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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 $T:C^1[0, 1]\rightarrow C[0, 1]$ 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?

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up vote 3 down vote accepted

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.

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I agree, your proof is all right =D Thanks a lot @Norbert – PtF Nov 18 '12 at 19:03
Why does this proof that the map is open? – Davide Giraudo Nov 18 '12 at 19:04
Because $T\in\mathcal{B}(E,F)$ is an open mapping iff $$\exists C>0\quad\forall y\in F\quad\exists x\in E:\quad T(x)=y\quad\wedge\quad\Vert x\Vert\leq C\Vert y\Vert$$ – Norbert Nov 18 '12 at 19:07
Dion this is wrong consider zero operator – Norbert Nov 18 '12 at 19:08
One can indeed use the open mapping theorem: $C^1[0,1]$ with the norm $\|f\|= \|f\|_\infty+\|f'\|_\infty$ is Banach and the identity $(C^1[0,1],\|\cdot\|_\infty) \to(C^1[0,1],\|\cdot\|)$ is open. – Jochen Nov 19 '12 at 11:25

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