Is the differentiation operator an open mapping? Consider the vector spaces $C([0, 1])$ and $C^1([0, 1])$ with norm,
\begin{align*}
\displaystyle \|f\|_\infty=\sup_{x\in [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? 
 A: Yes it is. Take arbitrary $g\in C([0,1])$ and consuder $f(t)=\int_{0}^tg(\tau)d\tau$. Then $T(f)=g$ and 
$$
\Vert f\Vert_\infty
=\max_{t\in[0,1]}\left|\int_{0}^tg(\tau)d\tau\right|
\leq\max_{t\in[0,1]}\int_{0}^t|g(\tau)|d\tau
\leq\int_{0}^1|g(\tau)|d\tau
\leq\int_{0}^1\max_{\tau\in[0,1]}|g(\tau)|d\tau=\Vert g\Vert_\infty
$$
Thus there exist $C=1$ such that for each $g\in C([0,1])$ there exist $f\in C^1([0,1])$ with $T(f)=g$ and $\Vert f\Vert\leq C\Vert g\Vert$. 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 $(C^1([0,1]),\Vert\cdot\Vert_\infty)$ 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 $T\in\mathcal{B}(E,F)$, 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 $y\in F$ there exist $x\in E$ with $T(x)=y$ and $\Vert x\Vert\leq C\Vert y\Vert$.
Proof. $1)\implies 2)$ Let condition $(1)$ holds, then $T(\mathrm{Ball}_E(0,1))$ is open. Obviously $0\in T(\mathrm{Ball}_E(0,1))$, 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 $y\in F$ and consider $\hat{y}=r\Vert y\Vert^{-1}y$. Then $\hat{y}\in\mathrm{Ball}_E(0,r)$. From $(1)$ we get some $\hat{x}\in\mathrm{Ball}_E(0,1)$ such that $T(\hat{x})=\hat{y}$. Define $x=r^{-1}\Vert y\Vert \hat{x}$. It is easy to check that $T(x)=y$ and $\Vert x\Vert\leq C\Vert y\Vert$. 
$3)\implies 1)$ Let $U\subset E$ be an open set. Take arbitrary $y_0\in T(U)$, then there exist $x_0\in U$ 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 $y\in \mathrm{Ball}(y_0,r)$, then there exist $\hat{x}\in E$ such that $T(\hat{x})=y-y_0$ and $\Vert \hat{x}\Vert\leq C\Vert y-y_0\Vert\leq R/2$. Consider $x=\hat{x}+x_0$. It is easy to check that $T(x)=y$ and $x\in\mathrm{Ball}_E(x_0,R)$. Thus for each $y\in\mathrm{Ball}_F(y_0,r)$ there exist $x\in\mathrm{Ball}_E(x_0,R)$ 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 $y_0\in T(U)$ 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.
