)$ is open. Since $U$ is arbitrary open set in $E$, then $T$ is open.